Square Root of Complex Number in Cartesian Form/Examples/-15-8i/Proof 2

Example of Square Root of Complex Number in Cartesian Form

$\sqrt {-15 - 8 i} = \pm \paren {1 - 4 i}$

Proof

Let the required square roots be $p = i q$ for real $p$ and $q$.

Then:

 $\ds \paren {p + i q}^2$ $=$ $\ds p^2 - q^2 + 2 p q i$ $\ds$ $=$ $\ds -15 + 8 i$ $\text {(1)}: \quad$ $\ds \leadsto \ \$ $\ds p^2 - q^2$ $=$ $\ds -15$ $\text {(2)}: \quad$ $\ds p q$ $=$ $\ds 4$ $\ds \leadsto \ \$ $\ds p^2 - \dfrac {16} p^2$ $=$ $\ds -15$ substituting $q = -\dfrac 4 p$ from $(2)$ into $(1)$ $\ds \leadsto \ \$ $\ds p^4 + 15 p^2 - 16$ $=$ $\ds 0$ $\ds \leadsto \ \$ $\ds \paren {p^2 + 16} \paren {p^2 - 1}$ $=$ $\ds 0$ $\ds \leadsto \ \$ $\ds p$ $=$ $\ds \pm 1$ as $p$ is real

The result follows by substituting $+1$ and $-1$ into $2$, in turn.

$\blacksquare$