Square Root of Complex Number in Cartesian Form/Examples/-15-8i/Proof 1

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Example of Square Root of Complex Number in Cartesian Form

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


Proof

We have that:

$-15 - 8 i = 17 \map \cis {\theta + 2 k \pi}$

where:

$\cos \theta = -\dfrac {15} {17}$
$\sin \theta = -\dfrac 8 {17}$


Then the square roots of $-15 - 8 i$ are:

\(\text {(1)}: \quad\) \(\ds z_1\) \(=\) \(\ds \sqrt {17} \cis \dfrac \theta 2\)
\(\text {(2)}: \quad\) \(\ds z_2\) \(=\) \(\ds \sqrt {17} \, \map \cis {\dfrac \theta 2 + \pi}\)
\(\ds \) \(=\) \(\ds -\sqrt {17} \cis \dfrac \theta 2\)


We have that:

\(\ds \cos \dfrac \theta 2\) \(=\) \(\ds \pm \sqrt {\dfrac {1 + \cos \theta} 2}\)
\(\ds \) \(=\) \(\ds \pm \sqrt {\dfrac {1 - 15 / 17} 2}\)
\(\ds \) \(=\) \(\ds \pm \dfrac 1 {\sqrt {17} }\)


and:

\(\ds \sin \dfrac \theta 2\) \(=\) \(\ds \pm \sqrt {\dfrac {1 - \cos \theta} 2}\)
\(\ds \) \(=\) \(\ds \pm \sqrt {\dfrac {1 + 15 / 17} 2}\)
\(\ds \) \(=\) \(\ds \pm \dfrac 4 {\sqrt {17} }\)


As $\theta$ is in the third quadrant, $\dfrac \theta 2$ is in the second quadrant.

Hence:

\(\ds \cos \dfrac \theta 2\) \(=\) \(\ds -\dfrac 1 {\sqrt {17} }\)
\(\ds \sin \dfrac \theta 2\) \(=\) \(\ds \dfrac 4 {\sqrt {17} }\)


Hence from $(1)$ and $(2)$:

\(\ds z_1\) \(=\) \(\ds -1 + 4 i\)
\(\ds z_2\) \(=\) \(\ds 1 - 4 i\)

$\blacksquare$


Sources

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