Show that i^i = e^(-pi/2).

This question is very similar to differentiating x^x and the method for tackling it is quite common when powers are involved. We want to show that i^i=e^(-pi/2). Start by writing z=i^i (act like we do not know what i^i is yet). A number raised to power of i is very strange. So take logarithms on both sides:z=i^i --> ln(z)=ln(i^i)=iln(i) using properties of logarithms.Now we have to figure out what ln(i) is. Recalling Euler's identity e^ix=cos(x)+isin(x) we remember that i=e^(ipi/2). ln is the inverse of exponential so it makes sense that ln(i)=ln(e^(ipi/2))=ipi/2. So ln(z)=i*(i*pi/2)=-pi/2. So z=e^(-pi/2). Thus i^i=e^(-pi/2).