Negative imaginary unit

Question

Suppose we are talking about non-principal square roots and we know that $\sqrt{4} = 2$, and also $\sqrt{4} = -2$. I am wondering how this works in the imaginary domain? Does the square root of a negative number also have two solutions? $\sqrt{-4} = 2i$ and $\sqrt{-4} = -2i$. Is this possible?

I have seen this here in the first answer, but it does not make sense to me.

Answer 1

Yes, that's how it happens (except for $0$, which has only one square root: $0$). If $a,b\in\mathbb R$ and $b\neq 0$, then both numbers$$\pm\left(\sqrt{\frac{a+\sqrt{a^2+b^2}}2}+\frac{\lvert b\rvert}b\sqrt{\frac{-a+\sqrt{a^2+b^2}}2}\,i\right)$$are square roots pf $a+bi$.

Answer 2

To expand on @drhab's comment, $\sqrt{x}$ is defined for $x\ge0$ as the unique $y\ge0$ with $y^2=x$, but for general $x\in\Bbb C\setminus[0,\,\infty)$ we don't have a natural definition of $\sqrt{x}$. (This will give you a flavour as to why.) In particular, we can't consistently choose one of two solutions to $y^2=x$ to preserve identities such as $\sqrt{zw}=\sqrt{z}\sqrt{w}$.

Answer 3

Yes.

We denote $$i=e^{i(2k+\frac 12)\pi},k\in\Bbb Z$$ But $$-i =\bar{i}=e^{i(2k-\frac 12)\pi}, k\in\Bbb Z$$

When plotted on the Argand Diagram of a unit circle, they are placed on opposite ends.

Observe that while $\forall t\in \Bbb Z; i^{2t}=(-i)^{2t}$ holds, we also have that $i^{2t+1}=-(-i)^{2t+1}$ so $-i$ is needed for that.

Answer 4

Using the factorization of $x^2+y^2=(x+iy)(x-iy)$ we can see that $x^2+4$ factors as $(x+2i)(x-2i)$ and thus the roots of $x^2=-4$ are $x= \pm 2i$.