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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:

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

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1 Answer 1

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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 $$\Re (z^w) = e^{c \log r - d\varphi} \cos(d\log r + c\varphi)$$ $$\Im(z^w)= e^{c\log r- d\varphi}\sin(d\log r+c \varphi)$$

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  • $\begingroup$ While you answered my question. I was hoping I could geometrically understand complex exponentiation as I do in addition and multiplication (shifting and rotation). It looks like it's just not possible with this. $\endgroup$ May 3, 2019 at 16:19

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