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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.

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    $\begingroup$ Are you allowed to use the Cauchy residue theorem (en.wikipedia.org/wiki/Residue_theorem)? $\endgroup$ – TZakrevskiy Apr 11 '16 at 9:29
  • $\begingroup$ You say "$\;z\;$ is outside the circle" . Are you given $\;\partial\Bbb E\;$ is a circle? And if so, what circle is it? I'm also guessing you haven't yet learned Cauchy's Integral theorem, residues and stuff, right? $\endgroup$ – DonAntonio Apr 11 '16 at 9:30
  • $\begingroup$ @TZakrevskiy No, we are not allowed to use it. Cauchy's integral formula is fine. $\endgroup$ – Matriz Apr 11 '16 at 9:39
  • $\begingroup$ @Joanpemo $\mathbb{E}$ is the complex unit circle and so $\partial \mathbb{E}$ is the boundary of the circle. We had Cauchy's integral theorem but not the Residue theorem so far. $\endgroup$ – Matriz Apr 11 '16 at 9:40
  • $\begingroup$ @Matriz Thank you, I see.. $\endgroup$ – DonAntonio Apr 11 '16 at 9:44
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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

$$\frac1{2\pi i}\int_{|z|=1}\frac{\frac1{\theta-z}}\theta\, d\theta=\left(\frac1{\theta-z}\right)_{\theta=0}=-\frac1z$$

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  • $\begingroup$ Hi, thank you very much. Unfortunatetly I don't understand the second last equality. You are using the CIF to calculate the integral, but I don't understand why it is $(\frac{1}{\theta - z})$ and why $\theta=0$? $\endgroup$ – Matriz Apr 11 '16 at 18:33
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    $\begingroup$ The CIF says that if $\;f\;$ is anayltic on a domain enclosed by a nice path $\;C\;$ , then for any point $\;a\;$ interior to the domain, we have $$f(a)=\frac1{2\pi i}\oint_C\frac{f(z)}{z-a}dz$$ IN your case, $$C= \partial\Bbb E\;,\;\;f(\theta)=\frac1{\theta-z}\;\;\text{and}\;\;a=z$$ $\endgroup$ – DonAntonio Apr 11 '16 at 19:19
  • $\begingroup$ ok thank you very much. $\endgroup$ – Matriz Apr 11 '16 at 19:44

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