# How to prove this? Let z = a + ib be a complex number. Show that a square root of z is given by the expression w=sqrt((|z|+a)/2)+iσ*sqrt((|z|-a)/2) where σ = 1 if b ≥ 0 and σ = −1 if b < 0. Do this by verifying that w^2=z ?

## w=sqrt((|z|+a)/2)+iσ*sqrt((|z|-a)/2) ${w}^{2} = z$

Mar 18, 2018

See below.

#### Explanation:

Calling

$\sqrt{z} = u + i v$ we have

${\left(u + i v\right)}^{2} = z = a + i b$ or

$\left\{\begin{matrix}{u}^{2} - {v}^{2} = a \\ 2 u v = b\end{matrix}\right.$

now solving for $u , v$

{(u = sqrt[(sqrt[ a^2 + b^2]+a)/2]), (v =b/(sqrt sqrt[sqrt[a^2 + b^2]+a])):}

but

$\frac{b}{\sqrt{2} \sqrt{\sqrt{{a}^{2} + {b}^{2}} + a}} = \frac{b \sqrt{\sqrt{{a}^{2} + {b}^{2}} - a}}{\sqrt{2} \sqrt{{a}^{2} + {b}^{2} - {a}^{2}}} = \frac{b}{\sqrt{{b}^{2}}} \sqrt{\frac{\sqrt{{a}^{2} + {b}^{2}} - a}{2}} = \sigma \left(b\right) \sqrt{\frac{\sqrt{{a}^{2} + {b}^{2}} - a}{2}}$

where

$\sigma \left(b\right) = \frac{b}{\left\mid b \right\mid}$

Finally, considering $\left\mid z \right\mid = \sqrt{{a}^{2} + {b}^{2}}$ we have

w=sqrt((|z|+a)/2)+iσ(b)sqrt((|z|-a)/2)