Posted by **jon smith** on Monday, January 3, 2011 at 8:41am.

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

b) Polar form

Any help with this question would appreciated

- maths/ physics -
**Reiny**, Monday, January 3, 2011 at 9:11am
let the angle be Ų

then r = √(3^2+4^2) = 5

tanŲ = 4/3, Ų = .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)

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