Complex solutions of the equation $x^{\frac{1}{2}}+1=0$

Question

How can I find complex solutions to the equation $$x^{\frac{1}{2}}+1=0$$ Squaring gives x=1 but it's not the solution

Answer 1

HINT: If you don't want a trigonometric representation, tak $w=a+bi$. Then $x=w^2=-1$, which gives a system of two equations to solve.

(Certainly, it is a genereal approach. If you know, what is $i$, in this case there is (near) nothing to do).

Answer 2

Hint:$\text{}\text{}\text{}\text{}\text{}\text{}$ $$e^{i \pi}+1=0$$

Answer 3

Squaring gives $\,x=1\,$ but it's not the solution

So you proved that $\,x^{1/2} = -1 \implies\, \big(x^{1/2}\big)^2 = (-1)^2 \implies x=1\,$, but $\,x=1\,$ does not verify the original equation (assuming that $\,x^{1/2}\,$ means the principal value of the complex square root). Therefore the original equation has no solutions, and there is nothing to further prove or look for.

Answer 4

What does $x^{\frac 12}$ mean exactly? Is $9^{\frac 12} = 3$ or $\{\pm 3\}$. What is $(-1 + \sqrt 3i)^{\frac 12}$ if $(-1 - \sqrt 3i)^2 = -1+\sqrt 3i$?

Maybe $1^{\frac 12} = -1$ because $(-1)^2 = 1$.

Or maybe given that $(\pm 1)^2 = 1$ and only one of them can be equal to $1^{\frac 12}$ and $1^{\frac 12} = 1$ and there is nothing so that $x^{\frac 12} = -1$. What that be so bad?

If $x^{\frac 12} = -1$ then $x = (x^{\frac 12})^2 = (-1)^2= 1$ so if there is any solution it must be that the solution is $x = 1$. But if there isn't any solution, would that be so bad?

So the question is: Does $x^{\frac 12} = $ the solution set to all square roots of $x$. (In which case if $x^{\frac 12} = k$ then $x = k^2$; that's all there is too it. And $1^{\frac 12} =\{\pm 1\}$ and $x^{\frac 12} = -1\implies x = 1$).

Or Does $x^{\frac 12} = $ the principal root of $x$? (In which case, no number has a negative number as a principal square root and so $x^{\frac 12} = -1$ has no solution and that is all there is to it.)

==== Old and not actually correct answer =====

Note "$x^{\frac 12} = k$" is not equivalent to $k^2 = x$. For example $(-3)^2 = 9$ abut $9^{\frac 12} \ne -3$.

It's a one way street. $x^{\frac 12} = k \implies (x^{\frac 12})^2 = k^2 \implies x = k^2$ but does NOT go the other way.

So if you have $(\pm k)^2 = x$ you can only have one $x^{\frac 12} = k$ or $x^{\frac 12} = -k$ be true. Which one?

In other words $(\pm 3)^2 = 9$ but which of the following statements are true $9^{\frac 12} = 3$ or $9^{\frac 12} = -3$? Well, by convention it is $9^{\frac 12} = 3$ and by convention $9^{\frac 12} \ne -3$.

[By convention; if $z = r*e^{i\theta}; 0 \le \theta < 2\pi; r> 0$ then $z^{\frac 1k} = \sqrt[k]r*e^{i\frac \theta k}$. If $w = z^{\frac 1k}$ then the the angular argument of $w$ must be less then $\frac {2\pi}k$! So $w^{\frac 12} = r*e^{i*\eta}$ where $\pi \le \eta < 2\pi$ is impposible.]

So if $9^{\frac 12}$ is NOT $-3$, what $what^{\frac 12}$ is $-3$.

Well, nothing... If $what^{\frac 12}$ is $-3$ than squaring we'd have $what = 9$ But $9^{\frac 12}= 3 \ne -3$.

Who said there had to be a solution in the first place?

Here's a simple statment: $z \to z^{\frac 12}$ is not surjective. I.E there are some $w \in \mathbb C$ so that $z^{\frac 12} =w$ will not have solutions. That is how the world is. If I put it like that.... well, it's hardly surprising, is it? I mean, if I had asked: What is the largest number that has $42$ as its largest prime factor it shouldn't be surprising that the answer is, no number has $42$ as a prime factor.

So if we ask: what complex number has -1 as its principal square root should we have been surprised the answer is... none?