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I was trying to find the square root of a complex number $w=u+iv$, I assumed $z^2=(x+iy)^2=w$ Now solving this equation I got two values of $x$ and two values of $y$ in terms of $u$ and $v$. So, I get a possible of 4 solutions. But fundamental theorem of algebra says it has exactly two roots. What are these two roots?

My work $x=\pm \sqrt {\sqrt {u^2+v^2}+u}$ and $y=\pm \sqrt {\sqrt {u^2+v^2}-u}$

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See here. – JavaMan Feb 24 '13 at 17:44
Do you know about the trigonetric/exponential form of a complex number? – Stefan Feb 24 '13 at 17:44
@Waqas Ali Azhar You've asked 20 questions so far and accepted only 1.Please consider going through your questions and accepting answers to your questions if any of the answers to each question satisfies you. – Git Gud Feb 24 '13 at 17:45
It is unlikely that anyone will be able to tell you what you did wrong unless you show us what you did. – MJD Feb 24 '13 at 17:46
No, $x$ and $y$ do not necessarily have the same sign. To obtain which signs are needed, one can without loss of generality assume $x\ge0$ and find $y$'s sign, then tack on a $\pm$ in the end. Or one can use geometric considerations. Oftentimes equations that require multiple squarings to solve will have ghost solutions, which occur since $a=b\implies a^2=b^2$ but $a^2=b^2\not\Rightarrow a=b$; that is, squaring an equation will result in a logically weaker claim about the relevant variables, hence in general can have more solutions. – anon Feb 24 '13 at 18:56

To choose two needed solutions, you have to check whether $-2xy=v$.

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The following way is elementary but it works. Let you want to find $\sqrt{x+yi}$. Set $$\sqrt{x+yi}=a+bi$$ Then $$x+iy=(a+ib)^2=a^2-b^2-2iab$$ so $$x=a^2-b^2,~~~y=-2ab$$ by solving the equations simultaneously, we have: $$a=\sqrt{\frac{\sqrt{x^2+y^2}+x}{2}},~~b=\sqrt{\frac{\sqrt{x^2+y^2}-x}{2}}$$. Note that $\sqrt{x+yi}=\pm(a+bi)$

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This seems to skip over the crucial point for this question, which is the signs in front of the $\sqrt{x^2+y^2}$ terms and what constrains them to be the same, or where the illusory solutions come from if there is no such constraint. – MJD Feb 24 '13 at 17:56
@MJD: I edited the answer. I don't exactly know what the OP are looking for then. I added what I knew. If it has defects note me to remove that. Thanks. – S. Snape Feb 24 '13 at 19:26

If you work with the system

$x^2-y^2 = u \\ 2xy = v$

you obtain, provided $x \neq 0$, that $y = v/2x$ and $4x^4-4ux^2-v^2 = 0$

From $4x^4-4u^2x^2-v^2 = 0$ you obtain two complex and two real values for $x$. Each real value for $x$ corresponds to only one real value for $y$ ($y = v/2x$), and those give the two square roots of $u+iv$.

This means that in your expressions, you have to pick the same signs if $v>0$, and pick opposite signs if $v<0$.

(and also your expressions seem to lack a division by $2$ somewhere).

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