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:
\[ i^i=e^{-\frac{\pi}{2}}\]