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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.

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  • $\begingroup$ 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 $a$. $\endgroup$
    – Andreas Blass
    Jun 16, 2013 at 23:26
  • $\begingroup$ 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 $\theta = arg(z)$. $\endgroup$
    – yarnamc
    Jun 16, 2013 at 23:37
  • $\begingroup$ 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. $\endgroup$
    – yarnamc
    Jun 16, 2013 at 23:38

1 Answer 1

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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.

This Arg is well defined exactly when the image is simply connected.

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  • $\begingroup$ Thank you for your comment! $\endgroup$
    – yarnamc
    Jun 24, 2013 at 20:36

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