Complex Analysis — Proofs & PYQ

\(f=x^2+y^2+0i\). \(u=x^2+y^2,v=0\). CR: \(u_x=2x,v_y=0\). Need \(u_x=v_y\Rightarrow2x=0\). Only at \(x=0\). And \(u_y=2y=-v_x=0\Rightarrow y=0\). CR satisfied only at origin. But CR is not continuous-with-partials elsewhere to give differentiability. \(f\) is differentiable only at \(z=0\), not analytic anywhere.

\(u_x=3x^2-3y^2,u_y=-6xy\). CR: \(v_y=u_x=3x^2-3y^2\Rightarrow v=3x^2y-y^3+\phi(x)\). \(v_x=6xy+\phi'(x)=-u_y=6xy\Rightarrow\phi'(x)=0\Rightarrow\phi=C\). \(v=3x^2y-y^3+C\). \(f(z)=(x^3-3xy^2)+i(3x^2y-y^3)=x^3+3ix^2y-3xy^2-iy^3=(x+iy)^3=z^3\). \(\blacksquare\)

Let \(v=e^x\sin y+xy\). \(v_x=e^x\sin y+y\), \(v_y=e^x\cos y+x\). By CR: \(u_x=v_y=e^x\cos y+x\), \(u_y=-v_x=-(e^x\sin y+y)\). Check: these are consistent (verify \(u_{xy}=u_{yx}\)). Using Milne-Thomson: \(f(z)=f'(z)dz\)... method: \(f'(z)=u_x+iv_x=(e^x\cos y+x)+i(e^x\sin y+y)\). At \(y=0\): \(f'(x+i0)=(e^x+x)+ie^x\cdot0=e^x+x\). In complex form: \(f'(z)=e^z+z\). Integrate: \(f(z)=e^z+z^2/2+C\). Verify: \(e^z=e^x\cos y+ie^x\sin y\), \(z^2/2=(x^2-y^2)/2+ixy\). \(\text{Im}(f)=e^x\sin y+xy\). Matches! \(\blacksquare\)

\(u_x=e^x\cos y,u_{xx}=e^x\cos y\). \(u_y=-e^x\sin y,u_{yy}=-e^x\cos y\). \(u_{xx}+u_{yy}=0\). Harmonic. \(v_y=u_x=e^x\cos y\Rightarrow v=e^x\sin y+\phi(x)\). \(v_x=e^x\sin y+\phi'(x)=-u_y=e^x\sin y\Rightarrow\phi'=0\). \(f(z)=e^x\cos y+ie^x\sin y=e^x(\cos y+i\sin y)=e^xe^{iy}=e^{x+iy}=e^z\). \(\blacksquare\)

\(u_{xx}=2a,u_{yy}=2\). Harmonic requires \(u_{xx}+u_{yy}=0\Rightarrow2a+2=0\Rightarrow a=-1\). \(b\) can be anything (since \(u_{xy}=b\) and Laplace doesn't directly constrain \(b\)). But \(b=0\) is typical unless stated otherwise. So \(a=-1\). Then \(u=-x^2+y^2=-(x^2-y^2)\). This is \(-\text{Re}(z^2)\), consistent with \(f(z)=-z^2+C\) (analytic).

Singularities at \(z=0\) and \(z=1\), both inside \(|z|=2\). By residue theorem (or CIF): \(\oint=2\pi i(\text{Res at }0+\text{Res at }1)\). Res at 0: \(\lim_{z\to0}z\cdot\dfrac{e^z}{z(z-1)}=\dfrac{e^0}{0-1}=-1\). Res at 1: \(\lim_{z\to1}(z-1)\cdot\dfrac{e^z}{z(z-1)}=\dfrac{e^1}{1}=e\). Total: \(2\pi i(-1+e)=2\pi i(e-1)\).

\(g(z)=\sin z/z\): the singularity at \(z=0\) is removable (Laurent series: \(\sin z/z=1-z^2/6+\cdots\) — no negative powers). By extended Cauchy theorem: \(\oint g\,dz=2\pi i\cdot\text{Res}(g,0)=2\pi i\cdot0=0\). (Alternatively: since \(g\) is entire after defining \(g(0)=1\), Cauchy gives 0.) Answer: \(\mathbf{0}\).

Singularities: pole of order 2 at \(z=1\), simple pole at \(z=-2\). Both inside \(|z|=3\).

Res at \(-2\): \(\dfrac{\cos(-2\pi)}{(-2-1)^2}=\dfrac{1}{9}\). Res at \(1\): \(\dfrac{d}{dz}\!\left[\dfrac{\cos\pi z}{z+2}\right]_{z=1}=\dfrac{-\pi\sin\pi z(z+2)-\cos\pi z}{(z+2)^2}\bigg|_{z=1}=\dfrac{-\pi\sin\pi\cdot3-\cos\pi}{9}=\dfrac{0+1}{9}=\dfrac{1}{9}\). Total: \(2\pi i\left(\dfrac{1}{9}+\dfrac{1}{9}\right)=\dfrac{4\pi i}{9}\).

Partial fractions: \(f=\dfrac{-1/2}{z}+\dfrac{1}{z-1}+\dfrac{-1/2}{z-2}\). In region \(1<|z|<2\): (i) \(1/z\): already in Laurent form, negative power of \(z\). (ii) \(1/(z-1)=\dfrac{1}{z}\cdot\dfrac{1}{1-1/z}=\dfrac{1}{z}\sum_{n=0}^\infty z^{-n}=\sum_{n=0}^\infty z^{-n-1}=\sum_{n=1}^\infty z^{-n}\) (since \(|1/z|<1\)). (iii) \(-1/(z-2)=\dfrac{1}{2}\cdot\dfrac{1}{1-z/2}=\dfrac{1}{2}\sum_{n=0}^\infty(z/2)^n=\sum_{n=0}^\infty\dfrac{z^n}{2^{n+1}}\) (since \(|z/2|<1\)). Combining: \(f=\sum_{n=0}^\infty\dfrac{z^n}{2^{n+1}}-\dfrac{1}{2z}+\sum_{n=1}^\infty z^{-n}-\dfrac{1}{z}\).

\(\sin z=z-z^3/6+z^5/120-\cdots\). So \(f(z)=\dfrac{1}{z^2}-\dfrac{1}{6}+\dfrac{z^2}{120}-\cdots\). Laurent series around \(z=0\): lowest power is \(z^{-2}\). So \(z=0\) is a pole of order 2. \(\text{Res}(f,0)=\) coefficient of \(z^{-1}\) = 0 (there is no \(z^{-1}\) term). \(\blacksquare\)

\(e^{1/z}=\sum_{n=0}^\infty\dfrac{1}{n!\,z^n}=1+\dfrac{1}{z}+\dfrac{1}{2z^2}+\cdots\). Infinitely many negative powers → essential singularity. Residue = coefficient of \(z^{-1}\) = 1. \(\blacksquare\)

Poles: \(z=-1\) (order 2), \(z=i\) (simple), \(z=-i\) (simple). Res at \(i\): \(\dfrac{i^2}{(i+1)^2(2i)}=\dfrac{-1}{(2i)(-1+1+2i)}=\dfrac{-1}{2i(2i)}=\dfrac{-1}{-4}=\dfrac{1}{4}\). Res at \(-i\): \(\dfrac{-1}{(-i+1)^2(-2i)}=\cdots=\dfrac{1}{4}\). Res at \(-1\): \(\dfrac{d}{dz}\!\left[\dfrac{z^2}{z^2+1}\right]_{z=-1}=\dfrac{2z(z^2+1)-z^2\cdot2z}{(z^2+1)^2}\bigg|_{-1}=\dfrac{-2\cdot2+(-1)\cdot(-2)}{4}=\dfrac{-4+2}{4}=\dfrac{-1}{2}\).

Let \(z=e^{i\theta}\): \(\cos\theta=\dfrac{z+z^{-1}}{2}\), \(d\theta=dz/(iz)\). \(I=\oint_{|z|=1}\dfrac{1}{5-2(z+z^{-1})}\cdot\dfrac{dz}{iz}=\oint\dfrac{dz}{iz(5-2z-2/z)}=\oint\dfrac{dz}{i(5z-2z^2-2)}=\oint\dfrac{dz}{-i(2z^2-5z+2)}\). Factor: \(2z^2-5z+2=(2z-1)(z-2)\). Poles at \(z=1/2\) (inside) and \(z=2\) (outside). Res at \(z=1/2\): \(\dfrac{1}{-i\cdot2\cdot(1/2-2)}=\dfrac{1}{-i\cdot2\cdot(-3/2)}=\dfrac{1}{3i}\). \(I=2\pi i\cdot\dfrac{1}{3i}=\dfrac{2\pi}{3}\). \(\blacksquare\)

Use semicircular contour (upper half plane). Poles in UHP: \(z=i\) and \(z=2i\). Res at \(i\): \(\dfrac{i^2}{(2i)(i^2+4)}=\dfrac{-1}{2i\cdot3}=\dfrac{-1}{6i}\). Res at \(2i\): \(\dfrac{(2i)^2}{(2i+1\cdot1)... }\) use formula: \(\dfrac{(2i)^2}{(4i^2+1)(4i)}=\dfrac{-4}{(-3)(4i)}=\dfrac{-4}{-12i}=\dfrac{1}{3i}\). Total residue: \(\dfrac{-1}{6i}+\dfrac{1}{3i}=\dfrac{-1+2}{6i}=\dfrac{1}{6i}\). \(I=2\pi i\cdot\dfrac{1}{6i}=\dfrac{\pi}{3}\).

Consider \(\oint_C\dfrac{e^{iz}}{z^2+a^2}dz\) over upper semicircle. Pole in UHP: \(z=ia\). Res: \(\dfrac{e^{i(ia)}}{2ia}=\dfrac{e^{-a}}{2ia}\). By Jordan's lemma: integral over arc \(\to0\). So \(\int_{-\infty}^\infty\dfrac{e^{ix}}{x^2+a^2}dx=2\pi i\cdot\dfrac{e^{-a}}{2ia}=\dfrac{\pi e^{-a}}{a}\). Taking real part: \(\int_{-\infty}^\infty\dfrac{\cos x}{x^2+a^2}dx=\dfrac{\pi e^{-a}}{a}\). Since integrand is even: \(\int_0^\infty\dfrac{\cos x}{x^2+a^2}dx=\dfrac{\pi e^{-a}}{2a}\).

Standard substitution: \(z=e^{i\theta}\). Denominator becomes \(a+b(z+z^{-1})/2=\dfrac{bz^2+2az+b}{2z}\). Integral becomes \(\oint_{|z|=1}\dfrac{2\,dz}{i(bz^2+2az+b)}\). Roots of \(bz^2+2az+b=0\): \(z=\dfrac{-a\pm\sqrt{a^2-b^2}}{b}\). Since \(a>b>0\): \(z_1=\dfrac{-a+\sqrt{a^2-b^2}}{b}\) has \(|z_1|<1\). Res at \(z_1\): \(\dfrac{2}{i\cdot(2bz_1+2a)}=\dfrac{1}{i(bz_1+a)}=\dfrac{1}{i\sqrt{a^2-b^2}}\). \(I=2\pi i\cdot\dfrac{1}{i\sqrt{a^2-b^2}}=\dfrac{2\pi}{\sqrt{a^2-b^2}}\). \(\blacksquare\)

Pole in UHP: \(z=i\) (order 2). Res at \(i\): \(\dfrac{d}{dz}\!\left[\dfrac{1}{(z+i)^2}\right]_{z=i}=\dfrac{-2}{(2i)^3}=\dfrac{-2}{-8i}=\dfrac{1}{4i}\). \(I=2\pi i\cdot\dfrac{1}{4i}=\dfrac{\pi}{2}\). \(\blacksquare\)

Consider \(f(z)=e^{iz}/z\). Integrate over indented semicircle (upper half plane, indentation at 0 of radius \(\epsilon\), large circle of radius \(R\)). By Cauchy's theorem (no poles inside): \(\int_{-R}^{-\epsilon}+\int_{\gamma_\epsilon}+\int_\epsilon^R+\int_{\Gamma_R}=0\). As \(\epsilon\to0,R\to\infty\): arc \(\Gamma_R\to0\) (Jordan); indentation \(\gamma_\epsilon\to-\pi i\cdot\text{Res}(f,0)=-\pi i\cdot1=-\pi i\). The real-axis parts: \(2i\int_0^\infty\dfrac{\sin x}{x}dx\). So \(2i\int_0^\infty\dfrac{\sin x}{x}dx-\pi i=0\Rightarrow\int_0^\infty\dfrac{\sin x}{x}dx=\dfrac{\pi}{2}\). \(\blacksquare\)

Use cross-ratio: \(\dfrac{(w-w_1)(w_2-w_3)}{(w-w_3)(w_2-w_1)}=\dfrac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)}\). With \((z_1,z_2,z_3)=(0,1,\infty)\) and \((w_1,w_2,w_3)=(-1,0,1)\): \(\dfrac{(w+1)(0-1)}{(w-1)(0+1)}=\dfrac{(z-0)(1-\infty)}{(z-\infty)(1-0)}\). For \(z_3=\infty\): \(\dfrac{z-0}{z-\infty}\cdot\dfrac{1-\infty}{1-0}\to\dfrac{z}{1}=z\) (limit as \(z_3\to\infty\)). So \(\dfrac{-(w+1)}{w-1}=z\Rightarrow-(w+1)=z(w-1)\Rightarrow-w-1=zw-z\Rightarrow-w-zw=1-z\Rightarrow w(-1-z)=(1-z)\Rightarrow w=\dfrac{z-1}{z+1}\). Verify: \(z=0\to w=-1\), \(z=1\to w=0\), \(z\to\infty\to w=1\). \(\checkmark\) \(\blacksquare\)

Standard map: \(w=\dfrac{z-z_0}{z-\bar{z}_0}\) where \(z_0\) is in the UHP (maps \(z_0\mapsto0\)). Choosing \(z_0=i\): \(w=\dfrac{z-i}{z+i}\). Check: \(z=0\to w=-1\), \(z=1\to w=\dfrac{1-i}{1+i}=\dfrac{(1-i)^2}{2}=\dfrac{-2i}{2}=-i\) (on circle). \(z=i\to w=0\) (center). \(\blacksquare\)

If \(z=re^{i\theta}\) with \(0<\theta<\pi/4\): \(w=r^2e^{2i\theta}\). So \(\text{arg}(w)=2\theta\in(0,\pi/2)\). The image is the region \(0<\text{arg}(w)<\pi/2\), i.e., the first quadrant \(\{w:\text{Re}(w)>0,\text{Im}(w)>0\}\). \(\blacksquare\)

On \(|z|=1\): \(z=e^{i\theta}\). \(w=e^{i\theta}+e^{-i\theta}=2\cos\theta\in[-2,2]\). So the unit circle maps to the real segment \([-2,2]\). \(\blacksquare\)

Any bilinear transformation mapping disk to disk has the form \(w=e^{i\alpha}\dfrac{z-z_0}{1-\bar{z}_0z}\) for \(|z_0|<1\) and real \(\alpha\). Condition \(z=1\to w=1\): \(1=e^{i\alpha}\dfrac{1-z_0}{1-\bar{z}_0}\). Condition \(z=-1\to w=-1\): \(-1=e^{i\alpha}\dfrac{-1-z_0}{1+\bar{z}_0}\). Divide: \(-1=\dfrac{-1-z_0}{1+\bar{z}_0}\cdot\dfrac{1-\bar{z}_0}{1-z_0}\). Simplest solution: \(z_0=0,\alpha=0\Rightarrow w=z\) (identity). Another: \(w=-z\) doesn't satisfy \(z=1\to w=1\). So \(w=z\). \(\blacksquare\)

Write \(f=u+iv\) with \(u=x^2+y^2,\,v=0\). CR: \(u_x=2x=v_y=0\) and \(u_y=2y=-v_x=0\). Both hold iff \(x=y=0\), i.e.\ only at the origin. So \(f\) is differentiable only at \(z=0\), hence not analytic in any neighbourhood. \(\blacksquare\)

Here \(u=x,\,v=-y\). CR equations: \(u_x=1\neq v_y=-1\). The Cauchy–Riemann equations fail everywhere. Hence \(f\) is not differentiable at any point, so certainly nowhere analytic. \(\blacksquare\)

Check harmonicity: \(u_{xx}=6x,\,u_{yy}=-6x\), so \(\nabla^2 u=0\). By CR, \(v_y=u_x=3x^2-3y^2\) and \(v_x=-u_y=6xy\). Integrating \(v_y=3x^2-3y^2\) w.r.t.\ \(y\): \(v=3x^2y-y^3+\phi(x)\). Then \(v_x=6xy+\phi'(x)=6xy\Rightarrow\phi'=0\Rightarrow\phi=C\). So \(v=3x^2y-y^3+C\) and \(f(z)=x^3-3xy^2+i(3x^2y-y^3)=z^3+iC\). \(\blacksquare\)

CR: \(v_y=u_x=e^x\cos y\Rightarrow v=e^x\sin y+\phi(x)\). Then \(v_x=e^x\sin y+\phi'(x)=-u_y=e^x\sin y\Rightarrow\phi'=0\). So \(v=e^x\sin y+C\) and \(f(z)=e^x\cos y+ie^x\sin y=e^x e^{iy}=e^z\). \(\blacksquare\)

Use polar CR: \(u_r=\tfrac{1}{r}v_\theta,\;v_r=-\tfrac{1}{r}u_\theta\). Here \(v_\theta=r\cos\theta+2r^2\cos2\theta\), so \(u_r=\cos\theta+2r\cos2\theta\). Integrating w.r.t.\ \(r\): \(u=r\cos\theta+r^2\cos2\theta+\phi(\theta)\). Then \(u_\theta=-r\sin\theta-2r^2\sin2\theta+\phi'(\theta)=-rv_r=-r(\sin\theta+2r\sin2\theta)\), so \(\phi'=0\). Thus \(u=r\cos\theta+r^2\cos2\theta+C\) and \(f=u+iv=(r\cos\theta+ir\sin\theta)+(r^2\cos2\theta+ir^2\sin2\theta)=z+z^2\). \(\blacksquare\)

Here \(u=xy,\,v=y^2\). CR: \(u_x=y=v_y=2y\Rightarrow y=0\). And \(u_y=x=-v_x=0\Rightarrow x=0\). So CR holds only at the origin. Since analyticity requires CR in a neighbourhood, \(f\) is nowhere analytic. (It is differentiable only at \(z=0\).) \(\blacksquare\)

Compute \(u_x=3x^2-3y^2+3\) and \(u_y=-6xy\). Milne-Thomson: \(f'(z)=u_x(z,0)-iu_y(z,0)=3z^2+3-0=3z^2+3\). Integrate: \(f(z)=z^3+3z+C\). \(\blacksquare\)

CR: \(v_y=u_x=\cos x\cosh y\Rightarrow v=\cos x\sinh y+\phi(x)\). Then \(v_x=-\sin x\sinh y+\phi'(x)=-u_y=-\sin x\sinh y\Rightarrow\phi'=0\). So \(v=\cos x\sinh y+C\). Check: \(f(z)=\sin x\cosh y+i\cos x\sinh y=-i\sin(iz+\cdots)\). Note \(\sin z=\sin(x+iy)=\sin x\cosh y+i\cos x\sinh y\), so \(f(z)=\sin z\). \(\blacksquare\)

\(u_x=\tfrac{x}{x^2+y^2},\;u_{xx}=\tfrac{y^2-x^2}{(x^2+y^2)^2}\). Similarly \(u_{yy}=\tfrac{x^2-y^2}{(x^2+y^2)^2}\). So \(\nabla^2 u=0\) for \((x,y)\neq(0,0)\). CR: \(v_y=u_x=\tfrac{x}{r^2},\;v_x=-u_y=-\tfrac{y}{r^2}\). Integrating: \(v=\arctan(y/x)+C=\arg(z)+C\). Thus \(f(z)=\log|z|+i\arg z=\operatorname{Log}z\). \(\blacksquare\)

Here \(u=\sqrt{|xy|},\,v=0\). At origin: \(u_x(0,0)=\lim_{h\to0}\frac{\sqrt{|h\cdot0|}}{h}=0\) and similarly \(u_y=0,\,v_x=0,\,v_y=0\). CR holds. But directional limit: along \(y=x\), \(\frac{f(h+ih)-f(0)}{h+ih}=\frac{\sqrt{h^2}}{h(1+i)}=\frac{1}{1+i}\). Along \(y=0\), limit is \(0\). Limits differ, so \(f\) is not differentiable at origin despite CR holding. (CR necessary but not sufficient without extra regularity.) \(\blacksquare\)

On \(|z|<1\), \(f(z)=\frac{1}{1-z}\). The function \(g(z)=\frac{1}{1-z}\) is analytic on \(\mathbb{C}\setminus\{1\}\), which is much larger than \(|z|<1\). By the identity theorem, \(g\) is the unique analytic continuation of \(f\). The only singularity is the simple pole at \(z=1\). \(\blacksquare\)

Decompose \(w\) into: (i) translation \(z\mapsto z+b/a\), (ii) inversion \(z\mapsto 1/z\), (iii) scaling \(z\mapsto az\). Translations and scalings clearly map circles/lines to circles/lines. For inversion \(w=1/z\): a circle/line has equation \(A(x^2+y^2)+Bx+Cy+D=0\). Substituting \(x+iy=1/(u+iv)\) gives an equation of the same form in \(u,v\). Hence the image is again a circle or line. \(\blacksquare\)

Fixed points: \(z=\frac{z-1}{z+1}\Rightarrow z(z+1)=z-1\Rightarrow z^2+z=z-1\Rightarrow z^2=-1\Rightarrow z=\pm i\). The fixed points are \(z=i\) and \(z=-i\). \(\blacksquare\)

Use the cross-ratio formula \(\frac{(w-w_1)(w_2-w_3)}{(w-w_3)(w_2-w_1)}=\frac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)}\) with \((z_1,z_2,z_3)=(0,1,\infty)\), \((w_1,w_2,w_3)=(1,i,-1)\). As \(z_3\to\infty\), the RHS simplifies to \(\frac{(z-0)(1-z_3)}{(z-z_3)(1-0)}\to\frac{z\cdot(-z_3)}{(-z_3)}=z\). So \(\frac{(w-1)(i+1)}{(w+1)(i-1)}=z\). Solve: \((w-1)(1+i)=z(w+1)(i-1)\). \(w[(1+i)-z(i-1)]=z(i-1)+(1+i)\). \(w=\frac{z(i-1)+(1+i)}{(1+i)-z(i-1)}=\frac{(1+i)+(i-1)z}{(1+i)-(i-1)z}\). Simplify with \(i-1=-(1-i)\): \(w=\frac{(1+i)-(1-i)z}{(1+i)+(1-i)z}\). \(\blacksquare\)

Let \(T(z)=\frac{az+b}{cz+d}\). Then \(T(z_i)-T(z_j)=\frac{(az_i+b)(cz_j+d)-(az_j+b)(cz_i+d)}{(cz_i+d)(cz_j+d)}=\frac{(ad-bc)(z_i-z_j)}{(cz_i+d)(cz_j+d)}\). Substituting into the cross-ratio, the factors \((ad-bc)\) and \((cz_i+d)\) cancel in numerator and denominator, leaving the original cross-ratio unchanged. \(\blacksquare\)

Let \(z=re^{i\theta}\). Then \(w=re^{i\theta}+\frac{1}{r}e^{-i\theta}=(r+\frac{1}{r})\cos\theta+i(r-\frac{1}{r})\sin\theta\). This traces an ellipse with semi-axes \(a=r+\frac{1}{r}\) and \(b=|r-\frac{1}{r}|\). For \(r=1\): \(b=0\), collapses to real segment \([-2,2]\). For \(r\neq1\): an ellipse. As \(r\to\infty\) or \(0\), the ellipse grows. Critical points where \(dw/dz=1-1/z^2=0\): \(z=\pm1\), where conformality fails. \(\blacksquare\)

Write \(z=x+iy\). Then \(w=e^x e^{iy}\), so \(|w|=e^x\) and \(\arg w=y\). For \(a

For \(z=re^{i\theta}\), \(\log z=\ln r+i\theta\). Since \(\theta\) is multi-valued (defined up to \(2\pi k\)), we restrict to the principal argument \(\operatorname{Arg}(z)\in(-\pi,\pi]\). The principal value is \(\operatorname{Log}z=\ln|z|+i\operatorname{Arg}(z)\). This is analytic on \(\mathbb{C}\setminus(-\infty,0]\). The negative real axis is the branch cut, where \(\operatorname{Arg}z\) jumps discontinuously from \(\pi\) to \(-\pi\). Different choices of branch cut give different analytic branches of \(\log\). \(\blacksquare\)

\(z^{1/2}=\sqrt{r}e^{i\theta/2}\). Going around \(z=0\) once (\(\theta\to\theta+2\pi\)): \(e^{i(\theta+2\pi)/2}=e^{i\theta/2}e^{i\pi}=-e^{i\theta/2}\). The function changes sign, so \(z=0\) is a branch point. Similarly \(z=\infty\) is a branch point (check by \(z=1/w\)). A branch cut from \(0\) to \(\infty\) (e.g.\ the negative real axis) removes the ambiguity, giving a single-valued analytic function on \(\mathbb{C}\setminus(-\infty,0]\). \(\blacksquare\)

For \(\operatorname{Im}z>0\): \(|z-i|^2=(x)^2+(y-1)^2\) and \(|z+i|^2=x^2+(y+1)^2\). Since \(y>0\): \((y+1)^2>(y-1)^2\Rightarrow|z+i|>|z-i|\Rightarrow|w|<1\). On the real axis (\(y=0\)): \(|w|=\frac{|x-i|}{|x+i|}=1\). So the real axis maps to \(|w|=1\) and the upper half-plane maps to \(|w|<1\). The map is a Möbius transformation (hence conformal and bijective). \(\blacksquare\)

\(f(z)=z^2\) is entire (analytic everywhere). By the Cauchy-Goursat theorem, if \(f\) is analytic inside and on a simple closed contour \(C\), then \(\oint_C f(z)\,dz=0\). Since \(z^2\) is entire, this applies for any \(C\). Hence \(\oint_C z^2\,dz=0\). \(\blacksquare\)

The integrand \(f(z)=\frac{e^z}{z-1}\) has a simple pole at \(z=1\), which is inside \(|z|=2\). By Cauchy's integral formula, \(\oint_C\frac{f(z)}{z-a}dz=2\pi i\,f(a)\) with \(f(z)=e^z,\,a=1\). Result: \(2\pi i\cdot e^1=2\pi ie\). \(\blacksquare\)

By the generalised CIF: \(\oint_C\frac{f(z)}{(z-a)^2}dz=2\pi i\,f'(a)\) with \(f(z)=\sin z,\,a=\pi/2\). Here \(f'(z)=\cos z\) and \(f'(\pi/2)=\cos(\pi/2)=0\). Hence the integral equals \(2\pi i\cdot0=0\). \(\blacksquare\)

Poles at \(z=\pm i\), both inside \(|z|=2\). \(\operatorname{Res}_{z=i}\frac{z^2}{(z-i)(z+i)}=\frac{i^2}{2i}=\frac{-1}{2i}=\frac{i}{2}\). \(\operatorname{Res}_{z=-i}=\frac{(-i)^2}{-2i}=\frac{-1}{-2i}=\frac{-i}{2}\). Sum of residues \(=\frac{i}{2}-\frac{i}{2}=0\). Integral \(=2\pi i\cdot0=0\). \(\blacksquare\)

Factor: \(z^2-2z+5=(z-(1+2i))(z-(1-2i))\). Both poles \(1\pm2i\) lie inside \(|z|=3\) (modulus \(\sqrt{1+4}=\sqrt{5}<3\)). \(\operatorname{Res}_{z=1+2i}=\frac{2+2i}{4i}=\frac{1+i}{2i}=\frac{(1+i)(-i)}{2}=\frac{1-i+(-i+1)}{...}\). More directly: \(\frac{1+2i+1}{(1+2i)-(1-2i)}=\frac{2+2i}{4i}=\frac{2+2i}{4i}\cdot\frac{-i}{-i}=\frac{(-2i+2)}{4}=\frac{1-i}{2}\). Similarly \(\operatorname{Res}_{1-2i}=\frac{(1-2i)+1}{(1-2i)-(1+2i)}=\frac{2-2i}{-4i}=\frac{(2-2i)i}{4}=\frac{2i+2}{4}=\frac{1+i}{2}\). Sum \(=\frac{1-i+1+i}{2}=1\). Integral \(=2\pi i\). \(\blacksquare\)

Poles at 4th roots of unity: \(z=1,\,-1,\,i,\,-i\), all inside \(|z|=2\). For simple poles of \(\frac{1}{z^4-1}\): \(\operatorname{Res}_{z=a}=\frac{1}{4a^3}\). Sum: \(\frac{1}{4}+\frac{1}{4(-1)^3}+\frac{1}{4i^3}+\frac{1}{4(-i)^3}=\frac{1}{4}-\frac{1}{4}+\frac{1}{4(-i)}+\frac{1}{4i}=\frac{i}{4}-\frac{i}{4}=0\). Integral \(=2\pi i\cdot0=0\). \(\blacksquare\)

Substitute \(z=e^{i\theta}\), \(\sin\theta=\frac{z-z^{-1}}{2i}\), \(d\theta=\frac{dz}{iz}\). Integral becomes \(\oint_{|z|=1}\frac{1}{2+\frac{z-z^{-1}}{2i}}\frac{dz}{iz}=\oint\frac{2\,dz}{iz(4i+z-z^{-1})}=\oint\frac{2\,dz}{4iz^2 \cdot\frac{1}{z}+iz(z-z^{-1})}\). Simplify: \(2+\sin\theta=2+\frac{z-1/z}{2i}\), so integrand \(=\frac{1}{iz}\cdot\frac{1}{2+\frac{z-1/z}{2i}}=\frac{2}{iz(4i+z-1/z)}\cdot 1\). Multiply through: \(\frac{2}{i(4iz+z^2-1)}=\frac{2}{i(z^2+4iz-1)}\). Roots of \(z^2+4iz-1=0\): \(z=\frac{-4i\pm\sqrt{-16+4}}{2}=\frac{-4i\pm\sqrt{-12}}{2}=-2i\pm i\sqrt{3}\). Inside \(|z|=1\): \(|{-2i+i\sqrt3}|=|i(\sqrt3-2)|=2-\sqrt3<1\). So pole at \(z_0=i(\sqrt3-2)\). Residue \(=\frac{2}{i\cdot2z_0+4i\cdot i}=\frac{2}{i(2z_0+4i)}\). With \(2z_0=2i(\sqrt3-2)\): \(2z_0+4i=2i(\sqrt3-2)+4i=2i\sqrt3\). Res \(=\frac{2}{i\cdot2i\sqrt3}=\frac{2}{-2\sqrt3}=-\frac{1}{\sqrt3}\). Integral \(=2\pi i\cdot\left(-\frac{1}{\sqrt3}\right)\cdot\frac{1}{1}\). Careful with sign: the contour integral equals \(\oint\frac{2\,dz}{i(z^2+4iz-1)}\), so result \(=2\pi i\cdot\frac{2}{i}\cdot\operatorname{Res}=\frac{2\pi i}{i}\cdot\frac{-1}{\sqrt3}\cdot\frac{1}{1}...\) Let me redo cleanly. The contour integral is \(\frac{2}{i}\oint\frac{dz}{z^2+4iz-1}\). Residue at \(z_0=i(\sqrt3-2)\): \(\frac{1}{2z_0+4i}=\frac{1}{2i\sqrt3}\). Result \(=\frac{2}{i}\cdot2\pi i\cdot\frac{1}{2i\sqrt3}=\frac{2\cdot2\pi}{2\sqrt3}\cdot\frac{i}{i}\cdot\frac{1}{i}...\) Direct answer: \(\boxed{\dfrac{2\pi}{\sqrt{3}}}\). \(\blacksquare\)

Set \(z=e^{i\theta}\), \(\cos\theta=\frac{z+z^{-1}}{2}\), \(d\theta=\frac{dz}{iz}\). Integrand: \(\frac{1}{5-4\cdot\frac{z+z^{-1}}{2}}\cdot\frac{1}{iz}=\frac{1}{iz(5-2(z+z^{-1}))}=\frac{1}{iz}\cdot\frac{z}{-2z^2+5z-2}=\frac{1}{i(-2z^2+5z-2)}\). Factor: \(-2z^2+5z-2=-(2z^2-5z+2)=-(2z-1)(z-2)\). Poles at \(z=1/2\) (inside) and \(z=2\) (outside). Residue at \(z=1/2\): \(\frac{1}{i\cdot(-(z-2)\cdot4)}\big|_{z=1/2}=\frac{1}{i\cdot(-(-3/2)\cdot4)}... \). Using \(\frac{1}{i(-(2z-1)(z-2))}\), res at \(z=\frac{1}{2}\): \(\frac{1}{i\cdot(-(2\cdot\frac{1}{2}-1-...)\cdot(-3/2))}...\). Direct: factor \(-2z^2+5z-2\), res of \(\frac{1}{i(-2z^2+5z-2)}\) at \(z=\tfrac12\) is \(\frac{1}{i(-4z+5)}\big|_{z=1/2}=\frac{1}{i\cdot3}=\frac{1}{3i}\). Integral \(=2\pi i\cdot\frac{1}{3i}=\frac{2\pi}{3}\). \(\blacksquare\)

Consider \(\oint_C\frac{dz}{1+z^2}\) over upper semicircle of radius \(R\). Poles at \(z=\pm i\); only \(z=i\) is in upper half-plane. \(\operatorname{Res}_{z=i}\frac{1}{(z-i)(z+i)}=\frac{1}{2i}\). Integral \(=2\pi i\cdot\frac{1}{2i}=\pi\). On the semicircular arc, the integral \(\to0\) as \(R\to\infty\) (by ML lemma: \(|\frac{1}{1+z^2}|\leq\frac{1}{R^2-1}\cdot\pi R\to0\)). Hence \(\int_{-\infty}^\infty\frac{dx}{1+x^2}=\pi\). \(\blacksquare\)

Poles of \(\frac{1}{1+z^4}\): 4th roots of \(-1\): \(z_k=e^{i\pi(2k+1)/4}\) for \(k=0,1,2,3\). In upper half-plane: \(z_0=e^{i\pi/4}=\frac{1+i}{\sqrt2}\) and \(z_1=e^{3i\pi/4}=\frac{-1+i}{\sqrt2}\). For simple poles: \(\operatorname{Res}_{z=z_k}=\frac{1}{4z_k^3}\). Sum of residues in UHP: \(\frac{1}{4z_0^3}+\frac{1}{4z_1^3}=\frac{1}{4}(e^{-3i\pi/4}+e^{-9i\pi/4})=\frac{1}{4}(e^{-3i\pi/4}+e^{-i\pi/4})\). \(e^{-3i\pi/4}+e^{-i\pi/4}=\cos\frac{3\pi}{4}-i\sin\frac{3\pi}{4}+\cos\frac{\pi}{4}-i\sin\frac{\pi}{4}=-\frac{1}{\sqrt2}-\frac{i}{\sqrt2}+\frac{1}{\sqrt2}-\frac{i}{\sqrt2}=-\frac{2i}{\sqrt2}=-i\sqrt2\). Sum \(=\frac{-i\sqrt2}{4}\). Integral \(=2\pi i\cdot\frac{-i\sqrt2}{4}=\frac{2\pi\sqrt2}{4}=\frac{\pi}{\sqrt2}\). \(\blacksquare\)

Same poles as previous: \(z_0=e^{i\pi/4},\,z_1=e^{3i\pi/4}\) in UHP. \(\operatorname{Res}_{z=z_k}\frac{z^2}{1+z^4}=\frac{z_k^2}{4z_k^3}=\frac{1}{4z_k}\). Sum: \(\frac{1}{4}(e^{-i\pi/4}+e^{-3i\pi/4})=\frac{1}{4}(\frac{1-i}{\sqrt2}+\frac{-1-i}{\sqrt2})=\frac{1}{4}\cdot\frac{-2i}{\sqrt2}=\frac{-i}{2\sqrt2}\). Integral \(=2\pi i\cdot\frac{-i}{2\sqrt2}=\frac{2\pi}{2\sqrt2}=\frac{\pi}{\sqrt2}\). \(\blacksquare\)

Consider \(f(z)=\frac{e^{iz}}{z}\) and indent the real axis at the origin with a small semicircle \(\gamma_\epsilon\) (radius \(\epsilon\), upper half). By Cauchy-Goursat (no poles inside): \(\int_{-R}^{-\epsilon}+\int_{\gamma_\epsilon}+\int_\epsilon^R+\int_{\Gamma_R}=0\). On \(\Gamma_R\): Jordan's lemma gives \(\to0\). On \(\gamma_\epsilon\) (traversed negatively): \(\int_{\gamma_\epsilon}\frac{e^{iz}}{z}dz\to-\pi i\cdot\operatorname{Res}_{z=0}=- \pi i\). Real part: \(\int_{-R}^{-\epsilon}\frac{\cos x}{x}dx+\int_\epsilon^R\frac{\cos x}{x}dx\to0\) (odd function). Imaginary part: \(\int_{-R}^{-\epsilon}\frac{\sin x}{x}dx+\int_\epsilon^R\frac{\sin x}{x}dx-\pi=0\Rightarrow 2\int_0^\infty\frac{\sin x}{x}dx=\pi\Rightarrow\int_0^\infty\frac{\sin x}{x}dx=\frac{\pi}{2}\). \(\blacksquare\)

Consider \(\oint_{\text{UHP}}\frac{e^{iaz}}{1+z^2}dz\). Pole in UHP at \(z=i\); \(\operatorname{Res}_{z=i}=\frac{e^{ia\cdot i}}{2i}=\frac{e^{-a}}{2i}\). By Jordan's lemma (\(a>0\)), the arc integral \(\to0\). So \(\int_{-\infty}^\infty\frac{e^{iax}}{1+x^2}dx=2\pi i\cdot\frac{e^{-a}}{2i}=\pi e^{-a}\). Taking the real part: \(\int_{-\infty}^\infty\frac{\cos(ax)}{1+x^2}dx=\pi e^{-a}\). Hence \(\int_0^\infty\frac{\cos(ax)}{1+x^2}dx=\frac{\pi e^{-a}}{2}\). \(\blacksquare\)

Jordan's Lemma: If \(f(z)\to0\) uniformly as \(|z|\to\infty\) in the upper half-plane, and \(a>0\), then \(\lim_{R\to\infty}\int_{\Gamma_R}f(z)e^{iaz}dz=0\), where \(\Gamma_R\) is the upper semicircle \(|z|=R\). Proof idea: Parametrise \(z=Re^{i\theta}\), \(0\leq\theta\leq\pi\). \(|e^{iaz}|=e^{-aR\sin\theta}\). Use \(\sin\theta\geq\frac{2\theta}{\pi}\) for \(\theta\in[0,\pi/2]\) (Jordan's inequality). Then \(\int_0^\pi e^{-aR\sin\theta}d\theta\leq\frac{\pi}{aR}\to0\). Application: Allows evaluating \(\int_{-\infty}^\infty f(x)e^{iax}dx\) via residues even when \(|f|\sim 1/|z|\) (too slow for direct ML). \(\blacksquare\)

Consider \(f(z)=\frac{z^{p-1}}{1+z}\) with branch cut along positive real axis (\(z^{p-1}=e^{(p-1)\log z}\), \(\arg z\in[0,2\pi)\)). Use keyhole contour: large circle \(|z|=R\), small circle \(|z|=\epsilon\), upper edge (arg=0) and lower edge (arg=\(2\pi\)). Arc contributions vanish. Upper edge: \(\int_0^\infty\frac{x^{p-1}}{1+x}dx=I\). Lower edge: \(\int_\infty^0\frac{x^{p-1}e^{2\pi i(p-1)}}{1+x}dx=-e^{2\pi i(p-1)}I\). Residue at \(z=-1=e^{i\pi}\): \(\operatorname{Res}=e^{i\pi(p-1)}=e^{ip\pi}e^{-i\pi}=-e^{ip\pi}\). So \(I(1-e^{2\pi i(p-1)})=2\pi i\cdot(-e^{ip\pi})\). Simplify: \(e^{2\pi i(p-1)}=e^{2\pi ip}\), so \(I(1-e^{2\pi ip})=-2\pi ie^{ip\pi}\). \(I=\frac{-2\pi ie^{ip\pi}}{1-e^{2\pi ip}}=\frac{2\pi ie^{ip\pi}}{e^{2\pi ip}-1}=\frac{2\pi i}{e^{i\pi p}-e^{-i\pi p}}=\frac{2\pi i}{2i\sin(p\pi)}=\frac{\pi}{\sin(p\pi)}\). \(\blacksquare\)

The improper integral \(\int_{-\infty}^\infty\frac{dx}{x}\) diverges. The Cauchy principal value is \(\text{P.V.}\int_{-\infty}^\infty\frac{dx}{x}=\lim_{\epsilon\to0^+,R\to\infty}\left[\int_{-R}^{-\epsilon}+\int_\epsilon^R\right]\frac{dx}{x}=\lim[\ln\epsilon-\ln R+\ln R-\ln\epsilon]=0\). The symmetric cancellation gives P.V.\(=0\). This uses the odd symmetry of \(1/x\). \(\blacksquare\)

Factor: \((z^2-4)^2=(z-2)^2(z+2)^2\). Poles of order 2 at \(z=\pm2\), both inside \(|z|=3\). For a pole of order 2 at \(z=a\): \(\operatorname{Res}=\lim_{z\to a}\frac{d}{dz}\left[(z-a)^2f(z)\right]\). At \(z=2\): \(g(z)=\frac{4z^2-3}{(z+2)^2}\), \(g'(z)=\frac{8z(z+2)^2-2(z+2)(4z^2-3)}{(z+2)^4}=\frac{8z(z+2)-2(4z^2-3)}{(z+2)^3}\). At \(z=2\): \(\frac{16\cdot4-2\cdot13}{4^3}=\frac{64-26}{64}=\frac{38}{64}=\frac{19}{32}\). At \(z=-2\): \(g(z)=\frac{4z^2-3}{(z-2)^2}\), \(g'(-2)=\frac{8(-2)(-4)-2(4\cdot4-3)}{(-2-2)^3}=\frac{64-26}{-64}=-\frac{38}{64}=-\frac{19}{32}\). Sum of residues \(=0\). Integral \(=2\pi i\cdot0=0\). \(\blacksquare\)

This is the same as the PYQ 2016 Beta function integral. The answer is \(\dfrac{\pi}{\sin(a\pi)}\) by the keyhole contour calculation shown above. Equivalently, using the Beta function: \(B(a,1-a)=\Gamma(a)\Gamma(1-a)=\frac{\pi}{\sin a\pi}\) (Euler's reflection formula). \(\blacksquare\)

Consider \(f(z)=\frac{(\log z)^2}{1+z^2}\) over the upper semicircle with branch cut on positive real axis (\(\log z = \ln|z|+i\arg z,\;0\leq\arg z\leq\pi\)). Poles in UHP at \(z=i\): \(\operatorname{Res}_{z=i}=\frac{(\log i)^2}{2i}=\frac{(i\pi/2)^2}{2i}=\frac{-\pi^2/4}{2i}=\frac{i\pi^2}{8}\). Upper edge: \(\int_0^\infty\frac{(\ln x)^2}{1+x^2}dx\). Lower edge (returning, on real axis, same branch since we go around UHP): \(-\int_0^\infty\frac{(\ln x)^2}{1+x^2}dx\) cancels. Actually use \(f(z)=\frac{\log z}{1+z^2}\) and semi-circular indented contour. \(\operatorname{Res}_{z=i}\frac{\log z}{1+z^2}=\frac{\log i}{2i}=\frac{i\pi/2}{2i}=\frac{\pi}{4}\). Real part: \(\int_{-\infty}^\infty\frac{\ln|x|}{1+x^2}dx=2\pi i\cdot\frac{\pi}{4}\) — taking imaginary part: \(\int_{-\infty}^\infty\frac{\arg(x)}{1+x^2}dx=\frac{\pi^2}{2}\)... By symmetry and careful analysis, \(\int_0^\infty\frac{\ln x}{1+x^2}dx=0\). (The odd symmetry of \(\ln x\) about \(x=1\) under \(x\to1/x\) proves this: substitute \(x=1/t\), integral transforms to its negative.) \(\blacksquare\)

On \(|z|=2\): let \(f(z)=3z\) and \(g(z)=z^5-1\). Then \(|f(z)|=3|z|=6\) and \(|g(z)|\leq|z|^5+1=33\). That doesn't work. Try \(f(z)=z^5\), \(g(z)=3z-1\). On \(|z|=2\): \(|f(z)|=32\) and \(|g(z)|\leq3\cdot2+1=7<32\). By Rouché, \(z^5+3z-1\) and \(z^5\) have the same number of zeros inside \(|z|=2\), namely 5. \(\blacksquare\)

(a) \(e^z=e^i\sum_{n=0}^\infty\frac{(z-i)^n}{n!}\), valid for all \(z\). (b) Let \(w=z-\pi\). \(\sin z=\sin(\pi+w)=-\sin w=-\sum_{n=0}^\infty\frac{(-1)^n w^{2n+1}}{(2n+1)!}\). So \(\frac{\sin z}{z-\pi}=\frac{-\sin w}{w}=-\sum_{n=0}^\infty\frac{(-1)^n w^{2n}}{(2n+1)!}=\sum_{n=0}^\infty\frac{(-1)^{n+1}(z-\pi)^{2n}}{(2n+1)!}\). At \(z=\pi\): value \(=\frac{(-1)^1}{1}=-1\). So the singularity at \(z=\pi\) is removable with value \(-1\). \(\blacksquare\)

Partial fractions: \(\frac{1}{z(z-1)}=\frac{-1}{z}+\frac{1}{z-1}\).

(a) \(0<|z|<1\): \(\frac{1}{z-1}=\frac{-1}{1-z}=-\sum_{n=0}^\infty z^n\). So \(f=\frac{-1}{z}-\sum_{n=0}^\infty z^n=-z^{-1}-1-z-z^2-\cdots\)

(b) \(|z|>1\): \(\frac{1}{z-1}=\frac{1}{z}\cdot\frac{1}{1-1/z}=\frac{1}{z}\sum_{n=0}^\infty z^{-n}=\sum_{n=0}^\infty z^{-(n+1)}\). So \(f=-z^{-1}+\sum_{n=0}^\infty z^{-(n+1)}=z^{-2}+z^{-3}+\cdots=\sum_{n=2}^\infty z^{-n}\). \(\blacksquare\)

Same as above (b): \(\frac{1}{z(z-1)}=\sum_{n=2}^\infty \frac{1}{z^n}=\frac{1}{z^2}+\frac{1}{z^3}+\frac{1}{z^4}+\cdots\), valid for \(|z|>1\). The principal part (negative powers of \(z\)) is the entire series here, but since these are \(z^{-n}\) for \(n\geq2\), there is no \(z^{-1}\) term, so the residue at \(\infty\) is related to the \(z^{-1}\) coefficient, which is 0 in this expansion. \(\blacksquare\)

\(e^{1/z}=\sum_{n=0}^\infty\frac{1}{n!z^n}=1+\frac{1}{z}+\frac{1}{2!z^2}+\frac{1}{3!z^3}+\cdots\), valid for \(z\neq0\). The principal part has infinitely many terms (\(z^{-n}\) for all \(n\geq1\)), so \(z=0\) is an essential singularity. Residue: coefficient of \(z^{-1}\) is \(1\). By Picard's great theorem, \(e^{1/z}\) takes every complex value except possibly one in every punctured neighbourhood of \(0\). \(\blacksquare\)

\(\sin(1/z)=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!z^{2n+1}}=\frac{1}{z}-\frac{1}{6z^3}+\frac{1}{120z^5}-\cdots\) The principal part is infinite, so \(z=0\) is an essential singularity. Note also that \(\sin(1/z)=0\) has solutions \(z=1/(n\pi)\) for integer \(n\neq0\), accumulating at \(0\): another signature of an essential singularity. \(\blacksquare\)

\(\sin z=z-\frac{z^3}{6}+\frac{z^5}{120}-\cdots\), so \(\frac{\sin z}{z}=1-\frac{z^2}{6}+\frac{z^4}{120}-\cdots\) This is a power series with no negative powers, so the singularity at \(z=0\) is removable. Defining \(f(0)=1\) makes \(f\) analytic at \(0\). Alternatively, \(\lim_{z\to0}\frac{\sin z}{z}=1\) is finite, confirming a removable singularity. \(\blacksquare\)

At \(z=1\), the function has the form \((z-1)^{-2}\), so it is a pole of order 2. The Laurent series is just \(\frac{1}{(z-1)^2}\) itself, which has principal part \((z-1)^{-2}\) and no \((z-1)^{-1}\) term. Hence the residue (coefficient of \((z-1)^{-1}\)) is \(\mathbf{0}\). For a pole of order \(m\): \(\operatorname{Res}_{z=a}f=\frac{1}{(m-1)!}\lim_{z\to a}\frac{d^{m-1}}{dz^{m-1}}[(z-a)^m f(z)]\). Here \(m=2\): \(\frac{d}{dz}[1]=0\). \(\blacksquare\)

Singularities where \(\cos z=0\): \(z=(2n+1)\frac{\pi}{2}\) for \(n\in\mathbb{Z}\), i.e.\ \(z=\pm\frac{\pi}{2},\pm\frac{3\pi}{2},\ldots\) Each is a simple pole since \(\cos z\) has simple zeros there (check: \(\sin z\neq0\) at these points). Residue at \(z_n=(2n+1)\frac{\pi}{2}\): \(\operatorname{Res}=\frac{\sin z_n}{(\cos z)'|_{z_n}}=\frac{\sin z_n}{-\sin z_n}=-1\). So residue is \(-1\) at every pole. \(\blacksquare\)

\(e^z=\sum_{k=0}^\infty\frac{z^k}{k!}\), so \(\frac{e^z}{z^n}=\sum_{k=0}^\infty\frac{z^{k-n}}{k!}\). The residue is the coefficient of \(z^{-1}\), i.e.\ the term with \(k-n=-1\), i.e.\ \(k=n-1\). Residue \(=\frac{1}{(n-1)!}\). The singularity at \(z=0\) is a pole of order \(n\). \(\blacksquare\)

If \(f\) has a simple pole at \(z=a\), then near \(a\): \(f(z)=\frac{c_{-1}}{z-a}+c_0+\cdots\), so \((z-a)f(z)=c_{-1}+(z-a)(\cdots)\to c_{-1}\) as \(z\to a\). When \(f(z)=p(z)/q(z)\) with \(q(a)=0,\,q'(a)\neq0,\,p(a)\neq0\): \(\operatorname{Res}=\frac{p(a)}{q'(a)}\). Example: \(\operatorname{Res}_{z=0}\frac{e^z}{z}=e^0=1\). Example: \(\operatorname{Res}_{z=i}\frac{1}{z^2+1}=\frac{1}{2i}\). \(\blacksquare\)

Pole of order 3. Residue \(=\frac{1}{2!}\lim_{z\to\pi}\frac{d^2}{dz^2}\cos z=\frac{1}{2}(-\cos\pi)=\frac{1}{2}\cdot1=\frac{1}{2}\). Alternatively, expand \(\cos z\) near \(z=\pi\): let \(w=z-\pi\), \(\cos(\pi+w)=-\cos w=-1+\frac{w^2}{2}-\cdots\). So \(\frac{\cos z}{(z-\pi)^3}=\frac{-1+w^2/2-\cdots}{w^3}=\frac{-1}{w^3}+\frac{1}{2w}-\cdots\). Residue (coefficient of \(w^{-1}\)) \(=\frac{1}{2}\). \(\blacksquare\)

Mittag-Leffler: A meromorphic function with simple poles at \(a_n\) and residues \(b_n\) can be expressed as \(f(z)=f(0)+\sum_n b_n\left(\frac{1}{z-a_n}+\frac{1}{a_n}\right)\) (suitably convergent). For \(\pi\cot(\pi z)\): poles at all integers \(n\in\mathbb{Z}\), residue at each pole is \(1\). The expansion is: \(\pi\cot(\pi z)=\frac{1}{z}+\sum_{n=1}^\infty\left(\frac{1}{z-n}+\frac{1}{z+n}\right)=\frac{1}{z}+\sum_{n=1}^\infty\frac{2z}{z^2-n^2}\). \(\blacksquare\)

\(\sin z\) has zeros at \(z=n\pi\) for all integers \(n\). The Weierstrass factorization gives: \(\sin z=z\prod_{n=1}^\infty\left(1-\frac{z^2}{n^2\pi^2}\right)\). This can be verified by comparing the zeros and growth. Taking logarithmic derivatives: \(\frac{d}{dz}\log\sin z=\cot z=\frac{1}{z}+\sum_{n=1}^\infty\frac{2z}{z^2-n^2\pi^2}\), consistent with Mittag-Leffler. A famous consequence: setting \(z=\pi/2\), \(\frac{\pi}{4}=\prod_{n=1}^\infty\frac{(2n)(2n)}{(2n-1)(2n+1)}\) (Wallis). \(\blacksquare\)

Suppose \(f\) is analytic at \(a\) and \(f(a)=0\). If \(f\not\equiv0\) near \(a\), expand: \(f(z)=(z-a)^m g(z)\) where \(g(a)\neq0\) and \(m\geq1\). Since \(g\) is continuous and \(g(a)\neq0\), there exists \(\delta>0\) such that \(g(z)\neq0\) for \(|z-a|<\delta\). Thus \(f(z)\neq0\) for \(0<|z-a|<\delta\): the zero is isolated. If every neighbourhood of \(a\) contains another zero, by continuity \(f\equiv0\) in a neighbourhood, and by the identity theorem \(f\equiv0\) everywhere. \(\blacksquare\)

Argument Principle: If \(f\) is meromorphic inside and on a simple closed contour \(C\) (with no zeros or poles on \(C\)), and \(N\) = number of zeros, \(P\) = number of poles (counted with multiplicity) inside \(C\), then \(\frac{1}{2\pi i}\oint_C\frac{f'(z)}{f(z)}dz=N-P\). The LHS equals \(\frac{1}{2\pi}\Delta_C\arg f(z)\), the total change in argument of \(f\) as \(z\) traverses \(C\). Proof: near a zero of order \(m\), \(f'/f\) has a simple pole with residue \(+m\); near a pole of order \(p\), residue is \(-p\). Sum gives \(N-P\). \(\blacksquare\)

Rouché's Theorem: If \(f,g\) are analytic inside and on a simple closed contour \(C\), and \(|g(z)|<|f(z)|\) on \(C\), then \(f\) and \(f+g\) have the same number of zeros inside \(C\). Proof: Let \(h(z)=1+g(z)/f(z)\). On \(C\): \(|g/f|<1\), so \(h(C)\) lies in \(\text{Re}(w)>0\) (open right half-plane), which excludes \(0\) and the negative real axis. By the argument principle, number of zeros of \(f+g\) minus poles \(=\frac{1}{2\pi i}\oint\frac{(f+g)'}{f+g}=\frac{1}{2\pi i}\oint\frac{f'}{f}+\frac{1}{2\pi i}\oint\frac{h'}{h}\). The last integral is zero since \(h\) has no zeros/poles (i.e.\ the winding number of \(h(C)\) around \(0\) is 0). So zeros of \(f+g=\) zeros of \(f\). \(\blacksquare\)

Identity Theorem: If \(f\) and \(g\) are analytic on a connected open set \(\Omega\) and \(f(z_n)=g(z_n)\) for a sequence \(z_n\to z_0\in\Omega\) with \(z_n\neq z_0\), then \(f\equiv g\) on \(\Omega\). Proof: \(h=f-g\) is analytic and has \(z_0\) as a non-isolated zero, so \(h\equiv0\). Application: Since \(\sin^2z+\cos^2z=1\) holds on the real axis (a sequence accumulating everywhere), it holds for all \(z\in\mathbb{C}\). Similarly \(e^{z_1+z_2}=e^{z_1}e^{z_2}\) extends from reals to all of \(\mathbb{C}\). \(\blacksquare\)

Schwarz Reflection Principle: If \(f\) is analytic in a region \(\Omega\) in the upper half-plane, continuous on the portion of the real axis bounding \(\Omega\), and real-valued on that real boundary, then \(f\) extends analytically to the reflection \(\Omega^*=\{\bar{z}:z\in\Omega\}\) by \(f(\bar{z})=\overline{f(z)}\). More generally applies to reflection across any arc. Application: If \(f\) maps an arc of the real axis to the real axis, it extends across by the formula above. Used to extend power series from the real axis and in conformal mapping. \(\blacksquare\)

Weierstrass Factorization: Every entire function \(f\) with zeros \(a_1,a_2,\ldots\) (counted with multiplicity) can be written as \(f(z)=z^m e^{g(z)}\prod_{n=1}^\infty E_p\!\left(\frac{z}{a_n}\right)\) where \(E_p(u)=(1-u)e^{u+u^2/2+\cdots+u^p/p}\) are elementary factors and \(g\) is an entire function. Example: \(\cos z=\prod_{n=0}^\infty\!\left(1-\frac{4z^2}{(2n+1)^2\pi^2}\right)\). Another: \(\Gamma(z)^{-1}=ze^{\gamma z}\prod_{n=1}^\infty(1+z/n)e^{-z/n}\) where \(\gamma\) is the Euler-Mascheroni constant. \(\blacksquare\)

Liouville: A bounded entire function is constant. Extended: If \(|f(z)|\leq M|z|^n\) for large \(|z|\), then \(f\) is a polynomial of degree \(\leq n\). FTA via Liouville: Suppose \(p(z)\) is a non-constant polynomial with no zeros. Then \(1/p(z)\) is entire. Since \(|p(z)|\to\infty\), \(|1/p(z)|\to0\), so \(1/p\) is bounded. By Liouville, \(1/p\) is constant, hence \(p\) is constant — contradiction. So \(p\) has at least one zero. \(\blacksquare\)

Write \(z=re^{i\theta}\) with \(r>0,\,\theta\in(0,\pi)\) (since \(\operatorname{Im}z>0\)). Then \(w=\ln r+i\theta\), so \(\operatorname{Re}w=\ln r\in(-\infty,\infty)\) and \(\operatorname{Im}w=\theta\in(0,\pi)\). Hence the image is exactly the strip \(\{w:0<\operatorname{Im}w<\pi\}\). The real axis maps to \(\operatorname{Im}w=0\) (for \(x>0\)) and \(\operatorname{Im}w=\pi\) (for \(x<0\)). \(\blacksquare\)

If \(|z|<1\) and \(z\neq0\), then \(|w|=|1/z|=1/|z|>1\). Conversely if \(|w|>1\), let \(z=1/w\); then \(|z|<1\). So the map is a bijection from \(\{0<|z|<1\}\) to \(\{|w|>1\}\). The boundary \(|z|=1\) maps to \(|w|=1\). The map is an involution (\(1/(1/z)=z\)) and is conformal (since \(dw/dz=-1/z^2\neq0\) for \(z\neq0\)). \(\blacksquare\)

The Schwarz-Christoffel formula maps the upper half-plane conformally onto a polygon. If the polygon has vertices \(w_1,\ldots,w_n\) with interior angles \(\alpha_k\pi\), the mapping is \(f'(z)=A\prod_{k=1}^n(z-x_k)^{\alpha_k-1}\) where \(x_k\) are the preimages on the real axis and \(A\) is a complex constant. Integrating gives \(f\). Application: maps UHP to rectangles (elliptic integrals), triangles, slits. For a rectangle with angles \(\pi/2\): involves the Schwarz-Christoffel integral \(\int(1-t^2)^{-1/2}(1-k^2t^2)^{-1/2}dt\) — an elliptic integral. \(\blacksquare\)

Maps from UHP to itself have the form \(w=\frac{az+b}{cz+d}\) with \(a,b,c,d\) real and \(ad-bc>0\). Fixed points: \(z=\frac{az+b}{cz+d}\Rightarrow cz^2+(d-a)z-b=0\). Fixing \(z=0\): \(0=\frac{a\cdot0+b}{c\cdot0+d}=\frac{b}{d}\Rightarrow b=0\). Then \(w=\frac{az}{cz+d}\). Fixing \(z=i\): \(i=\frac{ai}{ci+d}\Rightarrow i(ci+d)=ai\Rightarrow-c+di=ai\). Real part: \(-c=0\Rightarrow c=0\). Imaginary: \(d=a\). So \(w=\frac{az}{a}=z\) (identity). The only bilinear map fixing both \(0\) and \(i\) in the UHP is the identity. \(\blacksquare\)

Write \(z=re^{i\theta}\) with \(r>0,\,\theta\in(0,\pi/2)\). Then \(w=r^2e^{2i\theta}\), with \(\arg w=2\theta\in(0,\pi)\). So \(\operatorname{Im}w=r^2\sin(2\theta)>0\) (since \(2\theta\in(0,\pi)\)). The positive real axis (\(\theta=0\)) maps to the positive real axis, and the positive imaginary axis (\(\theta=\pi/2\)) maps to the negative real axis. The map is 2-to-1 on the full plane but 1-to-1 on the first quadrant. \(\blacksquare\)

Write \(z=x+iy\) with \(|x|<\pi/2\). Then \(w=\sin(x+iy)=\sin x\cosh y+i\cos x\sinh y\). Set \(u=\sin x\cosh y,\,v=\cos x\sinh y\). For fixed \(x\neq0\): \(\frac{u^2}{\sin^2x}-\frac{v^2}{\cos^2 x}=\cosh^2y-\sinh^2y=1\) (hyperbola in \(w\)-plane). For fixed \(y\): \(\frac{u^2}{\cosh^2y}+\frac{v^2}{\sinh^2y}=1\) (ellipse). The strip maps onto the entire \(w\)-plane. The boundary \(x=\pi/2\): \(u=\cosh y,\,v=0\), giving \([1,\infty)\) on the real axis (a slit). So \(\sin z\) maps the open strip conformally onto \(\mathbb{C}\setminus((-\infty,-1]\cup[1,\infty))\). \(\blacksquare\)

Let \(f\) be analytic at \(z_0\) with \(f'(z_0)\neq0\). Two smooth curves \(\gamma_1,\gamma_2\) meet at \(z_0\) with tangent directions \(\tau_1,\tau_2\). Under \(f\), the image curves have tangent directions \(f'(z_0)\tau_1\) and \(f'(z_0)\tau_2\). The angle between image curves is \(\arg\frac{f'(z_0)\tau_1}{f'(z_0)\tau_2}=\arg\frac{\tau_1}{\tau_2}\) — unchanged. So angles and their orientation are preserved. Where \(f'(z_0)=0\), angles are multiplied by the order of the zero: the map is not conformal there. \(\blacksquare\)

For \(z=\rho e^{i\theta}\) with \(\rho\in(1,r)\) and \(\theta\in(-\pi,\pi]\): \(w=\ln\rho+i\theta\). So \(\operatorname{Re}w=\ln\rho\in(0,\ln r)\) and \(\operatorname{Im}w=\theta\in(-\pi,\pi)\). The image is the rectangle \(\{w:0<\operatorname{Re}w<\ln r,\,-\pi<\operatorname{Im}w<\pi\}\). The inner circle \(|z|=1\) maps to the left edge \(\operatorname{Re}w=0\), the outer circle to \(\operatorname{Re}w=\ln r\). The branch cut (negative real axis) is the seam. \(\blacksquare\)

Riemann Mapping Theorem: Every simply connected proper open subset \(\Omega\subsetneq\mathbb{C}\) is conformally equivalent to the open unit disk \(\mathbb{D}=\{|z|<1\}\). That is, there exists a bijective analytic function \(f:\Omega\to\mathbb{D}\) with analytic inverse. Moreover, if \(z_0\in\Omega\), the map is unique if we require \(f(z_0)=0\) and \(f'(z_0)>0\). Note: \(\mathbb{C}\) itself is excluded (Liouville: any bounded entire function is constant, so no biholomorphism \(\mathbb{C}\to\mathbb{D}\) exists). \(\blacksquare\)

Let \(\Omega\) be a domain and \(E\subset\partial\Omega\). The harmonic measure \(\omega(z,E,\Omega)\) is the harmonic function on \(\Omega\) with boundary values \(1\) on \(E\) and \(0\) on \(\partial\Omega\setminus E\). Interpretation: \(\omega(z,E,\Omega)\) is the probability that Brownian motion starting at \(z\) first hits \(\partial\Omega\) in \(E\). For the Dirichlet problem with boundary data \(\phi\) on \(\partial\Omega\): \(u(z)=\int_{\partial\Omega}\phi(\zeta)\,d\omega(z,\zeta,\Omega)\). For the disk: the harmonic measure is given by the Poisson kernel. \(\blacksquare\)

Strategy: (1) Find a conformal map \(f:\Omega\to\mathbb{D}\) (Riemann mapping theorem). (2) Solve the Dirichlet problem in \(\mathbb{D}\) with the transformed boundary data using the Poisson integral formula: \(u(re^{i\theta})=\frac{1}{2\pi}\int_0^{2\pi}\frac{(1-r^2)\phi(e^{it})}{1-2r\cos(\theta-t)+r^2}dt\). (3) Pull back: \(\tilde{u}=u\circ f\) is harmonic in \(\Omega\) (harmonic functions compose with conformal maps). Boundary values are preserved. This works because harmonicity is invariant under conformal mappings. \(\blacksquare\)

The complex potential for uniform flow past a cylinder of radius \(a\) is \(F(z)=U(z+a^2/z)\) (superposition of uniform flow and doublet). The Joukowski map \(w=z+c^2/z\) maps the cylinder \(|z|=a\) to an airfoil shape in the \(w\)-plane. The complex velocity in the \(w\)-plane is \(dF/dw=(dF/dz)/(dw/dz)\), giving the flow field around the airfoil. The Kutta condition (smooth flow at trailing edge) determines the circulation, and the Kutta-Joukowski theorem gives lift \(=\rho U\Gamma\) per unit span. \(\blacksquare\)

The electrostatic potential \(\phi\) satisfies Laplace's equation \(\nabla^2\phi=0\). If \(w=f(z)\) is conformal, the composition \(\phi\circ f^{-1}\) is also harmonic. Strategy: map the given electrode configuration to a simpler geometry (e.g.\ parallel plates), solve there, then pull back. Example: Two concentric cylinders \(r=a,b\) — map to a strip via \(w=\log z\), where the potential is linear. Pull back gives \(\phi=A\log r+B\) (familiar coaxial result). \(\blacksquare\)

Critical points: \(dw/dz=1-1/z^2=0\Rightarrow z=\pm1\). At these points conformality fails. For the circle \(|z|=r\): \(w=(r+1/r)\cos\theta+i(r-1/r)\sin\theta\). This is an ellipse with semi-axes \(a=r+1/r,\,b=|r-1/r|\). For \(r>1\): \(a>b>0\), a non-degenerate ellipse. For \(r=1\): \(b=0\), the segment \([-2,2]\). For \(r<1\): same ellipse as for \(1/r\) (the map satisfies \(w(1/z)=w(z)\)). \(\blacksquare\)

\(\cosh(iz)=\frac{e^{iz}+e^{-iz}}{2}=\cos z\) by Euler's formula. \(\sinh(iz)=\frac{e^{iz}-e^{-iz}}{2}=\frac{e^{iz}-e^{-iz}}{2}=i\cdot\frac{e^{iz}-e^{-iz}}{2i}=i\sin z\). Conversely: \(\cos(iz)=\cosh z\) and \(\sin(iz)=i\sinh z\). These identities allow trigonometric identities to be converted to hyperbolic ones: e.g.\ \(\cos^2z+\sin^2z=1\Rightarrow\cosh^2z-\sinh^2z=1\). \(\blacksquare\)

\(\sin z=\sin(x+iy)=\sin x\cos(iy)+\cos x\sin(iy)=\sin x\cosh y+i\cos x\sinh y\). So \(|\sin z|^2=\sin^2x\cosh^2y+\cos^2x\sinh^2y=\sin^2x\cosh^2y+(1-\sin^2x)\sinh^2y=\sin^2x(\cosh^2y-\sinh^2y)+\sinh^2y=\sin^2x\cdot1+\sinh^2y=\sin^2x+\sinh^2y\). \(\blacksquare\)

By Euler's formula \(e^{i\theta}=\cos\theta+i\sin\theta\). Then \((\cos\theta+i\sin\theta)^n=(e^{i\theta})^n=e^{in\theta}=\cos(n\theta)+i\sin(n\theta)\). This holds for all integer \(n\) and can be extended to rational \(n\) (with care for multi-valuedness). Application: \(\cos3\theta=\text{Re}(e^{3i\theta})=\cos^3\theta-3\cos\theta\sin^2\theta=4\cos^3\theta-3\cos\theta\). \(\blacksquare\)

The \(n\)-th roots of unity are \(\omega_k=e^{2\pi ik/n}\) for \(k=0,1,\ldots,n-1\). They are the roots of \(z^n-1=0\). By Vieta's formulas: sum \(=0\) (coefficient of \(z^{n-1}\) in \(z^n-1\) is \(0\)). Product \(=(-1)^n\cdot(-1)=(-1)^{n+1}\)... More carefully: product of roots of \(z^n-1\) is \((-1)^n\cdot(-1)=(-1)^{n+1}\)... Actually by Vieta, product \(=(-1)^n\cdot\frac{-1}{1}=(-1)^{n+1}\). For even \(n\): product \(=(-1)^{n+1}=-1\). For odd \(n\): product \(=1\). Alternatively: \(\prod_{k=0}^{n-1}e^{2\pi ik/n}=e^{2\pi i\cdot n(n-1)/(2n)}=e^{i\pi(n-1)}\). For even \(n\): \(e^{i\pi(n-1)}=e^{-i\pi}=-1\). For odd: \(e^{i\pi(n-1)}=1\). \(\blacksquare\)

Direct substitution: let \(u=\cos\theta\), \(du=-\sin\theta\,d\theta\). Limits: \(\theta=0\Rightarrow u=1\), \(\theta=\pi\Rightarrow u=-1\). Integral \(=\int_1^{-1}\frac{-du}{5-4u}=\int_{-1}^1\frac{du}{5-4u}=\left[-\frac{1}{4}\ln|5-4u|\right]_{-1}^1=-\frac{1}{4}[\ln1-\ln9]=\frac{\ln9}{4}=\frac{\ln3}{2}\). Alternatively via contour: \(z=e^{i\theta}\), \(\sin\theta=\frac{z-z^{-1}}{2i}\), use residues on \(|z|=1\). \(\blacksquare\)

Consider \(\oint_C\frac{ze^{iz}}{z^2+a^2}dz\) over upper semicircle. Pole in UHP at \(z=ia\): \(\operatorname{Res}_{z=ia}=\frac{iae^{i(ia)}}{2ia}=\frac{e^{-a}}{2}\). By Jordan's lemma, arc \(\to0\). So \(\int_{-\infty}^\infty\frac{xe^{ix}}{x^2+a^2}dx=2\pi i\cdot\frac{e^{-a}}{2}=\pi ie^{-a}\). Equating imaginary parts: \(\int_{-\infty}^\infty\frac{x\sin x}{x^2+a^2}dx=\pi e^{-a}\). (Real part gives \(\int\frac{x\cos x}{x^2+a^2}dx=0\) by the odd-function structure, consistent.) \(\blacksquare\)

Partial fractions: \(f(z)=\dfrac{-1}{z}+\dfrac{1}{z-1}\). For \(|z|>1\): \[\frac{1}{z-1}=\frac{1}{z\left(1-\tfrac{1}{z}\right)}=\frac{1}{z}\sum_{n=0}^\infty\frac{1}{z^n}=\sum_{n=0}^\infty z^{-(n+1)}.\] Hence \[f(z)=\frac{-1}{z}+\sum_{n=0}^\infty z^{-(n+1)}=\frac{-1}{z}+\frac{1}{z}+\frac{1}{z^2}+\frac{1}{z^3}+\cdots=\sum_{n=2}^\infty z^{-n}=\frac{1}{z^2}+\frac{1}{z^3}+\cdots,\quad|z|>1.\] \(\blacksquare\)

Substitute \(z=e^{i\theta}\), \(d\theta=dz/(iz)\), \(\cos\theta=\tfrac{1}{2}(z+z^{-1})\): \[I=\oint_{|z|=1}\frac{1}{2+\frac{z+z^{-1}}{2}}\cdot\frac{dz}{iz}=\oint_{|z|=1}\frac{2\,dz}{i(z^2+4z+1)}.\] Roots of \(z^2+4z+1=0\): \(z=-2\pm\sqrt{3}\). Only \(z_1=-2+\sqrt{3}\approx-0.27\) lies inside \(|z|=1\). Residue at \(z_1\): \[\operatorname{Res}_{z=z_1}\frac{2}{i(z^2+4z+1)}=\frac{2}{i(2z_1+4)}=\frac{2}{i\cdot2(2-\sqrt{3}+2)\cdot\frac{1}{1}}=\frac{2}{i\cdot2\sqrt{3}\cdot\frac{(2+\sqrt{3})}{(2+\sqrt{3})}}.\] More directly: \(2z_1+4=2(-2+\sqrt{3})+4=2\sqrt{3}\), so \(\operatorname{Res}=\dfrac{2}{i\cdot 2\sqrt{3}}=\dfrac{1}{i\sqrt{3}}\). By the residue theorem: \[I=2\pi i\cdot\frac{1}{i\sqrt{3}}=\frac{2\pi}{\sqrt{3}}.\] \(\blacksquare\)

Poles: \(z=1\) (order 2), \(z=-2\) (simple).
At \(z=-2\) (simple pole): \[\operatorname{Res}_{z=-2}f=\lim_{z\to-2}(z+2)f(z)=\frac{(-2)^2}{(-2-1)^2}=\frac{4}{9}.\] At \(z=1\) (order-2 pole): \[\operatorname{Res}_{z=1}f=\lim_{z\to1}\frac{d}{dz}\!\left[(z-1)^2 f(z)\right]=\lim_{z\to1}\frac{d}{dz}\!\left[\frac{z^2}{z+2}\right]=\lim_{z\to1}\frac{2z(z+2)-z^2}{(z+2)^2}=\frac{2\cdot3-1}{9}=\frac{5}{9}.\] Verification: For a contour enclosing both poles, \(\oint f\,dz=2\pi i\!\left(\tfrac{4}{9}+\tfrac{5}{9}\right)=2\pi i\). \(\blacksquare\)

CR equations: \(u_x=v_y\) and \(u_y=-v_x\).
Proof: If \(u=v\), substitute into CR: \[u_x=v_y=u_y\quad\text{and}\quad u_y=-v_x=-u_x.\] From the second equation \(u_y=-u_x\); from the first \(u_x=u_y\). Combining: \(u_x=-u_x\Rightarrow u_x=0\), and then \(u_y=0\). Hence \(\nabla u=\mathbf{0}\) on the domain, so \(u\) (and \(v=u\)) is constant. Therefore \(f=u+iv=u+iu=(1+i)u\) is a real constant times \((1+i)\), i.e.\ \(f\) is constant. \(\blacksquare\)

Consider \(\oint_C\dfrac{z^2}{(z^2+1)(z^2+4)}\,dz\) over the upper semicircle \(C\). By Jordan/standard estimates the arc vanishes. Poles in UHP: \(z=i\) (from \(z^2+1=0\)) and \(z=2i\) (from \(z^2+4=0\)).
Res at \(z=i\): \[\operatorname{Res}_{z=i}=\frac{i^2}{(i+i)(i^2+4)}=\frac{-1}{2i\cdot3}=\frac{-1}{6i}.\] Res at \(z=2i\): \[\operatorname{Res}_{z=2i}=\frac{(2i)^2}{(2i)^2+1)\cdot(2i+2i)}\bigg|_{\text{simple}}=\frac{-4}{(-3)\cdot4i}=\frac{1}{3i}.\] Sum of residues: \(\dfrac{-1}{6i}+\dfrac{1}{3i}=\dfrac{1}{6i}\). \[I=2\pi i\cdot\frac{1}{6i}=\frac{\pi}{3}.\] \(\blacksquare\)

Partial fractions: \(\dfrac{z}{(z-1)(z-2)}=\dfrac{-1}{z-1}+\dfrac{2}{z-2}\).
For \(|z|>1\): \(\dfrac{1}{z-1}=\dfrac{1}{z}\cdot\dfrac{1}{1-1/z}=\displaystyle\sum_{n=0}^\infty z^{-(n+1)}\).
For \(|z|<2\): \(\dfrac{1}{z-2}=\dfrac{-1}{2}\cdot\dfrac{1}{1-z/2}=-\displaystyle\sum_{n=0}^\infty\frac{z^n}{2^{n+1}}\).
Combining: \[f(z)=-\sum_{n=0}^\infty z^{-(n+1)}+2\cdot\left(-\sum_{n=0}^\infty\frac{z^n}{2^{n+1}}\right)=-\sum_{n=0}^\infty z^{-(n+1)}-\sum_{n=0}^\infty\frac{z^n}{2^n},\quad 1<|z|<2.\] Explicitly: \[f(z)=\cdots-\frac{1}{z^2}-\frac{1}{z}-1-\frac{z}{2}-\frac{z^2}{4}-\cdots\] The principal part is \(\displaystyle-\sum_{n=1}^\infty z^{-n}\) and the analytic part is \(-\displaystyle\sum_{n=0}^\infty\frac{z^n}{2^n}\). \(\blacksquare\)