Square Root of Complex Number in Cartesian Form/Examples/2+2i

Example of Square Root of Complex Number in Cartesian Form

$\sqrt {2 + 2 i} = \pm \left({\sqrt {1 + \sqrt 2} + i \sqrt {\sqrt 2 - 1}}\right)$

Proof

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

\(=\)

\(\ds 2 + 2 i\)

\(\ds \leadsto \ \ \)

\(\ds x^2\)

\(=\)

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

Square Root of Complex Number in Cartesian Form

\(\ds \)

\(=\)

\(\ds 1 + \sqrt 2\)

\(\ds \leadsto \ \ \)

\(\ds x\)

\(=\)

\(\ds \pm \sqrt {1 + \sqrt 2}\)

\(\ds \leadsto \ \ \)

\(\ds y^2\)

\(=\)

\(\ds \dfrac {-2 + \sqrt {2^2 + 2^2} } 2\)

Square Root of Complex Number in Cartesian Form

\(\ds \)

\(=\)

\(\ds -1 + \sqrt 2\)

\(\ds \leadsto \ \ \)

\(\ds y\)

\(=\)

\(\ds \pm \sqrt {\sqrt 2 - 1}\)

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

$\sqrt {1 + \sqrt 2} + i \sqrt {\sqrt 2 - 1}$

$-\sqrt {1 + \sqrt 2} - i \sqrt {\sqrt 2 - 1}$

Sources

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