Determine the two square roots of 3+j4 in both a)Cartesian form

b) Polar form

Any help with this question would appreciated

1 answer

let the angle be Ø
then r = √(3^2+4^2) = 5
tanØ = 4/3, Ø = .9273

3 + 4j = 5(cos .9273 + jsin .9273)
by De Moivre's theorem

(3 + 4j)^(1/2) = √5(cos ((1/2).9273) + jsin ((1/2).9273))

= √5( cos .4636 + jsin .4636)

or 2 + j if expanded.

check:
if 2+j is the square root of 3+4j, then
(2+j)^2 should equal 3+4j

Left side = (2+j)^2
= 4 + 4j + j^2
= 4 + 4j - 1
= 3 + 4j

so √(3+4j) = 2+j or √5(cos .4636 + jsin .4636)