Square Root of Complex Number in Cartesian Form/Examples/i

Example of Square Root of Complex Number in Cartesian Form

$\sqrt i = \pm \left({\dfrac {\sqrt 2} 2 + \dfrac {\sqrt 2} 2 i}\right)$

Proof

\(\ds \left({x + i y}\right)^2\)

\(=\)

\(\ds i\)

\(\ds \leadsto \ \ \)

\(\ds x^2\)

\(=\)

\(\ds \dfrac {0 + \sqrt {0 + 1^2} } 2\)

Square Root of Complex Number in Cartesian Form

\(\ds \)

\(=\)

\(\ds \dfrac 1 2\)

\(\ds \leadsto \ \ \)

\(\ds x\)

\(=\)

\(\ds \pm \dfrac {\sqrt 2} 2\)

\(\ds \leadsto \ \ \)

\(\ds y\)

\(=\)

\(\ds \pm \dfrac 1 {2 \times \frac {\sqrt 2} 2}\)

\(\ds \)

\(=\)

\(\ds \pm \dfrac {\sqrt 2} 2\)

As $2 x y = 1$ it follows that the two solutions are:

$\dfrac {\sqrt 2} 2 + \dfrac {\sqrt 2} 2 i$

$-\dfrac {\sqrt 2} 2 - \dfrac {\sqrt 2} 2 i$

Sources

• 1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 1$. Algebraic Theory of Complex Numbers: Exercise $6 \ \text {(i)}$