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#

1 Answer
Cesareo R.
Mar 18, 2018

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)#

Related questions
Impact of this question
2198 views around the world
You can reuse this answer
Creative Commons License