What is the real and imaginary parts for the complex function $f(z)=z^z$ I know:
$f(z)=z^z
=|z|^ze^{iz\theta}
$
and 
$=|z|^z(\cos(zθ) + i\sin(zθ))$
But how do I continue to get the results for $\Re(z^z)$ and $\Im(z^z)$?
$$\text {         }$$
Thanks.
 A: The log of $z$ is
$$
\log(x+iy)=\log\left(\textstyle{\sqrt{x^2+y^2}}\right)+2i\arctan\left(\frac{y}{x+\sqrt{x^2+y^2}}\right)
$$
The log of $z^z$ is
$$
\begin{align}
(x+iy)\log(x+iy)&=\left[x\log\left(\textstyle{\sqrt{x^2+y^2}}\right)-2y\arctan\left(\frac{y}{x+\sqrt{x^2+y^2}}\right)\right]\\
&+i\left[y\log\left(\textstyle{\sqrt{x^2+y^2}}\right)+2x\arctan\left(\frac{y}{x+\sqrt{x^2+y^2}}\right)\right]
\end{align}
$$
The real part of $z^z$ is
$$
\exp\left[x\log\left(\textstyle{\sqrt{x^2+y^2}}\right)-2y\arctan\left(\frac{y}{x+\sqrt{x^2+y^2}}\right)\right]\\
\times\cos\left[y\log\left(\textstyle{\sqrt{x^2+y^2}}\right)+2x\arctan\left(\frac{y}{x+\sqrt{x^2+y^2}}\right)\right]
$$
The imaginary part of $z^z$ is
$$
\exp\left[x\log\left(\textstyle{\sqrt{x^2+y^2}}\right)-2y\arctan\left(\frac{y}{x+\sqrt{x^2+y^2}}\right)\right]\\
\times\sin\left[y\log\left(\textstyle{\sqrt{x^2+y^2}}\right)+2x\arctan\left(\frac{y}{x+\sqrt{x^2+y^2}}\right)\right]
$$
A: We start with @Cameron Williams suggestion and express:
$$z^w = \exp(w\log z)$$
Let
$$z=\rho \exp(i \phi+i2k\pi)=\rho(\cos\phi+i \sin\phi) \text{   (k is integer)}$$
Then
$$\ln z=\ln\rho+i(\phi +2k\pi)$$
The factor $2k\pi$ is copied from MJD's solution.
$$z \ln z=\rho (\cos\phi+i \sin\phi)(\ln\rho+i(\phi+2k\pi))=\rho\ln\rho\cos\phi-\rho(\phi+2k\pi)\sin\phi+i((\phi+2k\pi)\rho\cos\phi+\rho\ln\rho\sin\phi)=:u(\rho,\phi)+i v(\rho,\phi)$$
We can now substitute this (Re+i Im) into the following expression:
$$z^z = \exp(z\ln z)=\exp(u(\rho,\phi))\exp(i v(\rho,\phi))$$
$$=\exp(u(\rho,\phi))\cos( v(\rho,\phi))+i\exp(u(\rho,\phi))\sin( v(\rho,\phi)):=\Re(z^z)+i \Im(z^z)$$
A: By definition, one computes $z^z$ as $e^{z\log z}$. For ease of computation, write
$$z = re^{it}$$ where $r>0$ and $t$ is real. Then the possible values of $\log z$ are
$$\log z = \log r +it_k$$ where $$\boxed{t_k = t + 2k\pi}$$ for integral $k$.
That means the possible values of $z^z$ are
$$z^z = e^{z\log z} = e^{re^{it}\cdot(\log r + it_k)} = e^{(r\cos t + ir\sin t)(\log r + it_k)}$$
$$=e^{r\cdot\log r\cdot \cos t - rt_k\sin t}\cdot e^{i(r\cdot\log r\cdot\sin t + rt_k\cos t)}$$
$$=(r^r)^{\cos t}e^{-rt_k\sin t}\cdot e^{i(\log (r^r)^{\sin t} + rt_k\cos t)}
$$
The possible real and imaginary components of $z^z = R_k + iI_k$, therefore, are
$$\boxed{R_k = (r^r)^{\cos t}e^{-rt_k\sin t}\cos\left(\log (r^r)^{\sin t} + rt_k\cos t\right)}
$$
and
$$\boxed{I_k = (r^r)^{\cos t}e^{-rt_k\sin t}\sin\left(\log (r^r)^{\sin t} + rt_k\cos t\right)}
$$
respectively, for integral $k$.
Note 1: In the logarithmic expressions above, the exponent $\sin t$ is carried by $r^r$, not by $\log r^r$.
Note 2: This could be rewritten in terms of $z$ as
$$R_k = |z|^{\operatorname{Re}z}e^{-\operatorname{Im}z\operatorname{Arg}z}\cos\left(\log|z|^{\operatorname{Im}z} + \operatorname{Re}z\operatorname{Arg}z\right)
$$
$$I_k = |z|^{\operatorname{Re}z}e^{-\operatorname{Im}z\operatorname{Arg}z}\sin\left(\log|z|^{\operatorname{Im}z} + \operatorname{Re}z\operatorname{Arg}z\right)
$$
where $\operatorname{Arg}z$ takes on the various values $t_k$.
Note 3: We have essentially fleshed out the facts that
$$\left|z^z\right| = (r^r)^{\cos t}e^{-rt_k\sin t}
$$
and $$\operatorname{Arg}z^z = \log (r^r)^{\sin t} + rt_k\cos t
$$
