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Why is $\operatorname{Log} z^2$ not equal to $2\operatorname{Log} z$ where $z$ is a complex number.

$\operatorname{Log} z$ here refers to just the principal Log.

Detailed explanation would be greatly appreciated, thanks!

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3 Answers 3

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By definition, the principal value $\text{Log}\; z$ is the complex number $w$ with $e^w = z$ and $-\pi < \text{Im}(w) \le \pi$. If $w = \text{Log}\; z$, $2w$ is a logarithm of $z^2$ (i.e. $e^{2w} = (e^w)^2 = z^2$), but you have no right to expect $-\pi < \text{Im}(2w) \le \pi$. Indeed that will only be the case if $-\pi/2 < \text{Im}(w) \le \pi/2$.

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If the argument of $z$ is between $\pi/2$ and $\pi$, then you can see that $\log(z^2)$ can't be $2\log z$, since $2\log z$ will have too big an imaginary part to be the principal value of a logarithm.

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For any $z\in\mathbb{C}$, the principal value of the logarithm is $$\log z = \log |z| + i \arg z,$$ where $\arg z \in (-\pi, \pi]$. Note that $\arg (z_1 z_2)=\arg z_1 + \arg z_2 +2\pi n$, where $n$ must be chosen to keep the result in $(-\pi, \pi]$: more precisely, $$ \arg(z_1z_2)=\arg z_1 + \arg z_2 - 2\pi\left[\frac{\arg z_1 + \arg z_2}{2\pi}\right], $$ where $[x]$ is the nearest integer to $x$.

So $$ \log z^2=\log |z^2| + i \arg z^2=2\log|z|+i\arg z^2\\=2\log z+i\left(\arg z^2 - 2\arg z\right)\\=2\log z -2\pi i \left[\frac{\arg z}{\pi}\right]. $$ In particular, $\log z^2$ can be either $2\log z$ or $2\log z\pm 2\pi i$, depending on which quadrant of the complex plane contains $z$.

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