Evaluate the line integral $\oint_C \frac{z+1}{z^3-2z^2} dz$ along the given curves

Question

Curve 1: $|z|=1$

Curve 2: $|z-2-i|=2$

So I figured out how to solve the first curve using Cauchy's Integral Formula. My answer matched the given one,$-\frac{3}{2}\pi i$, so we are good. However, the answer to the 2nd problem is $\frac{3}{2}\pi i$. This leaves me with the question: what difference does the curve make in using the formula that would remove the negative sign?

For reference, this is the formula I used:

$$\oint_C \frac{f(z)dz}{(z-z_0)^n} = \frac{2\pi i}{(n-1)!}f^{n-1}(z_0)$$

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Thinking about it some more, I feel as though it has everything to do with the $z_0$ in the formula. Given curve 2, $z=2$ and $z=0$ both lie outside it. If this is the case, then what should I use in the formula for $z_0$?

Answer 1

For $|z|<1$:

$z=0$ is singular point so we will define $f$ wit

$$\oint_C \frac{z+1}{z^3-2z^2} dz=\oint_C \frac{z+1}{z^2(z-2)} dz$$

Let $f(z):=\frac{z+1}{(z-2)} $

and $f'(z)=-\frac{3}{(z-2)^2}$

so the answer for the first $2\pi i \cdot f'(0)=-\frac{3\pi i}{2}$

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For $|z-2-i|=2:$

The singular point is $z=2$

So let $g(z):=\frac{z+1}{z^2}$ so the answer $2\pi i \cdot g(2)=\frac{3\pi i}{2}$

Answer 2

Write your integral as $$ \oint_C \frac{g(z)}{z - 2} dz $$ for a suitable function $g$ and take it from there. Drawing the curve and the two poles of the function in the complex plane will also help clarify things.