How can $e^{i\pi}+1$ be zero?

Question

So, most of us are familliar with Euler’s equation stating that $e^{i\pi}+1=0$. But I was wondering: how can an irrational number to the power of another irrational number equal a whole integer? And if that works, then how can $e^{i\pi}+1=0$ if $i$ isn’t even real?

Answer 1

How can an irrational number to the power of another irrational number equal a whole integer?

You mean like $e^{\ln 2}=2$? Mind you, that's still a positive integer, unlike $-1$.

And if that works, then how can $e^{i\pi}+1=0$ if isn’t even real?

Well, $i$ is the reason the exponential doesn't have to be positive.

Answer 2

It all boils down to Euler's identity

$$e^{i\theta}=\cos \theta + i \sin \theta\implies e^{i\pi}=\cos \pi + i \sin \pi=-1$$

Answer 3

Since $e^{i\pi}$ = $\cos(\pi) + i\sin(\pi)$ and $\sin(\pi) = 1$, $\cos(\pi) = -1$ you get

\begin{equation} e^{i\pi}+1 = 0 \end{equation}

Answer 4

Taking a slightly different approach to the other answers, because I think your confusion is elsewhere. You asked

How can an irrational number to the power of another irrational number equal a whole integer?

But this is entirely possible, even for real numbers. For instance, both $a = \sqrt{10}$ and $b = \log_{10} 4$ are irrational, but $a^b = \sqrt{10}^{\log_{10} 4}=\sqrt{10}^{2 \log_{10} 2} = 10^{\log_{10}2} = 2$, which is an integer.