# Thread: [SOLVED] The complex logarithm of the square root of i.

1. ## [SOLVED] The complex logarithm of the square root of i.

Simple question but I can't seem to get it.

Show that the set of values of $log(i^{1/2})$ is $(n + 1/4)\pi\\i$ for any integer n.

But using the definition of a complex logarithm and the polar form of i. I get:

$i = e^{i\pi/2}$
So, $i^{1/2} = e^{i\pi/4}$

Since $log(z) = ln(r) + i(\theta + 2n\pi)$
So, $log(i^{1/2}) = ln(1) + i(\pi/4 + 2n\pi) = i(\pi/4 + 2n\pi)$

which is not the correct answer, is there something i'm missing?

2. Originally Posted by Haven
Simple question but I can't seem to get it.

Show that the set of values of $log(i^{1/2})$ is $(n + 1/4)\pi\\i$ for any integer n.

But using the definition of a complex logarithm and the polar form of i. I get:

$i = e^{i\pi/2}$
So, $i^{1/2} = e^{i\pi/4}$

Since $log(z) = ln(r) + i(\theta + 2n\pi)$
So, $log(i^{1/2}) = ln(1) + i(\pi/4 + 2n\pi) = i(\pi/4 + 2n\pi)$

which is not the correct answer, is there something i'm missing?

Note that $i=e^{i\left(\frac{\pi}{2}+2n\pi\right)}$.

So $i^{\frac{1}{2}}=e^{i\left(\frac{\pi}{4}+n\pi\right )}$.

Therefore, $\log\left(i^{\frac{1}{2}}\right)=\log e^{i\left(\frac{\pi}{4}+n\pi\right)}=i\left(\tfrac {\pi}{4}+n\pi\right)$.

Does this make sense?

3. Dear lord, that makes a lot of sense. It's so simple now. Thanks alot