Another 'find the real and imaginary parts of this expression' question. I see that this happens quite a bit, but none of the questions/answers I found helped me with my case.
I have $x$ and $y$ which are both real numbers, and $y$ is positive.
How do I express $x^y$ as some $a + bi$?
This seems like it should be really simple, but I've not had much luck, and general internet search results are helpfully filled with pages talking about only the integer or rational cases of (complex) exponentiation.
Best Answer
After reading up on complex logarithms, I think I may have figured it out.
Complex logarithms let me rewrite
$$x^y$$
as
$$e^{y \ln (x)}$$
$\ln (x)$ is still a complex number (actually many of them). The real and imaginary parts of this number are $\ln (|x|)$ (which is now just a standard natural logarithm of a positive real) and $\arg (x)$ (still a multivalued function) respectively.
That lets me write my expression as
$$e^{y (\ln (|x|) + i \arg (x))}$$
which I can further expand to
$$e^{y \ln (|x|)} e^{i y \arg (x)}$$
Referring to the identity suggested by @blakedylanmusic, I can observe that
$$r = e^{y \ln (|x|)}$$
$$\theta = y \arg(x)$$
Now applying the identity, I rewrite the expression as
$$e^{y \ln (|x|)} (\cos (y \arg (x)) + i \sin (y \arg (x)))$$
Now
$$a = e^{y \ln (|x|)} \cos (y \arg (x))$$
$$b = e^{y \ln (|x|)} \sin (y \arg (x))$$
If I understand the multiplicity of complex logarithm correctly, the multiplicity of the complex logarithm is captured by $\arg(x)$ returning many possible angles. That multiplicity is then collapsed away when we pass that value back into $\sin ()$ and $\cos ()$, so this is expression actually produces a unique complex number - the exact result of the initial exponentiation.