Does the complex modulus satisfy the power identity $|z^r|= |z|^r$?

Question

Can we "split the modulus" of complex numbers?

Let $z\in\mathbb{C}$. Then, does $$|z^{r}|=|z|^r$$ hold, where $r\in\mathbb{R}$. Is this true even for $r\in\mathbb C$ ?

Also, can we show this? I am able to show that it splits over power of natural numbers, but cant show it for $r\in \mathbb{R}$ or for $r\in\mathbb{Q}$.

Thank you.

Answer 1

The claim holds for all complex numbers $z\neq0$ and real exponents $r$ in the following precise sense:1) we treat the power $z^r$ as multivalued, because that's what complex powers are unless the exponent is an integer. 2) we treat $|z|^r$ as a real power, which is a well-defined (= uniquely valued) real number when the base is $\neq0$.

Given all of the above the claim follows from the definition of complex powers. Let's write $z$ in polar form $$ z=|z|e^{i\phi+i2n\pi}=e^{\ln|z|+i(\phi+2n\pi)}, $$ where $n$ is an arbitrary integer. Then $$ z^r=e^{r\ln|z|+i(r\phi+2nr\pi)}=|z|^re^{i(r\phi+2nr\pi)}. $$ When $r$ is real the latter factor here lies on the unit circle irrespective of choice of the integer parameter $n$, so $$ |z^r|=|z|^r. $$ Note that when $r=x+iy$ is non-real, the factor $u_n:=e^{i(r\phi+2nr\pi)}$ has absolute value $e^{-(\phi+2n\pi)y}$ which 1) depends on $n$, and 2) is not equal to one for any $n$ unless $\phi$ is an integer multiple of $2\pi$.

Answer 2

When $r$ is not an integer, the situation is slightly delicate: For $z \in \Bbb C - \{0\}$, we define $$z^r := \exp(r \log z),$$ but we can see immediately that this appears to depend on a choice of branch cut of the logarithm function. Indeed, depending on our choice of branch cut, $(-1)^{1 / 2}$ has value either $+i$ or $-i$.

Not all is lost, however: For any $z \in \Bbb C - \{0\}$ and any two choices $\log$, $\widetilde \log$ of branch cut of the logarithm, we have $$\widetilde \log z = \log z + 2 \pi i k$$ for some integer $k$. Because this difference is imaginary, for real $r$ the quantity $|z^r|$ is independent of the choice of branch cut: \begin{align*} \left|\exp (r \widetilde \log z)\right| &= \left|\exp [r (\log z + 2 \pi i k)]\right| \\ &= \left|\exp (r \log z) \exp(2 \pi i k r) \right| \\ &= \left|\exp (r \log z)\right| \left|\exp(2 \pi i k r) \right| \\ &= \left|\exp (r \log z)\right| . \end{align*} Of course, $|z|$ does not depend on any choice, so provided we either interpret $|z|^r$ as a real exponentiation or choose a branch cut so that $z \mapsto z^r$ coincides with real exponentiation for positive $z$, $|z|^r$ does not depend on any choice either. In short, we can show that the desired identity $$|z^r| = |z|^r$$ holds for any choice of branch cut (and again, real $r$) provided we can show it for some choice of branch cut.

Indeed: We can write any complex number $z \in \Bbb C - \{0\}$ as $\rho e^{i \theta}$ for some real $\rho, \theta$, and so we have $|z| = \rho$ and, for some branch of the logarithm function, $\log z = \log \rho + i \theta$ $$|z^r| = |\exp r \log z| = |\exp r (\log \rho + i \theta)| = |\exp (r \log \rho)| |\exp (i r \theta)| = \rho^r = |z|^r$$ as desired.

This argument fails immediately for nonreal $r$; indeed, $|z^r|$ is always real but for nonreal $r$ in general $|z|^r$ is not real.

Answer 3

We will show that $$ |z^r|=|z|^r\;\;\Longleftrightarrow\;\;r\in\Bbb R $$

Given $z$ a nonzero complex number, we can write it in its polar form $z=se^{i\theta}$. Let then $r=a+ib$. Thus \begin{align*} \left|z^r\right| =&\left|(se^{i\theta})^{a+ib}\right|\\ =&\left|s^as^{ib}e^{i\theta a}e^{-\theta b}\right|\\ =&s^ae^{-\theta b}\left|\left(e^{\theta a}s^b\right)^i\right| \end{align*}

Let's now call $\alpha:=e^{\theta a}s^b\in\Bbb R$; thus $$ \alpha^i=e^{\log \alpha^i}=e^{i\log\alpha} $$

hence

$$ \left|\left(e^{\theta a}s^b\right)^i\right|=1 $$

so we can conclude that $$ |z^r|=|z|^{\Re r}e^{-\arg z\cdot\Im r} $$ from which we can conclude.

Note: the logarithm taken is the principal logarithm: $\log w=\log |w|+i\arg w$ i.e. is a particular solution of $e^z=w$. Precisely, the one in which the argument is the principal argument, i.e. $\arg z\in[0,2\pi[$ and the general solution $\log|w|+i(\arg w+2k\pi),\;k\in\Bbb Z$ is choosen with $k=1$.