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How to compute $\int_C f'(z)/f(z) dz$, when $C$ is positively oriented disk $\{ z: |z|=4\}$, when $f(z)=\frac{(z-8)^2z^3}{(z-5)^4(z+2)^2(z-1)^5}$? Hint. $ \frac{f'(z)}{f(z)} = \frac{d}{dz} log f(z) = \frac{d}{dz}(Log|f(z)|+i arg f(z))$ I'm interested especially about the fact that $\int_C \frac{f'(z)}{f(z)} dz = (Log |f(z_2)| + i arg f(z_2)) - (Log |f(z_1)| + i arg f(z_1)) = \Delta_C Log |f| + i \Delta_C arg f = i \Delta_C arg f =(*) 2 \pi i(2 \cdot 0 + 3 \cdot 1 - 4 \cdot 0 - 2 \cdot 1 - 5 \cdot 1) = -8\pi$

But now question is how do you get (*) because if $i arg f(z)=2arg(z-8)+3arg z - 4 arg(z-5) - 2 arg(z+2) - 5 arg(z-1)$ then $i\Delta_C arg f=2 \pi i(2 \cdot 0 + 3 \cdot 1 - 4 \cdot 0 - 2 \cdot 1 - 5 \cdot 1)$?

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Do you know about the argument principle? –  mrf Feb 7 '13 at 12:48
    
I know that if $f$ is analytic and $\neq 0$ in every point in path C( simple, closed and positively oriented) and also meromorphic inside C, then $\frac{1}{2 \pi i} \int_C \frac{f'(z)}{f(z)}dz = N_0(f)-N_p(f)$, where $N_0(f)$ is number of zeros of f and $N_p(f)$ number of poles of f inside the path C. I understand this, but I do not understand how for example $2 arg(z-8) \rightarrow 2 \cdot 0$ because you don't know $\theta$ of $\arg(f(z))$ –  laovultai Feb 7 '13 at 13:11

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up vote 1 down vote accepted

Argument Principle says that, for a function $f$ meromorphic inside and analytic on a positively-oriented contour $C$:

$$\oint_C dz \: \frac{f'(z)}{f(z)} = i 2 \pi (N-P)$$

where $N$ is the number of zeros of $f$ inside $C$, and $P$ is the number of poles of $f$ inside $C$. (Note that I am assuming a winding number of 1 here.) $f$ should have no poles on $C$. Also note that a zero or pole of order $m$ is counted $m$ times.

In this case, $N=3$ and $P = 2 + 5 = 7$. ($z=5$ and $z=8$ lie outside of $|z|=4$.) Thus the integral is equal to $-i 8 \pi$.

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