The Square Root of -i

Question

I am having difficulty understanding $\sqrt -i$ from other sources (such as Wolfram Alpha) I have found it to equal $-(-1)^\frac{3}{4}$ but do not understand the steps in to reach this conclusion. I know it may be trivial but any assistance would be greatly appreciated.

Answer 1

Note that $(1-i)^2=1^2-2i+i^2=-2i$. Therefore, a square root of $-i$ is $\frac1{\sqrt{2}}(1-i)$. The other one is, of course, $-\frac1{\sqrt{2}}(1-i)$.

Answer 2

Another way to see it is to start with $-i =e^{3i\pi\over 2}$. The square root has as module the square root of the module i.e $1$ and as argument half the argument $\pmod\pi$ because the argument is defined $\pmod {2\pi}$. So there are two square roots

$$\begin{align} e^{3i\pi\over 4}=&\cos{3\pi\over 4}+i \sin{3\pi\over 4}=-{\sqrt{2}\over 2}+i{\sqrt{2}\over 2}\\ e^{-i\pi\over 4}=&\cos{-\pi\over 4}+i \sin{-\pi\over 4}={\sqrt{2}\over 2}-i{\sqrt{2}\over 2} \end{align}$$

Answer 3

First the notation $\sqrt{z}$ for a complex number $z$ should not be used, as there's no general way to distinguish between both square roots of $z$, as can be done with real numbers.

The simplest way here to obtain the square roots of $-i$ is with the exponential notation: as $-i=\mathrm e^{-\tfrac{i\pi}2}$, a square of $-i$ has the form $\;\mathrm e^{i\theta},\;$ where $$2i\theta\equiv-\frac{i\pi}2\mod 2i\pi\iff \theta\equiv -\frac\pi4\mod\pi\iff\theta\equiv\begin{cases}-\frac\pi4 \\[1ex] \frac{3\pi}4\end{cases}\bmod 2\pi $$ Hence the square roots of $-i$ are $$\Bigl\{\:\mathrm e^{-\tfrac{i\pi}4},\mathrm e^{\tfrac{3i\pi}4}\:\Bigr\}.$$

Note the latter square root could be (but shouldn't) be written as $\;\bigl(\mathrm e^{i\pi}\bigr)^{\tfrac34} $, i.e. $\;\bigl(-1\bigr)^{\tfrac34} $.

Answer 4

The roots of $-1$.

Solve:

$z^2+1=0.$

$z^2=-1=e^{i(3π/2+2πk)}$ for

$k=0,1, 2, 3,.....$.

$2$ distinct roots (mod $(2π)$) for $k=0,1.$

$z_1= e^{i(3π/4)}$, $k=0$.

$z_2=e^{i(3π/4+π)}= e^{-iπ/4},$ $k=1$.