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

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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.

$\blacksquare$


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