Evaluate the integral $\int_\gamma f(z)dz,$ where $\gamma(t)=e^{it}$, and $0\leqslant t\leqslant2\pi$. For $f(z)$ equal to: $$\dfrac{e^z}{z^3},\quad\dfrac1{z^2\sin z},\quad\tanh z,\quad\dfrac1{\cos(2z)}.$$
I think I need to find the points where the functions are undefined?? So for the first one $z=0$, however, not too sure where to go from there?
I have solved;
i) The Laurent series is; $1/z^3 + 1/z^2 + \dots$
hence residue $= z(1/z^3 +1/z^2+ 1/z + \dots) $ at $z=0$ , so residue$=1/2$
therefore $\int_\gamma f(z)dz=\pi i$
For the first one, I get $$ \frac{e^z}{z^3} = \frac{1}{z^3} + \frac{1}{z^2} + \frac{1}{2z} + \cdots $$ So I don't follow your expansion were you go from $\frac{1}{z^3}$ to $\frac{1}{6z}$. However, the residue is $1/2$ so $$ \oint f(z) dz = \pi i $$ For the second problem, we have $$ \frac{1}{z^2\sin(z)} = \frac{1}{z^3(1 - z^2/(3!) + \cdots)} $$ so we have a pole at $z = 0$ of order three. For three, we can write $\tanh(z)$ as $$ \tanh(z) = \frac{\sinh(z)}{\cosh(z)}\text{ or as } \frac{e^{z} - e^{-z}}{e^{z} + e^{-z}} $$ If we use the second identity for $\tanh(z)$, we have a pole when $e^z+e^{-z} = 0\Rightarrow e^{2z} = -1 = e^{i\pi(1 + 2k)}$. We have poles when $2z = i\pi(1 + 2k)\Rightarrow z = \frac{i\pi}{2}(1+2k)$. Then $$ 2\pi i\lim_{z\to\pi i/2}(z-\pi i/2)\tanh(z) = 2\pi i $$ For the final one, I would write $\cos(2z) = \frac{e^{2iz} + e^{-2iz}}{2}$. Then let $w = e^{iz}$ so $dw = ie^{iz}dz = iwdz$. $$ \int_C\frac{dz}{\cos(2z)} = \int_C\frac{2dw}{iw(w^2 + w^{-2})} = \int_C\frac{2dw}{i(w^3 + w^{-1})} = \int_C\frac{2wdw}{i(w^4 + 1)} $$ Let $w = re^{i\theta}$. Then \begin{align} r^4e^{4i\theta} &= -1\\ &= e^{\pi i(1 + 2k)} \end{align} so $r =1$ and $\theta = \frac{\pi}{4}(1+2k)$ where are where the poles occur for $k = 0,1,2,3$.