How to calculate $\ \sqrt{-i}$ without trigonometric form

Question

I'm trying to calculate $\ \sqrt{-i}$. I've seen other answers using the trigonometric form, but I was trying this way:

$\ \sqrt{-i}$ = $\ \sqrt{-1} * \sqrt{i} $ = $\ ({-1})^{1/2} * ({({-1})^{1/2}})^{1/2} $ = $\ ({-1})^{3/4} $

The result i get from wolfram is $\ -({-1})^{3/4} $.

Answer 1

The rules for the way the surd ($\sqrt{x}$) behaves don't carry over to the complex numbers. For instance,

$$4 = \sqrt{16} = \sqrt{(-4)(-4)} = \sqrt{-4}\sqrt{-4} = (2i)(2i) =-4$$

doesn't work. The step at the 3rd equal sign isn't valid in the complex numbers. And what you've done is essentially the same thing in your first step.

You might do this geometrically. In the complex plane, $-i$ is at the point $(0,-1)$ which vector is $270$ degrees from the positive $x$-axis. So one of its square roots has angle $270/2 = 135$ degrees. I don't know if you count dealing with a 45-45 right triangle as "trigonometry", but it's easy to compute the square root as $(-1+i)/\sqrt{2}$ from here.

Answer 2

If $(a+bi)^2=-i$ with $a,b\in\mathbb R$, then $a^2-b^2=0$ and $2ab=-1$.

Thus $a=\pm b$, but $a=b$ wouldn't work with $2ab=-1$, so $a=-b$.

That means $-2a^2=-1$, so $a=\pm\dfrac1{\sqrt2}$ and $b=-a$ are solutions.

Answer 3

Half its argument in the complex plane is $\dfrac{3\pi}{4}$

Accordingly in its exponential form

$$ z= e ^{i \dfrac{3\pi}{4}}$$