Find w=a+bi=√(z), where a and b are real numbers √(2i)

Question

Find w=a+bi=
`√(z)`, where a and b are real numbers


`√(2i)`

Answer 1

I assume you're talking about finding the square roots of 2i.

We have

• modulus: |2i| = √(2²) = 2

• argument: arg(2i) = π/2

so that the polar form of 2i is

2i = 2 exp(i π/2)

By DeMoivre's theorem,

√(2i) = √2 exp(i (π/2 + 2πk)/2)

where k = 0 or k = 1.

For k = 0,

√(2i) = √2 exp(i (π/2)/2)

… = √2 exp(i π/4)

… = √2 (cos(π/4) + i sin(π/4))

… = √2 (1/√2 + 1/√2 i)

… = 1 + i

For k = 1,

√(2i) = √2 exp(i (π/2 + 2π)/2)

… = √2 exp(i 5π/4)

… = √2 (cos(5π/4) + i sin(5π/4))

… = √2 (-1/√2 - 1/√2 i)

… = -1 - i