In my class, my instructor told us that the square root of a complex number is in general not a function because it is multi-valued. For example, e^(ipi/4) could have a square root of e^(ipi/8) or e^(i*9pi/8). He then added that if we shortened the domain of the polar angle to (-pi,pi), the square root then becomes a function. I don't see how this works. The square root has a period of pi, over which it repeats itself. For eq, both e^(-ipi/4) and e^(ipi/4) qualify as square roots of e^(i*pi/2), and they both lie in the shortened domain. I don't see where I am going wrong.
Turn the square root of a complex number into a function
complex numbers
Best Answer
$e^{i\pi/4}$ is a square root of $e^{i\pi/2} = i$ but $e^{-i\pi/4}$ is not: its square is $-i$.