## i^i using Euler's Equation

Problem: Using Euler's Equation, evaluate $$i^i$$. Euler's Equation is given by: $e^{ix}=\cos x +i\sin x$ (a proof of this equation can be found here)

Solution:

Notice that if we let $$x=\ln i$$, then we get: $e^{i(\ln i)}=\cos (\ln i) +i\sin (\ln i)$

$\left(e^{(\ln i)}\right)^i=\cos (\ln i) +i\sin (\ln i)$

$i^i=\cos (\ln i) +i\sin (\ln i)$

Now perhaps if we evaluate $$\ln i$$, we can get a cleaner answer. Consider what we get if we $$ln$$ both sides of Euler's Equation:

$\ln\left(e^{ix}\right)=\ln(\cos x +i\sin x)$

$ix=\ln(\cos x +i\sin x)$

Now we want $$\ln i$$ , so notice that we can get that on the RHS if we let $$x=\frac{\pi}{2}$$:

$i\frac{\pi}{2}=\ln\left(\cos \frac{\pi}{2} +i\sin \frac{\pi}{2}\right)$

$i\frac{\pi}{2}=\ln(i)$

So $$\ln i = \frac{\pi i}{2}$$. We can substitute this into our equation for $$i^i$$ to give:

$i^i=\cos \left(\frac{\pi i}{2}\right) +i\sin \left(\frac{\pi i}{2}\right)$

But notice that by Euler's Equation:

$e^{i\frac{\pi i}{2}}=\cos \left(\frac{\pi i}{2}\right) +i\sin \left(\frac{\pi i}{2}\right)$

$e^{-\frac{\pi}{2}}=\cos \left(\frac{\pi i}{2}\right) +i\sin \left(\frac{\pi i}{2}\right)$

So we can substitute this into our equation for $$i^i$$ to give:

$i^i=e^{-\frac{\pi}{2}}$