Complex Number Question - $|z^{z}|$

Question

Find all possible values of $$\mid z^{z} \mid$$ using the polar for of $z$.

I have tried putting it into polar form but nothing comes out that seems easy to work with/looks like a reasonable simple answer.

Answer 1

Assuming $\;\arg z=\theta\;,\;\;x:=z+iy\notin\Bbb R_-\cup\{0\}\,,\;$ and we choose the principal branch of the logarithmic function:

$$z^z=e^{z\,\text{log}\,z}=e^{z\left(\log|z|+i\theta\right)}=e^{\frac x2\log(x^2+y^2)-y\theta+i\left(x\theta+\frac12y\log(x^2+y^2)\right)}\implies$$

$$\implies |z^z|=e^{\frac x2\log(x^2+y^2)-y\theta}$$

Using the polar form:

$$z=|z|e^{i\theta}\implies z^z=|z|^ze^{iz\theta}=e^{z\log|z|+iz\theta}$$

which is the same as before (look after the second equality sign above)

Answer 2

You can write $a^b = e^{b\text{Log}(a)}$. We need to have some understanding of notation before we go any further:

\begin{align} \text{Log}(a):& \text{Most general natural logarithm--gives all complex values}\\ \log(a):& \text{Principle logarithm--gives value using the arg from } [0, 2\pi) \text{ (may be same as ln)}\\ \ln(a):& \text{real valued natural logarithm. Returns a single, real value and it must be that } a \text{ is real and } a > 0 \end{align}

So now we can write $z^z = e^{z\text{Log}(z)}$. So we first need to find $\text{Log}(z)$:

$$ \text{Log}(z) = a + bi \\ z = e^{a + bi} = r_ze^{\theta_z i + 2\pi ni} \\ e^ae^{bi} = r_ze^{\theta_z i}\rightarrow r_z = e^a \rightarrow a = \ln(r_z) = \ln\left(|z|\right) \\ b = \theta_z + 2\pi n \\ \text{Log}(z) = \ln\left(r_z\right) + (\theta_z + 2\pi n)i $$

Now we just need to find $z\cdot \text{Log}(z)$. I'm going to stick with $r_z$ (which remember is really $|z|$) and $\theta_z$, so that we have:

\begin{align} z =& r_z\left(\cos(\theta_z) + i\sin(\theta_z)\right) \\ z\text{Log}(z) =& r_z\left(\cos(\theta_z) + i\sin(\theta_z)\right)\left(\ln\left(r_z\right) + (\theta_z + 2\pi n)i\right) \\ z\text{Log}(z) =&r_z\left(\cos(\theta_z)\ln(r_z) - \sin(\theta_z)(\theta_z + 2\pi n) + \text{imaginary terms}\right) \end{align}

Remember $\left|e^{a + bi}\right| = e^a$ because it really is $e^ae^{bi}$ and $\left|e^{bi}\right| = 1$. Thus the magnitude of $z^z$ would be:

\begin{align} \left|z^z\right| =& e^{r_z\left(\cos(\theta_z)\ln(r_z) - \sin(\theta_z)(\theta_z + 2\pi n)\right)} = \frac{e^{r_z\cos(\theta_z)\ln(r_z)}}{e^{r_z\sin(\theta_z)(\theta_z + 2\pi n)}} \\ \left|z^z\right| =& \frac{r_z^{r_z\cos(\theta_z)}}{e^{r_z\sin(\theta_z)(\theta_z + 2\pi n)}} \end{align}

I don't really want to write it any other way, I mean $r_z = |z|$, $r_z\cos(\theta_z) = \text{Re}(z)$, and $r_z\sin(\theta_z) = \text{Im}(z)$, but you still have a $\theta_z$ by itself and it's misleading to write that as $\theta_z = \tan^{-1}\left(\frac{\text{Im}(z)}{\text{Re}(z)}\right)$ because the inverse tangent won't give the correct $\theta_z$ in general. I could write $\theta_z = \arg(z)$, but then why not just write $\theta_z$?