Hi folks,

I am trying to find the square roots of the complex number $3 + 4i$

Suppose the square root is a + ib where a and b $\in \mathcal{\ R}$

$3 + 4i = (a + ib)^2$

$3 + 4i = a^2 - b^2 + 2abi$

so,

$3 = a^2 - b^2$ ......... (1)

and

$4 = 2ab$ ................(2)

sub (2) in (1)

$3 = (\dfrac{2}{b})^2 - b^2$

$3 = \dfrac{4}{b^2} - b^2$

$3b^2 = 4 - b^4$

$b^4 + 3b^2 - 4 = 0$

let $y = b^2$

$y^2 + 3y - 4 = 0$

$y = 1$ or $y = -4$

i.e.

$b^2 = 1 $ or $b^2 = -4$

$b = \pm 1$ or $b = \sqrt{4}\sqrt{-1} = \pm 2i$

so I have $b = +1 + 2i$ or $b = -1 + 2i$ or $b = +1 - 2i$ or $b = -1 - 2i$

from equation (2) for each of the 4 values of b, I get a corresponding value of a

$a = \dfrac{2}{b}$

$a = \dfrac{2}{1 + 2i}$

$a = \dfrac{2}{1 + 2i} . \dfrac{1 - 2i}{1 - 2i}$

$a = \dfrac{2 - 4i}{1 - 4i^2}$

$a = \dfrac{2}{5} - \dfrac{4}{5}i$

so

$a + ib = (1 + 2i) + i(\dfrac{2}{5} - \dfrac{4}{5}i)$

$a + ib = 1 + 2i + \dfrac{2}{5}i - \dfrac{4}{5}$

$a + ib = \dfrac{1}{5} + \dfrac{12}{5}i $

this is the first of 4 square roots

unfortunately these do not match the answers which are $\pm (2 + i)$

where am I going wrong?