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Calculate the following integral $$ \int_\gamma \frac{1}{z-i}dz $$ with $\gamma$ circle of radius 2 centered at the origin. Any suggestions please? Thank you very much.

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    $\begingroup$ cauchy integral formula, or more general the residue theorem $\endgroup$ – Dominic Michaelis Dec 28 '13 at 12:18
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The question is pretty basic so I assume you're beginning with this stuff and you probably want to go with basic line integrals, parametrization and stuff.

Parametrize thus...but with a twist

$$z-i=2e^{it}\;,\;\;0\le t\le 2\pi\implies dz=2ie^{it}\,dt\implies$$

$$\oint\limits_C\frac{dz}{z-i}=\int\limits_0^{2\pi}\frac{2ie^{it}dt}{2e^{it}}=\int\limits_0^{2\pi}i\,dt=2\pi i$$

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    $\begingroup$ Fine, but your circle is not centered at the origin. If the OP is beginning with this stuff. he's not going to know the principle of deformation of paths, is he? $\endgroup$ – bof Dec 28 '13 at 12:41
  • $\begingroup$ Well, that's the twisty thing...and the OP must have studied already, I presume, advanced real calculus if he's dealing now with complex analysis, mustn't he? Parametrizations like this one must be piece of cake in this case. "Beginner" here means he's probably not heard about residues, Cauchy's theorem and etc. $\endgroup$ – DonAntonio Dec 28 '13 at 12:42
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    $\begingroup$ Yes, but your $z=i+2e^{it}$ is a circle of radius $2$ centered at $i$, and the OP's contour $\gamma$ is a circle of radius $2$ centered at the origin. And if the OP has had advanced real calculus but hasn't had Cauchy's theorem yet, how would he know that those two paths of integration give the same result? $\endgroup$ – bof Dec 28 '13 at 12:48
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    $\begingroup$ @Bof, that a real analysis based result, not complex... $\endgroup$ – DonAntonio Dec 28 '13 at 12:49
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Hint: Cauchy's Integral Formula.

Alternatively note that $z\mapsto \dfrac 1{z-i}$ is holomorphic in the simply connected sets $\color{grey}{A:=}\mathbb C\setminus \{z\in \mathbb C\colon \Re(z-i)\leq 0 \land \Im(z-i)=0\}$ and $\color{grey}{B:=}\mathbb C\setminus \{z\in \mathbb C\colon \Re(z-i)\ge 0 \land \Im(z-i)=0\}$, therefore it has an anderivative on each of this sets.

Consider the following functions: $$\log_A\colon A\to \mathbb C, z\mapsto \log\left(\left|z-i\right|\right)+i\arg_A\left(z-i \right)$$ $$\log_B\colon B\to \mathbb C, z\mapsto \log\left(\left|z-i\right|\right)+i\arg_B\left(z-i \right)$$ where $\arg_A, \arg_B$ are functions such that $\text{im}(\arg_A)\subseteq]-\pi, \pi[$ and $\text{im}(\arg_B)\subseteq]0,2 \pi[$.

Since $\gamma=\gamma _A\lor \gamma _B$, where $$\gamma _A\colon \left[-\dfrac \pi 2, \dfrac \pi 2\right]\to \mathbb C, \theta \mapsto 2e^{i\theta},$$ $$\gamma _B\colon \left[\dfrac \pi 2, \dfrac {3\pi}2\right]\to \mathbb C, \theta \mapsto 2e^{i\theta},$$

since $\log _A$ and $\log _B$ are anderivatives of $z\mapsto \dfrac 1{z-i}$ in $A$ and $B$, respectively, it follows that

$$\begin{align} \int _{\gamma _A}\dfrac 1{z-i}\mathrm dz&=\log _A(z)\bigg|^{z =\gamma _A\left( \pi/ 2\right)}_{z=\gamma _A\left( -\pi/ 2\right)}\\ &=\log\left(|2i-i|\right)+i\dfrac \pi 2-\log (|-2i-i|)+i\dfrac \pi 2\\ &=-\log(3)+i\pi\end{align}$$ and $$\begin{align} \int _{\gamma _B}\dfrac 1{z-i}\mathrm dz&=\log _B(z)\bigg|^{z =\gamma _B\left( 3\pi/ 2\right)}_{z=\gamma _B\left( \pi/ 2\right)}\\ &=\log\left(|-2i-i|\right)+i\dfrac {3\pi} 2-\log (|2i-i|)-i\dfrac \pi 2\\ &=\log(3)+i\pi\end{align}.$$

Hence $$\int _{\gamma }\dfrac 1{z-i}\mathrm dz=\int _{\gamma _A}\dfrac 1{z-i}\mathrm dz+\int _{\gamma _B}\dfrac 1{z-i}\mathrm dz=2\pi i.$$

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By the Residue theorem we have

$$ \int_\gamma \frac{1}{z-i}dz=2i\pi Res(f,i)=2i\pi $$

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