Argument principle and change of argument

Question

Let's say $f\left( z \right) = {z^8} - {z^3} + 10$ and we wish to know the change of argument of $f(z)$ over $\gamma $, where $$\gamma = \left\{ {z = x{\rm{ }}|{\rm{ }}0 \le x \le R} \right\}$$.

One the one hand, it is clear that if $x \in \gamma $ then $f(x)$ is real and $f(x)>0$, thus the change in the argument is zero.

One the other hand, the following integral (which should compute the change in the argument) is not zero:

$${1 \over {2\pi i}}\int\limits_\gamma {{{f'(z)} \over {f(z)}}dz} = {1 \over {2\pi i}}\int\limits_0^R {{{f'(x)} \over {f(x)}}dx} = {1 \over {2\pi i}}\ln \left( {{{f\left( R \right)} \over {f\left( 0 \right)}}} \right) = {1 \over {2\pi i}}\ln \left( {{{{R^8} - {R^3} + 10} \over {10}}} \right) \ne 0$$

Please tell me where I was wrong. It is crystal clear to me that I am wrong on the second part, but I am not sure where exactly.

This integral only computes the change in argument if $\gamma$ is a closed contour. — Kenny Wong, Jun 05 '18 at 08:38
@KennyWong What you wrote is a complete answer. Why don't you post it as such? — José Carlos Santos, Jun 05 '18 at 08:50

score 2 · Accepted Answer · answered Jun 05 '18 at 09:04

We know that $Ln(z)=\ln|z|+i\arg(z)$ so that $$ \arg(f(z))=Im(Ln(f(z)). $$ The infinitesimal angle increment is thus $$ \frac{d}{dt}\arg(f(z))=Im\left(\frac{f'(z)}{f(z)}\frac{dz}{dt}\right) $$ or $$ angle(\gamma)=Im\left(\int_\gamma\frac{f'(z)}{f(z)}\,dz\right) $$ As your situation gives a fully real integral, there is no imaginary part, the angle is zero.

The formula you cite is for the winding number of a closed curve.

Argument principle and change of argument

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