What is the real and imaginary parts for the complex function $f(z)=z^z$

Question

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.

Answer 1

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 $$ \exp\left[x\log\left(\textstyle{\sqrt{x^2+y^2}}\right)-2y\arctan\left(\frac{y}{x+\sqrt{x^2+y^2}}\right)\right]\\ \times\sin\left[y\log\left(\textstyle{\sqrt{x^2+y^2}}\right)+2x\arctan\left(\frac{y}{x+\sqrt{x^2+y^2}}\right)\right] $$

Answer 2

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:

$$z^z = \exp(z\ln z)=\exp(u(\rho,\phi))\exp(i v(\rho,\phi))$$ $$=\exp(u(\rho,\phi))\cos( v(\rho,\phi))+i\exp(u(\rho,\phi))\sin( v(\rho,\phi)):=\Re(z^z)+i \Im(z^z)$$

Answer 3

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 $$\operatorname{Arg}z^z = \log (r^r)^{\sin t} + rt_k\cos t $$