Grade 11Algebra

find the square root of the complex number:
(3+4i)/(3-4i)

apurva

7 Years agoGrade 11

1 Answer

Arun

7 Years ago

Dear student

I am thinking that you are asking square root of any one.

Set the square root of this equal to a + bi. Square both sides. Simplify. Then compare the real and imaginary parts separately and solve for a and b.

√(3 + 4i) = a + bi
3 + 4i = (a+bi)^2
3 + 4i = a^2 + 2abi + (bi)^2
3 + 4i = a^2 + 2abi - b^2
So:
3 = a^2 - b^2, and 4i = 2abi, which means
3 = (a^2 - b^2) and 2 = ab

Replace a = 2/b in the first equation to get
3 = (2/b)^2 - b^2
3 = (4 / b^2) - b^2
3b^2 = 4 - b^4
b^4 + 3b^2 - 4 = 0
(b^2 + 4)(b^2 - 1) = 0

So b^2 = -4, 1. We're only looking for real values of b, so b = ±1. This means a = 2/b = ±2. So 2 + i and -2 - i are two solutions.