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.
$\blacksquare$
Sources
- 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Solved Problems: Roots of Complex Numbers: $30$