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.

Because of $i^2=-1$ , there are in fact two solutions of $z^2=-4$, namely $-2i$ and $2i$. I am not sure whether there is a convention which should be "the" square root in this case. — Peter, Feb 12 '20 at 12:26
IMV it is wrong to state things like $\sqrt4=2$ and also $\sqrt4=-2$. Even if you are informally speaking of $\sqrt4$. It implicitly suggests that $2=\sqrt4=-2$. — drhab, Feb 12 '20 at 12:31
My answer resolves that: $(-i)^2=(-1)^2(i)^2=i^2$. But $i^2\neq - i^2$ because $i^2\neq 0$. Be careful to be accurate with notation. — Rhys Hughes, Feb 12 '20 at 14:03
@MichaelMunta Be careful with the position of minus signs in relation to terms being squared. $-i^2$ reads to most people as $-(i^2)$ rather than $(-i)^2$. — JMoravitz, Feb 12 '20 at 14:20
For complex number $z^2= 4 $ there are 4 roots. Two real, two complex. — Narasimham, Feb 14 '20 at 19:38
@Narasimham That equation has 2 roots, not 4 (as it is a quadratic). — Radial Arm Saw, Jul 06 '20 at 21:18

Rhys Hughes · Answer 1 · 2020-02-12T14:08:52.923

1

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.

edited Feb 12 '20 at 14:08

answered Feb 12 '20 at 12:36

Rhys Hughes

12,842

score 0 · Answer 2 · answered Feb 12 '20 at 12:29

0

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 $a+bi$.

answered Feb 12 '20 at 12:29

José Carlos Santos

427,504

Where did you derive the expression in your answer from? – Michael Munta Feb 13 '20 at 10:27
SInce $(x+yi)^2=x^2-y^2+2xyi$, I solved the system$$\left{\begin{array}{l}x^2-y^2=a\2xy=b.\end{array}\right.$$ – José Carlos Santos Feb 13 '20 at 12:15

score 0 · Answer 3 · answered Feb 12 '20 at 12:34

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 $\sqrt{zw}=\sqrt{z}\sqrt{w}$.

score 0 · Answer 4 · answered Feb 12 '20 at 12:38

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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 $x= \pm 2i$.

answered Feb 12 '20 at 12:38

user289143

4,440

Negative imaginary unit

4 Answers4