# Variation of argument of a complex function

Variation of Argument :

Definition( Collect from my book ) : Let $f$ be analytic inside and on a sinple closed contour $C$ except possibly for poles inside $C$ and $f(z)\not=0$ on $C$. As $z$ describes $C$ once in the positive direction in the $z$-plane , the image point $w=f(z)$ describes a closed curve $\Gamma=f(C)$ in the $w$-plane in a particular direction which determines the orientation of the image curve $\Gamma$. Since $f(z)\not=0$ on $C$ , $\Gamma$ never passes through the origin in the $w$-plane. Let $w_0$ be the arbitrary fixed point on $\Gamma$ and let $\phi_0$ be the argument of $w_0$. Then let , $\arg z$ run continuously from $\phi_0$ , as the point begins at $w_0$ and traverses $\Gamma$ once in the direction of orientation assigned to it by $w=f(z)$. If $w$ returns to the staring point $w_0$ , then $\arg w$ assume a particular value of $\arg w_0$ which we denote by $\phi_1$. We define , $$\Delta_C\arg f(z)=\phi_1-\phi_0.$$

I am unable to understand the meaning of bold sentences. Can anyone explain these sentences with a proper example or by a rough figure such that I can realize what actually the variation of argument $\Delta_C f(z)$ over $C$?

If you can provide an example of a function and a simple closed contour $C$ explaining the value of $\Delta_C f(z)$ over $C$ then it is better to me.

• I agree that sentence is confusing, maybe wrong. I think the part "Then let, $\text{arg} \; z$ should read "Then let $\text{arg} \; f(z) \cdots$. The point here is that as $z$ goes around the unit circle, $f(z)$ traces out some curve that will go around the origin some number of times $= \Delta_C \text{arg} f(z) \; / \; 2 \pi$. – user226970 Mar 5 '16 at 6:06

Let $w_0$ be an arbitrary fixed point on $\Gamma$ and let $\phi_0$ be an argument of $w_0$. Let $\gamma:[0,1]\to C$ be a parametrization of $C$ starting and ending at some $z_0$ such that $f(z_0)=w_0$. Then let $\arg z$ run continuously from $\phi_0$ , as the point begins at $z_0=\gamma(0)$ and traverses $C$ once and ends at $z_0=\gamma(1)$. Now $f\circ\gamma$ starts and ends at $f(z_0)=w_0$. Then the the continuous choice in change of argument induced by $\gamma$ is denoted $\phi_1$.
Let us take the best example of this phenomenon. Let $C$ be the unit circle and $f(z)=z^n$ for some positive integer $n$. Then the geometry here is that $f(C)=C$ but the function "wraps" $C$ around itself $n$ times. More precisely, every point $z\in C$ of the range has $n$ pre-images under the map $f$ (Basically De Moivre's theorem). Let us pick $w_0=1$.
With this setting, let us talk about the $\arg$ function a little bit. For any point $z\in \mathbb C$, there are infinitely many choices for the argument, namely if $\theta$ is a value such that $z=re^{i\theta}$, then we also have that $z=re^{i(\theta+2\pi n)}$ as well so $\arg(z)=\{\theta+2\pi n |n\in \mathbb Z\}$. Next, arg cannot be globally defined in a continuous manner. The essential problem being that if you go around the unit circle once and try to continuously vary the argument as you go along, you will end up with a $2\pi$ difference in argument from the beginning and end. Let me demonstrate this concretely with our example.
Let us start at $w_0,z_0=1$. If $e^{2\pi i t}$ parametrizes $C$ clockwise as $0\leq t\leq 1$, with a continuous choice of argument along $\gamma$ being $2\pi t$, we see that the change in argument from $t=0$ to $t=1$ is $2\pi$. Now if I consider $f\circ \gamma(t)=e^{2\pi nt}$, then this a parametrization of $\Gamma$ and the induced continuous choice of argument is $2\pi n t$ which shows that $\phi_1=2\pi n$.
So, to conclude, $\Delta_C\arg f(z)=\phi_1-\phi_0=2\pi n$.