Square Root of Complex Number in Cartesian Form/Examples/3+4i
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Example of Square Root of Complex Number in Cartesian Form
- $\sqrt {3 + 4 i} = \pm \left({2 + i}\right)$
Proof
\(\ds \left({x + i y}\right)^2\) | \(=\) | \(\ds 3 + 4 i\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x^2\) | \(=\) | \(\ds \dfrac {3 + \sqrt {3^2 + 4^2} } 2\) | Square Root of Complex Number in Cartesian Form | ||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {3 + \sqrt {25} } 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {3 + 5} 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 4\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds \pm 2\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds \pm \dfrac 4 {2 \times 2}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \pm 1\) |
As $2 x y = 4$ it follows that the two solutions are:
- $2 + i$
- $-2 - i$
Sources
- 1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 1$. Algebraic Theory of Complex Numbers: Exercise $6 \ \text {(iii)}$