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Let $g(z)$ denote an analytic branch of $log z$ in a domain $D \subset\mathbb{C}$. Show that $g'(z) = 1/z$. Show also that if $h(z)$ is another analytic branch of $log z$ in $D$, then the function $$\frac{g(z)-h(z)}{\pi i}$$ is constant in $D$ and equal to an even integer.

Here is how I approached the problem:

$$z = exp(log\ z)$$

By differentiating both sides of the equation we get:

$$1 = exp(log\ z) * (log\ z)' = z*(log\ z)$$

By dividing both sides by $z$ we get:

$$(log\ z)' = \frac{1}{z}$$

For the second part of the problem:

$$\frac{g(z) - h(z)}{\pi i} = \frac{(log|z| + iArg\ z + 2k_1i\pi) - (log|z| + iArg\ z + 2k_2i\pi)}{\pi i} = \frac{2k_1i\pi - 2k_2i\pi}{\pi i}$$ $$= 2(k_1 - k_2)$$

I have been struggling with wrapping my head around branches and whatnot so I am a little unsure if I have done what was asked of me here. Does it look correct?

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Note that\begin{align}\exp\bigl(g(z)-h(z)\bigr)&=\frac{\exp\bigl(g(z)\bigr)}{\exp\bigl(h(z)\bigr)}\\&=\frac zz\\&=1.\end{align}So, for each $z\in D$, $g(z)-h(z)=2\pi in$, for som integer $n$. Since $D$ is connected, the range of $g-h$ must be connected too, and therefore there is some $n\in\mathbb Z$ such that, for each $z\in D$, $g(z)-h(z)=2\pi i n$.

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  • $\begingroup$ Sorry, I don't really understand what you mean. How did you get to $exp(g(z) - h(z))$? $\endgroup$ – Max Apr 1 '20 at 9:30
  • $\begingroup$ I wanted to prove that $g(z)-h(z)$ is constant and my first step was to prove that $\exp\bigl(g(z)-h(z)\bigr)$ is constant. $\endgroup$ – José Carlos Santos Apr 1 '20 at 9:33
  • $\begingroup$ Oh, I see. So the way I tried proving it is wrong? $\endgroup$ – Max Apr 1 '20 at 9:39
  • $\begingroup$ I didn't understand it. There is no $\log$ function. So, what is $\log'$? And what is $\operatorname{Arg}$? $\endgroup$ – José Carlos Santos Apr 1 '20 at 9:46
  • $\begingroup$ $g(z), h(z)$ are two analytical branches of $logz$. $log'(z)$ is how I noted it as a derivative (i.e. $g'(z)$). Arg is the principal argument $\endgroup$ – Max Apr 1 '20 at 9:51

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