Square Root of Complex Number in Cartesian Form/Examples/-15-8i
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Example of Square Root of Complex Number in Cartesian Form
- $\sqrt {-15 - 8 i} = \pm \paren {1 - 4 i}$
Proof 1
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 $(2)$:
| \(\ds z_1\) | \(=\) | \(\ds -1 + 4 i\) | ||||||||||||
| \(\ds z_2\) | \(=\) | \(\ds 1 - 4 i\) |
$\blacksquare$
Proof 2
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.
$\blacksquare$