$\sqrt{x}=-1$. How can I solve it?

Question

I am so curious about this equation: $\sqrt{x}=-1$

1. Does the $x$ where $x\in \mathbb{C}$ exist?
2. How can I solve it?

Answer 1

The equation $\sqrt{x}=-1$ implies that $x^2=1$, so necessarily $x=1$ or $x=-1$. Certainly $x=-1$ is not a solution, because $\sqrt{-1}$ is not a real number, hence different from $-1$. On the other hand, $x=1$ is only a solution if we choose the negative square root, i.e., $\sqrt{1}=-1$. Whether or not this is a valid solution, depends on the definition of $\sqrt{}$. For complex numbers, certainly it need not be the principal square root. And the OP asks for complex solutions.

Answer 2

This shows you how to work with this problems for a general $n$-th root Let $x = \rho (\cos \theta + i \sin \theta)$.

Then $$\sqrt x = \sqrt \rho (\cos \left(\frac{\theta + 2k\pi}{2}\right) + i \sin(\cos \left(\frac{\theta + 2k\pi}{2}\right)$$

for $k = 0,1$. That is $\sqrt x$ is not a single number but rather a set of two numbers, one for each value of $k$. Now, since you want $\sqrt x = -1 = 1 (\cos (\pi) + i \sin \pi)$, you need $\rho = 1$. Then you get either $\frac \theta 2 = \pi \implies \theta = 2\pi$ or $\frac{\theta}2 + \pi = \pi \implies \theta = 0$. Therefore the two solutions are

$$x = 1 (\cos 2\pi + i \sin 2\pi) = 1$$ $$x = 1 (\cos 0 + i \sin 0) = 1$$

So there is the unique solution in $\mathbb C$, that is $\sqrt 1 = -1$ Or better: $$ -1 \in \sqrt 1 = \{-1, 1\}$$ since the latter is a set

Answer 3

Let us assume the complex solution

$$\sqrt{x+iy}=-1.$$

By squaring,

$$x+iy=1=1\cdot e^{i0}$$

is the only possibility.

Then the principal value of the square root is

$$\sqrt{x+iy}=\sqrt1\cdot e^{i0/2}, $$ which doesn't fit.

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Had the principal branch be defined as

$$\sqrt z=\sqrt\rho e^{i\theta/2+i\pi},$$ then $x+iy=1$ would have been a solution.