$Log(z^2)=2Log(z)$ not always satisfied

Question

We have that the main value of the logarithm in complex numbers is given by: $$Ln(z)=log_e(r)+i\theta$$

with $-\pi<\theta\leq\pi$.

However, it can be seen that the relationship $Log(z^2)=2Log(z)$ is not always satisfied when considering the main branch.

For example, I realized that for $z=1+i$ the relationship is valid, but for $z_1=-1+i$ it is not. After a long time I realized that this happens because $Arg(z_1^2)$ is $\dfrac{-\pi}{2}$ while $Arg(z_1)$ is $\dfrac{3\pi}{ 4}$, that is, $Arg(z_1^2)$ was influenced by the main value of the logarithm due to $-\pi<\theta\leq\pi$.

Now, I'm not sure how to find every value of $z$ that satisfies this equality.

I believe that $z$ is in the first or fourth quadrants

Answer 1

Let $\DeclareMathOperator{\Arg}{Arg}\DeclareMathOperator{\Log}{Log} \Arg\colon \mathbb C\setminus\{0\}\to (-\pi,\pi]$ be the principle value of the multi-valued argument function and let $\Log(z) = \ln|z| + i \Arg(z)$ be the principle branch of logarithm. Then, we have

\begin{align}\Log(z^2) = 2\Log(z) &\iff \ln|z^2| + i \Arg(z^2) = 2\ln|z| + 2i\Arg(z)\\ &\iff \Arg(z^2) = 2\Arg(z).\end{align}

Now, if $\Arg(z) = \theta$, then $z = re^{i\theta}$ and $z^2 = r^2e^{2i\theta} = r^2e^{i(2\theta + 2n\pi)}$, $n\in\mathbb Z$, so we conclude that $\Arg(z^2) = 2\theta + 2k\pi$, where $k$ is the unique integer such that $2\theta + 2k\pi\in (-\pi,\pi]$. From here we can conclude that $$\Arg(z^2) = 2\Arg(z) \iff 2\theta \in (-\pi,\pi].$$

Answer 2

As you note, the only problem is with the angle. And the discontinuity in the number we assign to the angle happens along the negative $x$-axis.

I suggest you start with a $z$ that makes a zero angle with the positive $x$-axis, and track what happens to the angle made by both $z$ and $z^2$ as you continuously increase the angle it makes. You should be able to see when things break down.

Then you can also do the same thing as the angle gets increasingly negative to figure out the whole range.