Prove that the solution of $z=\dfrac{\sqrt{-21+20i}-\sqrt{-21-20i}}{\sqrt{-21+20i}+\sqrt{-21-20i}}$ are $\dfrac{5i}{2}$ and $\dfrac{-2i}{5}$
I know that If I write $\sqrt{-21+20i}=\pm(2+5i)$ and $\sqrt{-21-20i}=\pm(2-5i)$, then I will get the required solutions. But, when I do the following :
\begin{align} z&=\frac{\sqrt{-21+20i}-\sqrt{-21-20i}}{\sqrt{-21+20i}+\sqrt{-21-20i}}\\ &=\frac{\sqrt{-21+20i}-\sqrt{-21-20i}}{\sqrt{-21+20i}+\sqrt{-21-20i}}.\frac{\sqrt{-21+20i}-\sqrt{-21-20i}}{\sqrt{-21+20i}-\sqrt{-21-20i}}\\ &=\frac{(\sqrt{-21+20i}-\sqrt{-21-20i})^2}{(\sqrt{-21+20i})^2-(\sqrt{-21-20i})^2}\\ &=\frac{-21+20i-21-20i-2\sqrt{21^2+20^2=841}}{-21+20i+21+20i}\\ &=\frac{-21-\sqrt{841}}{20i}=\frac{-21-29}{20i}=\frac{5i}{2} \end{align}
Why am I missing the solution $\dfrac{-2i}{5}$ in my attempt ?
If I were to take $\sqrt{841}=-29$ I think I'll get the other solution but, I always considered $\sqrt{x}$ as the +ve square root and $-\sqrt{x}$ the -ve square root of the number $x$.
Remember that $841$ has two square roots.