Complex Integral Cauchy Integral Formula

Question

Im trying to calculate

$$\int_{\gamma}\frac{e^{iz}}{z^2} dz$$ where $\gamma(t)=e^{it}, 0\le t \le 2\pi$

My problem is that $z^2$ in the denominator.

If the function was $\frac{e^{iz}}{z}$ as $f(z)=e^{iz}$ is holomorphic

$\int{_{\gamma}\frac{e^{iz}}{z}dz}=2\pi if(0)=2\pi i$

Obs: I dont know Residue Theorem

Thanks!

Answer 1

Note that from the Cauchy Integral Formula

$$f(z)=\frac1{2\pi i}\oint_{|z'-z|=1}\frac{f(z')}{z'-z}\,dz'$$

we can express the first derivative of $f(z)$, $f'(z)$, as

$$f'(z)=\frac{1}{2\pi i}\oint_{|z'-z|=1}\frac{f(z')}{(z'-z)^2}\,dz'\tag 1$$

Setting $z=0$ in $(1)$ yields

$$f'(0)=\frac{1}{2\pi i}\oint_{|z'|=1}\frac{f(z')}{z'^2}\,dz'\tag2$$

Now, letting $f(z)=e^{iz}$ with $f'(0)=i$ in $(2)$ reveals

$$i=\frac{1}{2\pi i}\oint_{|z'|=1}\frac{e^{iz'}}{z'^2}\,dz'$$

whereupon solving for the integral of interest we find

$$\bbox[5px,border:2px solid #C0A000]{\oint_{|z'|=1}\frac{e^{iz'}}{z'^2}\,dz'=-2\pi}$$

Answer 2

Doing a Taylor expansion of $e^{iz},$ resulting in a Laurent expansion of $\frac{e^{iz}}{z^2},$ we have $$ \oint_\gamma \frac{e^{iz}}{z^2} dz = \oint_\gamma \frac{1+iz+O(z^2)}{z^2} dz = \oint_\gamma \left( \frac{1}{z^2} + i\frac{1}{z} + O(z) \right) dz $$ Here $O(z)$ is holomorphic.

It's now easy to verify now that $$\oint_\gamma \frac{1}{z^2} dz = 0, \quad \oint_\gamma \frac{1}{z} dz = i\ 2\pi \quad \oint_\gamma O(z) dz = 0$$

Thus, $$\oint_\gamma \frac{e^{iz}}{z^2} dz = i \cdot i\ 2\pi = -2\pi$$