Question

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`

Answer 1

See below.

Explanation:

Calling

`sqrtz = u + i v` we have

`(u+iv)^2 = z = a+ib` or

`{(u^2-v^2=a),(2uv=b):}`

now solving for `u,v`

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

but

`b/(sqrt[2] sqrt[sqrt[a^2 + b^2]+a]) = (bsqrt[sqrt[a^2 + b^2]-a])/(sqrt[2] sqrt(a^2+b^2-a^2)) = b/sqrt(b^2)sqrt[(sqrt[a^2 + b^2]-a)/2]=sigma(b)sqrt[(sqrt[a^2 + b^2]-a)/2]`

where

`sigma(b) = b/abs(b)`

Finally, considering `absz = sqrt(a^2+b^2)` we have

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