Simplification of complex exponentiation of a number

Question

Can someone help me find the simplification (like $a+bi$) to the expression $x^i$, where $x$ is a number? How do I proceed?

For example.. what are the $a$ and $b$ of the number $5^i$?

Thank you :)

Answer 1

Let $r=\sqrt{a^2+b^2}$ and $\theta=\text{arctan2}(\frac{b}a)$, then$$a+bi=re^{i\theta}=r^{-i\cdot i}e^{i\theta}=(e^{\theta}(\cos\ln r-i\sin\ln r))^i$$ with the correct choice of branch.

EDIT:

I was doing the opposite:)

Every $x$ can be represented uniquely as $re^{i\theta}$ where $r>0$ and $\theta\in[0,2\pi)$.

Thus, $$x^i=e^{i\ln r}e^{-\theta}=e^{-\theta}(\cos\ln r+i\sin\ln r)$$

However, you can actually choose a different branch to obtain a different result.

Indeed, $$x=re^{i\theta+2\pi ik}=e^{\text{Log} (r)+i\theta+2\pi i k}$$ for any integer $k$.

Therefore, $$x^i= e^{2\pi k-\theta}(\cos\text{Log}( r)+i\sin\text{Log}( r))$$

Answer 2

Let $x=u+iv$. Then

$$(u+iv)^i=e^{i\log(u+iv)}=e^{i(\log\sqrt{u^2+v^2}+i\arctan v/u})=e^{-\arctan v/u}\left(\cos\log\sqrt{u^2+v^2}+i\sin\log\sqrt{u^2+v^2}\right).$$

Hence,

$$a=e^{-\arctan v/u}\cos\log\sqrt{u^2+v^2}=e^{-\angle x}\cos\log|x|, \\b=e^{-\arctan v/u}\sin\log\sqrt{u^2+v^2}=e^{-\angle x}\sin\log|x|.$$

Answer 3

This will always be multiple-valued.

One generally defines $$z^w \stackrel{\small{\textrm{def}}}{=} e^{w\log z}$$ and the multiple-valuedness of the logarithm produces multiple values for this expression.

If you write $z=re^{it}$ where $r>0$, and $w=u+iv$ with $u$ and $v$ real, then the possible values are precisely (where "$\ln$" is the real-valued function of a real variable)

$$e^{(u+iv)(\ln r + i(t + 2k\pi))}$$ $$=e^{u\ln r - v(t+2k\pi) + i(v\ln r + u(t+2k\pi))}$$ $$=r^ue^{-v(t+2k\pi)}\cdot e^{i(\ln r^v + u(t+2k\pi))}$$ $$=r^ue^{-v(t+2k\pi)}\cdot e^{i\ln (r^v e^{ u(t+2k\pi)})}$$ $$=\boxed{r^ue^{-v(t+2k\pi)}\cos\ln(r^v e^{ u(t+2k\pi)}) + ir^ue^{-v(t+2k\pi)}\sin\ln(r^v e^{ u(t+2k\pi)})}$$

for integral $k$.

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Note: Taking $k=0$ yields a somewhat simpler "principal" value $$\boxed{r^ue^{-vt}\cos\ln(r^v e^{ ut}) + ir^ue^{-vt}\sin\ln(r^v e^{ ut})}$$

For your example, we write $z=5=5e^{0i}$, so $r=5$, $t=0$, $u=0$, and $v=1$. This gives the possible values

$$=\boxed{e^{-2k\pi}\cos\ln 5 + ie^{-2k\pi}\sin\ln 5}$$ with "principal" value (taking $k=0$) $$=\boxed{\cos\ln 5 + i\sin\ln 5}$$