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

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