If $\displaystyle z=x+yi$ determine the values of $\displaystyle x$ and $\displaystyle y$ such that $\displaystyle z=\sqrt{3+4i}$
Another approach:
$\displaystyle z^2 = 3 + 4i$
$\displaystyle \Rightarrow (x + iy)^2 = 3 + 4i$
$\displaystyle \Rightarrow (x^2 - y^2) + 2xy i = 3 + 4i$.
Therefore:
$\displaystyle x^2 - y^2 = 3$ .... (1)
$\displaystyle 2xy = 4 \Rightarrow xy = 2 \Rightarrow y = \frac{2}{x}$ .... (2)
Substitute (2) into (1): $\displaystyle x^2 - \frac{4}{x^2} = 3 \Rightarrow x^4 - 3x^2 - 4 = 0 \Rightarrow (x^2 - 4)(x^2 + 1) = 0$.
But since x is real, $\displaystyle x = \pm 2$.
x = 2: y = 1.
x = -2: y = -1.