I try to solve for $z\in\mathbb{C}\setminus\bar{\mathbb{E}}$ the following integrals (without the residue theorem):
a) $$\int\limits_{\partial\mathbb{E}}\frac{1}{\theta-z} d\theta$$ b) $$f(z)=\frac{1}{2\pi i}\int\limits_{\partial\mathbb{E}}\frac{1}{\theta(\theta-z)} d\theta$$
I have great troubles understanding the task mainly because $z$ is outside of the circle.
What I thought so far:
a) I'm not sure wether I should integrate from on $t\in[0,2\pi]$ around $\gamma(t):e^{it}$ counterclockwise since $z$ is not inside the circle.
But here is what I thought so far: $$\int\limits_{\partial\mathbb{E}}\frac{1}{\theta-z} d\theta=\int\limits_{0}^{2\pi} \frac{te^{it}}{e^{it}-z} dt=\int\limits_{0}^{2\pi} \frac{te^{it}-z + z}{e^{it}-z} dta=\int\limits_{0}^{2\pi}t dt + z\int\limits_{0}^{2\pi}\frac{1}{e^{it}-z} dt$$ But I don't know how to solve the last integral
b) $$\frac{1}{2\pi i}\int\limits_{\partial\mathbb{E}}\frac{1}{\theta(\theta-z)} d\theta=\frac{1}{2\pi i}\int\limits_{0}^{2\pi}\frac{te^{it}}{e^{it}(e^{it}-z)} dt=\frac{1}{2\pi i}\int\limits_{0}^{2\pi}\frac{t}{e^{it}-z} dt$$
And then I'm stucked again.
a) Take the function $\;f(z)=\frac1{\theta-z}\;$ , so for $\;|z|>1\;$ we get by CIT
$$\int_{|z|=1}\frac1{\theta-z}d\theta=0$$
since $\;f(z)\;$ analytic on $\;\Bbb E\cup\partial\Bbb E\;$ .
b) Directly, since $\;\frac1{\theta-z}\;$ is analytic on the unit circle included its perimeter for $\;|z|>1\;$ , we get from CIF
$$\frac1{2\pi i}\int_{|z|=1}\frac{\frac1{\theta-z}}\theta\, d\theta=\left(\frac1{\theta-z}\right)_{\theta=0}=-\frac1z$$