# How is $(\arg F)(z)$ of complex function calculated?

My book on complex analysis introduces $\arg(z)$ and $\arg(z-a)$ after the complex logarithm is introduced. It shows that the two are just the oriented angles between $z$ and the point $z_0$ of the complex logarithm that a contour makes that goes from $z_0$ to $z$.

Now it introduces also the complex logarithm of $F(z)$, which is a holomorphic function in $\Omega$ and which has no zero points in $\Omega$ by:

$$(\log F)(z) = w_0 + \int_{z_0}^z{\frac{f'(\zeta)}{f(\zeta)}d\zeta}$$

In this way they can also define $\arg F(z)$ by saying that this is the imaginary part of $\log F(z)$. They start using a similar property to that of $\arg(z)$, without specifying it. So I wonder if anyone can tell how this $\arg F(z)$ can be calculated, probably it is in the same way as $\arg(z)$ or something similar, but I don't see why that should be true.

• I don't understand "oriented angles between $z$ and the point $z_0$ of the complex logarithm that a contour makes that goes from $z$ to $z_0$." For me, oriented angles are between vectors or line segments, not points. I don't know what a "point of" the complex logarithm is, and I don't know how a contour "makes" a logarithm or a point or an angle (depending on how you try to parse the sentence). Furthermore, the quoted phrase is supposed to describe two things, one of which is $arg(z-a)$, yet the quoted phrase doesn't mention $a$. Jun 16, 2013 at 23:26
• Yes, I admit that it isn't formulated really mathematically, but I assumed that the reader knows where I was talking about. With $z_0$ I mean the point in the complex plane which is used in the definition of the complex logarithm. Then you take a contour which stays in $\Omega$ and that goes from $z_0$ to $z$. If we parametrize this contour by $\gamma(t) = Rexp(i\theta)$ (for the case of $arg(z)$), then you see that by carrying out the integration $arg(z)$ equals $\theta_0 - \theta + k2\pi$. With $\theta_0 = arg(z_0)$ and $\theta = arg(z)$. Jun 16, 2013 at 23:37
• And because the amount of $k$ equals the amount of times you went around the zero point, this can be interpreted as the oriented angle the two points make. Jun 16, 2013 at 23:38

Because $f$ has no zeros, it maps $\Omega$ to a portion of the complex plane that misses the origin. Then you can calculate it the same way as in the simpler case: fix a $z_0$ in the image, then take a path from $z_0$ to $z$. The argument is the oriented change in angle along that path, plus the argument you choose for the base point.