What's the process in rectangular form for deriving $(a+bi)^{a+bi}$?

Question

What's the process, using Euler's Formula, solving $(a+bi)^{a+bi}$ when outlined algebraically in rectangular form?

Edit: And also solving this in the same form and process, $(d cos(y)+i d sin(y))^{(d cos(y)+i d sin(y))}$, where $d = e^x$

Answer 1

$a+bi = \rho e^{\theta i}\\ (a+bi)^{a+bi}(\rho e^{\theta i})^{a+bi}\\ (\rho^a e^{a\theta i})(\rho^{bi}e^{-b\theta})\\ (\rho^a e^{a\theta i})(e^{(b\ln\rho)i}e^{-b\theta}\\ \rho^ae^{-b\theta}e^{(a\theta + b\ln\rho i)}\\ \rho^ae^{-b\theta}(\cos(a\theta + b\ln\rho) + i\sin(a\theta + b\ln\rho)\\ $

Answer 2

$(a+bi)^{(a+bi)}=e^{\log((a+bi)^{(a+bi)})}=e^{(a+bi)\log(a+bi)}$ with $\log(a+bi)=\ln|a+bi|+i*\arg(a+bi)$ when sometimes $\arg(a+bi)=\arctan(b/a)$ depending on the direction of the vector. Note that the outcomes of the log is multivalued, since the log of a complex number is multivalued. The second part you just write $d(\cos(y)+i\sin(y))=d*e^{y}=e^{x+y}$ and solve like above with using some logarithms and power product- and summation rules.

Answer 3

$$ z=a+bi = r e^{i\theta}$$ in its polar form.

$$ (a+bi)^{a+bi}= r e^{i\theta (a+bi)}= r e^{-b\theta +i a\theta}=$$

$$r e^{-b\theta}(\cos a\theta + i \sin a\theta)$$