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?

The $\pi$ is not relevant. What you really ask is, why is: $\sqrt{i} = \frac{1+i}{\sqrt{2}}$. Square the expression and use $i^2=-1$ to see why! — Winther, Sep 18 '14 at 12:17

mookid · Answer 1 · 2014-09-18T12:59:42.053

5

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

edited Sep 18 '14 at 12:59

answered Sep 18 '14 at 12:10

mookid

28,236

1

Such skillful character limit avoiding. +1 Also, its a pretty darn good answer, even though its short. – Asimov Sep 18 '14 at 12:29
1

You can pad the character length by inserting {}{}{} into one of your formulas. – MJD Sep 18 '14 at 12:41
1

The square of both sides might be equal, but that doesn't necessarily mean the two sides are equal. Don't you agree? – Sheheryar Zaidi Sep 18 '14 at 12:43
Perhaps that's why the answer is labeled "hint". – David K Sep 18 '14 at 12:49
@MJD: thanks!${}{}{}{}{}{}$ – mookid Sep 18 '14 at 13:00
if only the downvoters had to justify their vote here... – mookid Sep 18 '14 at 18:42

Ian · Answer 2 · 2014-09-18T12:22:07.690

3

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.

edited Sep 18 '14 at 12:22

answered Sep 18 '14 at 12:14

Ian

101,645

score 0 · Answer 3 · edited Sep 18 '14 at 13:01

0

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.

edited Sep 18 '14 at 13:01

Winther

24,478

answered Sep 18 '14 at 12:44

Thomas

49

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

3 Answers3