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

Example of Square Root of Complex Number

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

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


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