Square root of a negative number

Question

Correct me if I am wrong, $\sqrt{-4}=2i$.

But how do you explain it to a student?

We know $\sqrt{-1}=i$, but one cannot say $\sqrt{-4}=\sqrt{-1}\sqrt{4}=2i$ as the laws of indices can only be applied to real numbers.

Any advice would help.

Answer 1

Edit:

$i^2 = -1$

Therefore,

$4i^2 = -4$

and by taking square root on both sides, we have $2i = \sqrt{-4}$

Answer 2

The square root of a negative number is usually given by $$\begin{align} (-4)^{1/2} &=e^{\ln{(-4)}/2}\\ &=e^{\ln{(4e^{i\pi})}/2}\\ &=e^{\ln{(e^{i\pi+2\ln{(2)}})}/2}\\ &=e^{\ln{(2)}+i\pi/2}\\ &=2e^{i\pi/2}\\ &=2\left(\cos{\left(\frac\pi2\right)}+i\sin{\left(\frac\pi2\right)}\right)\\ &=2i\\ \end{align}$$ where the principal branch of the natural logarithm is taken instead of the 'positive' square root.

Answer 3

I would approach the topic dealing first with the $2 \pi$ periodicity of the angle.

Then I would pass to the convention used to split the $2 \pi$ angle: the "standard" is $-\pi < \alpha \le \pi$ thus the "principal" "standard" solution is $2i$.
But any other split can be adopted.