`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:

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

(0 Votes)