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$$ \begin{align} \gamma(t)=2cost + isint, t\in[0,2\pi] \end{align} $$

Could someone verify my thinking and my results for the following Integrals and if I have properly justified my thoughts: $$ \begin{align}1)I_1= \int_\gamma \sin{(z^2)} dz=0 \end{align} $$ $\sin{(z^2)}$ is analytic everywhere in and on $\gamma$ therefore due to Cauchy's Integral Theorem $I_1=0$. $$ \begin{align}2)I_2= \int_\gamma\frac{1}{z}dz=2\pi i \end{align} $$ If Γ is any positively-oriented loop in the complex plane and $z_0$ a point in the interior of Γ then

$\int_{\Gamma}\frac{1}{z-z_0}dz=2\pi i$ . In this case $z_0=0$ which lies in the interior of $\gamma$ therefore $I_2=2\pi i$.

$$ \begin{align}3)I_3= \int_\gamma\frac{1}{z^2+2iz}dz=2\pi i \end{align} $$ $$ \begin{align}3)I_3= \int_\gamma\frac{1}{z^2+2iz}dz= \int_\gamma\frac{1}{z(z+2i)}dz=\\= \int_\gamma\frac{i}{2(z+2i)}dz-\int_\gamma\frac{i}{2z}dz=\\ 0-\frac{i}{2}2\pi i=\pi \end{align} $$ because $z+2i=0$ in the exterior of $\gamma , \int_\gamma\frac{i}{2(z+2i)}dz=0$ due to Cauchy's Integral Theorem

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Hi again @helplessKirk.

$1)$ and $2)$ look fine to me. Although the result of $3)$ is also OK, let me write down for you how the residue theorem would work here.

Note that $f(z) = 1/(z^2 + 2 \mathrm{i} \, z)$ has two poles, each one of first order, given by: $z_1 = 0, \ z_2 = -2\mathrm{i} $. Note again that only $z_1$ lies in the region enclosed by your curve $\gamma$. Hence, the residue theorem tells us that ($n=1, \ c= 0$):

$$ \color{blue}{I_3 =\oint_\gamma f(z) \mathrm{d}z = 2 \pi \mathrm{i} \times 1 \times \lim_{z\to z_1}(z-z_1) \frac{1}{z^2 + 2 \mathrm{i} z} = 2 \pi \mathrm{i} \, \frac{1}{\underbrace{z_1-z_2}_{2 \mathrm{i}}} = \pi }$$

Cheers!

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  • $\begingroup$ thanks for the explanation! made things more clear. $\endgroup$ Aug 16, 2014 at 10:41
  • $\begingroup$ You're welcome @helplessKirk. Though I'm not much of an expert in complex integration, I'm happy to help! $\endgroup$
    – Dmoreno
    Aug 16, 2014 at 10:42

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