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Let $z_1, \dots ,z_n$ be complex numbers such that $\Re(z_k)>0$ and $\Re(z_1 \dots z_n)>0$ for $1 \leq k \leq n$. Show that

$$\log(z_1 \dots z_n) = \log(z_1)+\dots +\log(z_n)$$

where $\log$ is the principal branch of the logarithm. If the restrictions on the $z_k$ are removed, does the formula remain valid.

My Solution

Let $z_1, \dots ,z_n$ be complex numbers such that $\Re(z_k)>0$ and $\Re(z_1 \dots z_n)>0$ for $1 \leq k \leq n$. Consider

\begin{align*} \log(z_1 \dots z_n)&=\ln(|z_1||z_2|\dots|z_n|)+i\arg(z_1 \dots z_n)\\ &= \ln(|z_1|) + \dots + \ln(|z_n|) + i\arg(z_1)+\dots i\arg(z_n)\\ &=\log(z_1) + \dots + \log(z_n) \end{align*}

Now, clearly we must always exclude any $z_k=0$.

My Question

The log function loses continuity if I include the negative real axis (I'm pretty sure at least.) but I am having trouble determining if the function is still valid if the restriction $\Re(z_k)$ is removed. Thanks for your help!

Edit

As pointed out in the comments my solution is fallacious. I have found an example that shows it fails if $\Re(z)<0$. So, we are assuming that $-\dfrac{\pi}{2}<\arg(z_k)<\dfrac{\pi}{2}$. We need to show that $\arg(z_1 \dots z_n)=\arg(z_1)+ \dots+\arg(z_n)$ NOT $\arg(z_1 \dots z_n)=\arg(z_1)+ \dots+\arg(z_n) \quad \text{ mod }2 \pi$. I am not sure how to go about this from here.

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  • $\begingroup$ I am definitely not familiar with complex logarithm. But my suggestion is: why don't you choose another branch? Since you have only finitely many $z_i$, and none of them equals to $0$, then you can easily pick a $\theta$ s.t. the line $e^{i\theta}$ becomes the branch cut. Or you may just want to claim that there exists a simply connected domain which contains all $z_i$, and does not contains $0$. Then you can still have it holds. You may also want to take $z_{i_i} \dots z_{i_k}$, which is the product of an arbitrary subset of $z_1, \dots, z_n$ into account s.t. they won't fall on your line. $\endgroup$ – k99731 Feb 27 '16 at 2:09
  • $\begingroup$ Did you consider to use the representation $re^{i\theta}$? You will get I believe more than one potential outcome. $\endgroup$ – Moti Feb 27 '16 at 2:55
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In the first part, you are assuming the very thing you are supposed to prove: How do you know that $\arg(z_1...z_n) = \arg(z_1) + ... + \arg(z_n)$?

As for the second part, consider $\left(\frac{-1 + i}{\sqrt 2}\right)^2 = (e^{i(3\pi/4)})^2$.

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  • $\begingroup$ Because the argument of the product is the sum of the arguments. Let $z_i=r_ie^{i\theta_i}$. $\arg{\Pi z_i}=\sum\arg{z_i}$ $\endgroup$ – Aaron Zolotor Feb 27 '16 at 4:04
  • $\begingroup$ No, that is just you restating your assumption that this holds. You need to prove it. That is the point of this exercise. As as my example shows. It does NOT hold in general, when you restrict $\arg$ to the principle branch. $\endgroup$ – Paul Sinclair Feb 27 '16 at 4:08
  • $\begingroup$ I don't understand. That is just a fact about complex numbers. They can be represented in polar form. We look at the product. The magnitude is the product of the magnitude and the argument is the sum of the arguments? $\endgroup$ – Aaron Zolotor Feb 27 '16 at 4:10
  • $\begingroup$ Try my example. what is $\left(\frac{-1 + i}{\sqrt 2}\right)^2$? What is the argument of this number? Then what is twice the argument of $\frac{-1 + i}{\sqrt 2}$? (I am assuming by "principle branch", you mean arguments between $-\pi$ and $\pi$. If you mean $0$ to $2\pi$ instead, then a different example.is needed.) $\endgroup$ – Paul Sinclair Feb 27 '16 at 4:14
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    $\begingroup$ But you are applying to the logarithm, which is restricted in its codomain. For it to be true, you have to show that the RHS arguments sum up to the particular argument on the LHS, not some equivalent one. $\endgroup$ – Paul Sinclair Feb 27 '16 at 22:54

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