Calculate the following integrals of complex variable

Question

I am beginning to compute integrals of complex variable functions. As soon as I have them I will share my solutions, so that you can give me your opinion.

$$ i) \int_{|z|=1} \frac{e^z}{z^n} dz$$

$$ ii) \int_{|z|=2} \frac{dz}{z^2+1}$$

$$ iii) \int_{|z|=R} \frac{|dz|}{|z-a|^2}, |a| \ne R$$

$$ iv) \int_{|z|=R} \frac{|dz|}{|z-a|^4}, |a| \ne R$$

For (i) I used:

$f(z)=e^z$, then $f^{(n-1)}(z)=e^z$ and $f^{(n-1)}(0)=1$.

$z^n=(z-0)^{(n-1)+1}$

Winding number is 1. Then, with the Cauchy's Integral Formula:

$$ \int_{|z|=1} \frac{e^z}{z^n} dz= \frac{2 \pi i}{(n-1)!}$$

for (ii):

use partial fractions and, use Cauchy's Integral Formula.

Answer 1

For the third question, I do not quite agree with the previous author. Reparametrize $z(t)=Re^{it}$, where $t\in[0,2\pi]$. $\left|dz\right|=Rdt$ by a simple calculation. By techniques of partial fractions, we yield $$=\int_{|z|=R}\frac{-idz/z}{(z-a)(\bar{z}-\bar{a})}=-iR\int_{|z|=R}\frac{dz}{(z-a)(R^2-\bar{a}z)}=\frac{iR}{\bar{a}\left(a-\frac{R^2}{\bar{a}}\right)}\int_{|z|=R}\frac{z-R^2/\bar{a}-z+a}{(z-a)\left(z-\frac{R^2}{\bar{a}}\right)}dz=\frac{iR}{|a|^2-R^2}\left(\int_{|z|=R}\frac{dz}{z-a}-\int_{|z|=a}\frac{dz}{z-R^2/a}\right)$$ If $|a|<R$, $$\int_{|z-a|=\alpha}\frac{dz}{z-a}=2\pi i$$ where $0<\alpha<R-|a|$. Since $\frac{R^2}{\bar{a}}\notin\{z:|z|=R\}$, $$n\left(\gamma,\frac{R^2}{\bar{a}}\right)=\int_{|z|=a}\frac{dz}{z-R^2/a}=0$$ Hence, the integral equals to $$\frac{2\pi R}{R^2-|a|^2}$$ If $|a|>R$, it is just the opposite sign of the result of the first case.

For the fourth question, it is very similar to the method I use in the third one. $|dz|=iR\frac{dz}{z}$. (Why?) If $a=0$, we can reduce the integral to $$-iR\int_{|z|=R}\frac{dz}{R^4z}=\frac{2\pi}{R^3}$$ When $a\neq 0$, we need to consider the cases $|a|>R$ and $|a|<R$. When $|a|>R$, by some techniques, you can reduce the integral to $$-\frac{iR}{\bar{a}^2}\int_{|z|=R}\frac{zdz}{\left(z-\frac{R}{\bar{a}}\right)(z-a)^2}$$ Let $b=\frac{R}{\bar{a}}$. Since $\frac{z}{(z-a)^2}$ is analytic in $|z|=R$, the answer will be $$\frac{2\pi R}{\bar{a}^2}\frac{a+b}{(a-b)^3}$$ Something similar works for $|a|<R$.

Answer 2

$\textbf{Hint:}$ Use the following relationships

$$dz = iRe^{it}dt \implies |dz| = Rdt = \frac{R\cdot iRe^{it}dt}{iRe^{it}} = \frac{Rdz}{iz}$$

$$\bar{z} = \frac{R^2}{z}$$

to change the last two integrals into normal line integrals that we can use our theorems on. Can you take it from here?

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The third integral can be manipulated into

$$\int\limits_{|z|=R} \frac{|dz|}{|z|^2-\bar{a}z-a\bar{z}+|a|^2} = \int\limits_{|z|=R} \frac{Rdz}{R^2-\bar{a}z-\frac{aR^2}{z}+|a|^2}\cdot\frac{1}{iz}$$

$$ = \frac{1}{2\pi i}\int\limits_{|z|=R}\frac{-2\pi Rdz}{\bar{a}z^2-(R^2+|a|^2)z+aR^2}$$

which by quadratic formula has roots at

$$z = \frac{R^2+|a|^2}{2\bar{a}}\pm\sqrt{\frac{(R^2+|a|^2)^2}{4\bar{a}^2}-\frac{a}{\bar{a}}R^2}$$

which gives a residue of

$$= \frac{-2\pi R}{\sqrt{(R^2+|a|^2)^2-4|a|^2R^2}}$$