Evaluating the complex integral $\int_{|z|=2}^{}\frac{1}{z^2+1}dz$

Question

I tried finding the integral of $\int_{|z|=2}^{}\frac{1}{z^2+1}dz$ but not sure whether it is correct.

$\gamma(t)=2e^{it},t\in[0,2\pi]$

$$\int_{|z|=2}^{}\frac{1}{z^2+1}dz=\int_{0}^{2\pi}\frac{2ie^{it}}{4e^{2it}+1}dt$$

I cannot seem to take it forward from here. Is it possible to use Cauchy's integral formula to get an answer? Thanks

Answer 1

HINT: $\frac{1}{1+z^2}=\frac{1}{(z-i)(z+i)}$. Use partial fractions to expand this and then apply Cauchy Integral Formula. You should get the $0$ (As proved in the other answer).

Answer 2

You can use the residue theorem to solve the problem: $$I:=\int_{\left| z \right| = 2} {\frac{1}{{{z^2} + 1}}dz} = 2\pi \left[ {\operatorname{Residue} \left( {\frac{1}{{{z^2} + 1}};z = i} \right) + \operatorname{Residue} \left( {\frac{1}{{{z^2} + 1}};z = - i} \right)} \right].$$ We have $$\operatorname{Residue} \left( {\frac{1}{{{z^2} + 1}};z = i} \right) = \mathop {\lim }\limits_{z \to i} \left( {z - i} \right)\frac{1}{{{z^2} + 1}} = \mathop {\lim }\limits_{z \to i} \frac{1}{{z + i}} = \frac{1}{{2i}},$$ $$\operatorname{Residue} \left( {\frac{1}{{{z^2} + 1}};z = - i} \right) = \mathop {\lim }\limits_{z \to - i} \left( {z + i} \right)\frac{1}{{{z^2} + 1}} = \mathop {\lim }\limits_{z \to - i} \frac{1}{{z - i}} = - \frac{1}{{2i}}.$$ So $I=0$.