Finding the two square roots of $-i$

Question

I want to find the two square roots of $-i$. Rewriting $-i$ in its polar form produces $e^{(3\pi/2)i}$, however, I am not sure where to go from here. I ideally wanted to find the roots and draw them as vectors in the complex plane in order to intuitively understand complex numbers and their roots, yet, I am stuck.

Answer 1

Well, we also have the algebraic way.

$$\begin{align} \sqrt{-i} & =a+bi \\ -i & = (a+bi)^2 \\ -i & = a^2-b^2+2abi \\ \end{align}$$

Clearly, $a^2-b^2=0$ since $\Re(-i)=0$ and $2ab=-1$ since $\Im(-i)=-1$, giving us some equations to work with.

$$\begin{cases}0 =a^2-b^2 \\ -1 = 2ab \end{cases}\implies\begin{cases}a=\pm b \\ -1 = 2ab \end{cases}$$

$$-1=2(\pm b)b=\pm2b^2\implies b=\pm\sqrt{\pm1/2}$$

Since $b$ must be real by definition, then this simplifies a little

$$b=\pm\sqrt{1/2}$$

Which gives the solution for $a$

$$a=\pm(\pm\sqrt{1/2})$$

So, we have

$$\sqrt{-i}=\pm\left(\sqrt{1/2}+i\sqrt{1/2}\right)$$

$$\sqrt{-i}=\pm\left(\sqrt{1/2}-i\sqrt{1/2}\right)$$

After some simplifications and checking for extraneous solutions by squaring both sides tells us the final answer:

$$\boxed{\sqrt{-i}=\pm\frac{\sqrt2}2\left(1-i\right)}$$

Answer 2

Hint: $$\frac{1}{2}(1-i)^2=-i$$

Answer 3

This is like to ask for the square root of the complex number

$$z = 0 + (-1)i$$

Proceeding in the same way for general comple numbers you have

$$|z| = 1 ~~~~~~~ \theta = \arctan(-1) = -\frac{\pi}{4}$$

Hence

$$-i = e^{-\frac{i\pi}{4}}$$

So the square roots are

$$w_k = \sqrt[n]{|z|}\exp\left(i\left[\frac{\theta}{n} + \frac{2k\pi}{n}\right]\right)$$

Where $k = 0, 1$ because in your case $n = 2$.

Now you should be able to proceed.