Why is $i^i$ real but not $\left(\frac{1+i}{\sqrt{2}}\right)^i$

complex numbersexponentiation

I know that generally the exponent rule $\left(a^b\right)^c=a^{bc}$ does not apply over the field of complex numbers. From what I've read it is due to the periodicity of the complex exponential function causing $\ln{e^c}$ to be a multivalued function when $c$ is complex. But what I don't get is why $i^i$ can be found by doing $\left(e^{i\frac{\pi}{2}}\right)^i=e^{i^2\frac{\pi}{2}}=e^{\frac{-\pi}{2}}$ and still obeys normal exponentiation rules even though the exponent is complex but for a different base such as $\left(\frac{1+i}{\sqrt{2}}\right)^i$, according to Wolfram Alpha, this number is still complex. Can someone explain why changing the base in this case means that the exponent rule can no longer be followed??

For example, I tried to be a little more rigorous (I think it's more rigorous at least) in my calculation but still arrived at a real number for $\left(\frac{1+i}{\sqrt{2}}\right)^i$ as follows:

$\left(\frac{1+i}{\sqrt{2}}\right)^i=e^{i\ln{\frac{1+i}{\sqrt{2}}}}=e^{i(\ln{1}+i(\frac{\pi}{4}+2k\pi))}=e^{-\frac{\pi}{4}-2k\pi}$ which is still real. So could someone help explain what is going on?

Best Answer

On the unit circle $z=e^{i\theta}$ becomes real on $z^i$.

Off the unit circle, e.g. $z=1+i$, becomes $z=re^{i\theta}$, and $r^i$ is complex unless $|r|=1$.

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