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 $.