MHB Residue Calculation for Contour $|z-i|=1$

[Dustinsfl]

For the contour $|z| = 2$

$$
\int_C\frac{z + 1}{z^2 + 1}dz = \int_C\frac{z + 1}{(z + i)(z - i)}dz = 2\pi i\sum\text{Res}_{z = z_j}\frac{z + 1}{z^2 + 1}
$$
Let $g(z) = z^2 + 1$. The zeros of $g$ occur when $z = \pm i$. $g'(\pm i)\neq 0$ so the poles are simple for $1/g$. Let $f(z) = \dfrac{z + 1}{z^2 + 1}$. Then
$$
\text{Res}_{z = i}f(z) = \frac{i + 1}{2i}\quad\text{and}\quad\text{Res}_{z = -i}f(z) = \frac{i - 1}{2i}.
$$
So
$$
\int_C\frac{z + 1}{z^2 + 1}dz = 2\pi i\sum\text{Res}_{z = z_j}\frac{z + 1}{z^2 + 1} = 2\pi i.
$$

Correct?

For the contour $|z-i|=1$

For this contour, the only residue is when $z = i$.
So the
$$
\text{Res}_{z = i}\frac{z + 1}{z^2 + 1} = \frac{i + 1}{2i} \ \text{Res}_{z = i}\frac{1}{z - i}
$$
Then
$$
\int_Cf(z)= \pi(1 + i)
$$

Correct?

 

[Opalg]

Correct?

Yes! (Star)