# General solution to complex number to complex power in complex form

Given the form:

$$(a+bi)^{(c+di)}$$

Does there exist a generalized solution for the principle branch where:

$$(e+fi) = (a+bi)^{(c+di)}$$

I ask this because addition and multiplication (with subtraction and division counterparts) have generalized solutions:

$$(a+bi) + (c+di) = (e+fi)$$

$$e=a+b$$

$$f=b+d$$

and for multiplication:

$$(a+bi)*(c+di) = (e+fi)$$

$$e=a*c-b*d$$

$$f=b*c+a*d$$

I also realize that:

$$(a+bi)^{(c+di)} = e^{(c+di)log(a+bi)}$$

and that this can be converted to polar form to solve this problem; however, I'm not sure how to reduce this to a complex form afterwards. In any case, I'd like it solved in terms of e and f of the first form mentioned:

$$(e+fi) = (a+bi)^{(c+di)}$$

1. First, what is $$z^w$$ when $$z$$ and $$w$$ are complex numbers. It is not trivial at all to define a power in complex field. It is multivalued function, i.e. you will not receive a unique value $$z^w$$ for given $$z,w$$. In fact, even $$i^i$$ with $$i$$ is the imaginary unit is not really well-define, it can be $$e^{-\frac{\pi}{2}+2k\pi}$$ with $$k \in \mathbb{Z}$$. The reason is that we define $$z^w$$ by $$\exp(w \log z)$$ and $$log(z)$$ (as you noticed) is defined with several branches.
2. So which branch of $$\log$$ are we using here? As you intended, we will use principle branch. Briefly, the principle branch $$Log(z)$$ when we $$z$$ in the polar form $$|z|e^{i\arg(z)}, \arg(z) \in (-\pi,\pi)$$ is $$Log(z) = \log|z|+iArg(z)$$ I refer to wikipedia page https://en.wikipedia.org/wiki/Complex_logarithm and references therein for you to look further.
3. So, let return to your question $$e+fi=(a+bi)^{(c+di)}$$. Denote by $$z=a+bi,w=c+di$$. Thus $$e=\Re(z^w), f=\Im(z^w)$$. Let us write $$z$$ in polar form $$z=a+bi=re^{i\varphi}, r=\sqrt{a^2+b^2},\varphi=Arg(z)$$ How to calculate $$Arg(z)$$ from $$a$$ and $$b$$, please see https://en.wikipedia.org/wiki/Argument_(complex_analysis). Then $$\begin{array}{rcl} z^w &=& \exp(w Log(z)) \\\\ &=& \exp(w(\log r + i\varphi))\\\\ &=& \exp((c+id)(\log r + i\varphi))\\\\ &=&\exp(c\log r- d\varphi + i(d\log r +c \varphi)\\\\ &=& e^{(c\log r - d \varphi)} e^{i(d\log r+c\varphi)} \end{array}$$ Thus $$\Re (z^w) = e^{c \log r - d\varphi} \cos(d\log r + c\varphi)$$ $$\Im(z^w)= e^{c\log r- d\varphi}\sin(d\log r+c \varphi)$$