complex integrationcomplex-analysis
I am asked to calculate the integral $$\int_C \frac{1}{z-a}dz$$ where $C$ is the circle centered at the origin with radius $r$ and $|a|\neq r$
I parametrized the circle and got $$\int_0^{2\pi}\frac{ire^{it}}{re^{it}-a}dt=\text{log}(re^{2\pi i}-a)-\text{log}(r-a)=\text{log}(r-a)-\text{log}(r-a)=0$$
Because of how the complex logarithm is defined, I am pretty sure that the first equaility is wrong, it it is, what is the correct solution?
Hint:
$$\int_{a}^{b}\frac{f'(\theta)}{f(\theta)} \, d\theta = \ln(f(\theta))\Big|_{\theta = a}^{\theta=b} .$$
Consider the contour $\Gamma$ below.
This consists of two pieces: the circle $z=R$, $R \gt 1$, minus the dogbone inside the circle, consisting of two small circular arcs at $z=\pm 1$ of radius $\epsilon$, and joined by line segments above and below the real axis.
Now consider $f(z) = \sqrt{(z-1)(z+1)}$. On the upper line segment, $\arg{(z+1)} = 0$ and $\arg{(z-1)} = \pi$. On the lower line segment, $\arg{(z+1)} = 0$, but $\arg{(z-1)} = -\pi$. Thus, the contour integral about $\Gamma$ is, in the limit as $\epsilon \to 0$,
$$\oint_{\Gamma} dz \, \sqrt{z^2-1} = \int_{|z|=R} dz \, \sqrt{z^2-1} + i \int_1^{-1} dx \, \sqrt{1-x^2} - i \int_{-1}^1 dx \, \sqrt{1-x^2} $$
By Cauchy's theorem, the integral about $\Gamma$ is zero. Thus,
$$ \int_{|z|=R} dz \, \sqrt{z^2-1} = i 2 \int_{-1}^1 dx \, \sqrt{1-x^2} = i \pi$$
Best Answer
If $r<|a|$ the integral is indeed $0$, but if $r>|a|$, then using Cauchy integral formula (or more generally, Residue theorem), you get $2i\pi$.
You can also do what you did, but :
Notice that if $a\in \mathbb R$, $a>0$, then :
If $r>a$, then $$\ln(re^{2i\pi}-a)=\ln(e^{2i\pi}(r-a))=2i\pi+\ln(r-a),$$
If $r<a$, then $$\ln(re^{2i\theta }-a)-\ln(r-a)=\ln(e^{i\pi}(a-r))-\ln(e^{i\pi}(a-r))=0.$$
You can generalize this to the case $a\in \mathbb C$, and distinguish the case $r>|a|$ or $r<|a|$.