Can anyone tell me how $\frac{\pi}{\sqrt 2} = \frac{\pi + i\pi}{2\sqrt i}$

Question

I was working out a problem last night and got the result

$\frac{\pi + i\pi}{2\sqrt i}$

However, WolframAlpha gave the result

$\frac{\pi}{\sqrt 2}$

Upon closer inspection I found out that

$\frac{\pi}{\sqrt 2} = \frac{\pi + i\pi}{2\sqrt i}$

But I cant seem to derive it myself and it has been bugging me all day. How can this complex number be reduced to a real number?

Answer 1

Hint: $$(1+i)^2 = 2i{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$$

Answer 2

Rewriting everything in polar form, the numerator is $\sqrt{2} \pi e^{i \pi/4}$ and the denominator is, for the choice of branch of $\sqrt{}$ that Wolfram is using, $2 e^{i \pi/4}$. So the $e^{i \pi/4}$ terms cancel and you're left with a real number.

Answer 3

Consider the following: Let $z=a+bi$ where $a,b$ belong to the set of all real number.

We would like to solve for $a$ and $b$ such that $z^2=i$, that is $(a+bi)^2=0+i$

Then, you will know why $\sqrt{i}= \frac{1+i}{\sqrt{2}}$ and how to solve your problem.