Evaluate $1^{i+i}$

Question

I thought $1^{i+i}$ was equal to 1 but was told the answer was $e^{-4n \pi}$.

Why I was incorrect and what was used to find the correct answer?

Answer 1

I reckon that expressions like $z^w$, where $z$ and $w$ are complex numbers, are best avoided. But, it you really, really must use it, then the natural interpretation of $z^w$ is as $$z^w=\exp(w\ln z).$$ Of course the catch here is that $\ln z$ is "many-valued", that is if $u$ is a logarithm of $z$ then so is $u+2n\pi i$ where $n$ is your favourite integer. If you were being perverse, you might argue that $\ln 1$ should be $2\pi i$ rather than $0$ and so $$1^{1+i}=\exp((1+i)(2\pi i))=\exp(2\pi i-2\pi)=e^{-2\pi}$$ etc.

Answer 2

Because the complex power function is a multivalued function.

$1^{2i}=\exp(2i\cdot \mathrm{Ln}(1))=\exp(2i\cdot(\ln(1)+2n\pi i))=\exp(-4n\pi)$.

Answer 3

Note that $$1=e^{2k \pi i}$$

thus $$1^{2i} = (e^{2k\pi i})^{2i} = e^{-4k \pi}$$

Answer 4

$1^{i+i}$ needs to be evaluated from the perspective of complex analysis.

First note, that any complex number (which includes pure real and pure imaginary ones) can be presented as

$$z = r e^{i \theta}$$

Now, recall that $2i$ in polar form will be $2i = 2 e^{i \pi/2}$, because this number makes a $\pi/2$ angle with $x$-axis and has a length of $r = 2$ (it is just the absolute value of $2i$). So, now you have

$$1^{2i}=1^{2e^{i \pi/2}}= e^{2 i \log (1)} =e^{2 i(\mathrm{Log} (1)+2 i \pi n)} =e^{2 i(0+2 i \pi n)} =e^{-4 \pi n}$$