Calculate with Cauchy integral formula

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Can you please check if i did any mistakes.

a) $$\int_{\gamma_{B_2(0)}} \frac{z^7+41}{1-z}$$ $\frac{z^7+41}{1-z}$ = $ \frac{-z^7-41}{z-1}$ = $\frac{f(z)}{z-1}$ with the Cauchy integral formula i get $2 \pi i \cdot (-42) = -84\pi i$

b) $$\int_{\gamma_{B_r(\frac{i}{2})}} \frac{1}{z^2+1}$$

(1) Let $r = \frac{1}{4}$ then $\frac{1}{z^2+1}$ is holomorphic and there exists a holomorphic antiderivative on $\mathbb C\setminus\{i,-i\}$ so therefore the integral is $0$.

(2)Let $ r = 1$ then we have $ \frac{1}{z^2+1} = \frac{1}{(z+i)(z-i)}=\frac{\frac{1}{z+i}}{z-i} = \frac{f(z)}{z-i}$ . With the CIF i get $2\pi i \cdot f(i) = \pi i$

(3)Let $r=2$ then i get $\pi i$ aswell which is probably wrong

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1 Answer

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For $r=2$, both poles are enclosed. We can write

$$\frac{1}{z^2+1}=\frac{1/2i}{z-i}-\frac{1/2i}{z+i}$$

and apply Cauchy's Integral Formula as

$$\begin{align} \oint_{B_{2}(i/2)}\frac{1}{z^2+1}\,dz&=\oint_{B_{2}(i/2)}\left(\frac{1/2i}{z-i}-\frac{1/2i}{z+i}\right)\,dz\\\\ &=\oint_{B_{2}(i/2)}\frac{1/2i}{z-i}\,dz-\oint_{B_{2}(i/2)}\frac{1/2i}{z+i}\,dz\\\\ &=\frac1{2i}-\frac{1}{2i}\\\\ &=0 \end{align}$$

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