Complex solutions of the equation $x^{\frac{1}{2}}+1=0$ 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 
 A: Hint:$\text{}\text{}\text{}\text{}\text{}\text{}$
$$e^{i \pi}+1=0$$
A: 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).
A: 
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.
A: 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? 
