Complex Number Question - $|z^{z}|$ 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.
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)
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):&amp; \text{Most general natural logarithm--gives all complex values}\\
\log(a):&amp; \text{Principle logarithm--gives value using the arg from } [0, 2\pi) \text{ (may be same as ln)}\\
\ln(a):&amp; \text{real valued natural logarithm.  Returns a single, real value and it must be that } a \text{ is real and } a &gt; 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 =&amp; r_z\left(\cos(\theta_z) + i\sin(\theta_z)\right) \\
z\text{Log}(z) =&amp; 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) =&amp;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| =&amp; 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| =&amp; \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 $\theta_z$?