## Complex Roots

Suppose $\displaystyle z = a+bi$ and $\displaystyle w = u+iv$ and $\displaystyle a= \left(\frac{|w|+u}{2} \right)^{1/2}$ and $\displaystyle b = \left( \frac{|w|-u}{2} \right)^{1/2}$. Prove that $\displaystyle z^2 = w$ if $\displaystyle v \geq 0$ and that $\displaystyle (\bar{z})^{2} =w$ if $\displaystyle v \leq 0$. So every complex number has two complex roots with one exception.

So $\displaystyle z^2 = (a,b)(a,b) = (a^2-b^2, 2ab)$. Then just plug in values of $\displaystyle a$ and $\displaystyle b$ to verify the equality? Do the same for $\displaystyle (\bar{z})^{2}$? The exception is $\displaystyle 0$.