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For any $z1, z2$ in $\mathbb{C} \setminus {0}$, $\log(z_1 z_2)=\log(z_1)+\log(z_2)$, but in general $\text{Log}(z_1 z_2)\ne \text{Log}(z_1)+\text{Log}(z_2)$.

Is $\log(z^2)=2\log(z)$?

Mark Viola
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Cody Weaver
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2 Answers2

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The answer to the question in the OP is that in general $\displaystyle \log(z^2)\ne 2\log(z)$.

This might seem paradoxical given the relationship expressed as

$$\log(z_1z_2)=\log(z_1)+\log(z_2) \tag1$$

But $(1)$ is interpreted as a set equivalence. It means that any value of $\log(z_1z_2)$ can be expressed as the sum of some value of $\log(z_1)$ and some value of $\log(z_2)$. And conversely, it means that the sum of any value of $\log(z_1)$ and any value of $\log(z_2)$ can be expressed as some value of $\log(z_1z_2)$.

Note that $(1)$ is true since $\log(|z_1z_2|)=\log(|z_1|)+\log(|z_2|)$ and $\arg(z_1z_2)=\arg(z_1)+\arg(z_2)$.

EXAMPLE:

As an example, suppose that $z_1=z_2=-1$. Then, $z_1z_2=1$ and $\log(1)=i2n\pi$ for any integer $n$. For $n=0$, $\log(1) =0$. Then $(1)$ is certainly satisfied by $\log(z_1)=i\pi$ and $\log(z_2)=-i\pi$.

Note that although $z_1=z_2$ here, we needed to take two distinct values for $\log(z_1)$ and $\log(z_2)$. We were afforded that degree of freedom since we viewed $z_1$ and $z_2$ as independent.

In general,

$$\log(z^2)\ne 2\log(z) \tag 2$$

To see this, we note that for $z=re^{i\theta}$,

$$2\log(z)=2\log(r)+i2(\theta+2n\pi) \tag 3$$

while

$$\begin{align} \log(z^2)&=\log(r^2e^{i2\theta+i2n\pi})\\\\ &=2\log(r)+i2(\theta+n\pi)\tag 4 \end{align}$$

Comparing $(3)$ and $(4)$ we see that $\log(z^2)$ and $2\log(z)$ do not share the same set of values.

EXAMPLE:

As an example, suppose $z=i$. For the value $\log(i^2)=i3\pi$, there is no corresponding value of $2\log(z)$. Hence, $\log(i^2)\ne 2\log(i)$ in general.

Mark Viola
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  • Why then $\log(z^{\frac{1}{n}})=\frac{1}{n}\log(z)$. Following your logic $ \log(z^{\frac{1}{n}})= \frac{1}{n} \left( \log(z) +2\pi m n \right)$. But isn't $mn$ spoil the equality of the sets. Also $\log(z^{-1})=-\log(z)$ or $\log(z^{-1})=-\left( \log(z) -2\pi m \right)$. Again the negative $-2\pi m$ looks like a problem. – Alexander Cska Dec 07 '19 at 17:07
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    I assume the your $n\in\mathbb{N}$ with $n\ge 1$. Well, then we have

    $$\begin{align} \log(z^{1/n})&=\log(|z|^ne^{i\frac1n\arg(z)})\\ &=\log\left(|z|^ne^{i\frac1n(\text{Arg}(z)+2k\pi)}\right)\\ &=\frac1n \text{Log}(|z|)+i\left(\frac{\text{Arg}(z)+2k\pi}{n}+2p\pi\right)\\ &=\frac1n \text{Log}(|z|)+i\left(\frac{\text{Arg}(z)+2(pn+k)\pi}{n}\right) \end{align}$$

    where $k\in\mathbb{N}$, with $0\le k\le n-1$, and $p\in\mathbb{Z}$.

    … continued to the next comment

     – Mark Viola Dec 07 '19 at 17:45
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    @AlexanderCska … Continued from the previous comment.On the other hand,

    $$\frac1n \log(z)=\frac1n \text{Log}(|z|) +i \left(\frac{\text{Arg}(z)+2q\pi}{n}\right)$$

    where we assert that $\log(z^{1/n})=\frac1n \log(z)$ in terms of SET EQUIVALENCE. That is, corresponding to a value of $\log(z^{1/n})$ on the left-hand side, the appropriate value of $\log(z)$ is to be selected on the right-hand side, and conversely.

     – Mark Viola Dec 07 '19 at 17:45
  • Well in the first expression we have $\frac{2(pn+k)\pi}{n}$ and in the second $\frac{2q\pi}{n}$. This looks different, or since both $pn+k$ and $q$ are arbitrary integers, somehow the set equivalence holds? In terms of Cantor we can pair the sets, if I am not totally wrong. Last but not least thank you for the reply. – Alexander Cska Dec 07 '19 at 18:13
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    Yes indeed. SET EQUIVALENCE. For any value of $pn+k$ there is a value of $q=pn+k$. And for any value of $q$ there is a corresponding vale of $p$ and a $k$ with $0\le k\le n-1$ such that $pn+k=q$. And that is all there is to say. You have it now. Well done. – Mark Viola Dec 07 '19 at 18:18
  • Well this part is not so simple. $q$ can be any integer. $pn+k$ is a product of arbitrary integers.To show $x\in A \rightarrow x\in B$ and $x \in B \rightarrow x \in A$ is not easy for me. – Alexander Cska Dec 07 '19 at 18:44
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    @AlexanderCska Choose any $q$. Then you can find a $p$ and a $k$ with $0\le k\le n-1$ so that $pn+k=q$. Try it out. – Mark Viola Dec 07 '19 at 19:03
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If $p$ is a non-integer then $z^p$ is a complex multi-valued function and the principal value of $\ln z$ must lie in $- \pi < \Im \ln z< \pi$. From this it follows that \begin{align} \ln (z_1 z_2) &= \ln z_1 + \ln z_2 + 2 \pi i N_{+} \\ \ln \left( z_1 / z_2 \right) &= \ln z_1 - \ln z_2 + 2 \pi i N_{-}\\ \ln z^n &= n \ln z + 2 \pi i N_{n} \end{align} Where $N_{\pm} = 0, +1, -1$ and $$N_n = \frac{1}{2}+\left(\frac{n}{2 \pi}\right)\arg z$$