# Square root of complex

#### TheSpArK505

Joined Sep 25, 2013
111
Hello guys,

I wnat to take the square root of a complex over complex, i have set my calculator(casio fx-991ES plus)
to complex but when i try to calculate it gives me math error!!!! is there any set ups or its impossible to take square root of complex

#### t_n_k

Joined Mar 6, 2009
5,448
It's not that difficult a computation - if that's your main concern.
Starting with the original complex ratio value as a single value polar form, the square root is readily determined - as you presumably know. If your calculator can do the complex division and display the result in polar form the last step is relatively easy.
The polar form of the square root
$\sqrt {A \angle {\pm \theta}} =\sqrt {A} \angle\pm \frac{\theta}{2}$

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#### TheSpArK505

Joined Sep 25, 2013
111
Thanks it's easier now

#### MrAl

Joined Jun 17, 2014
6,450
Hi,

It might also help sometimes to remember:

sqrt(A/B)=sqrt(A)/sqrt(B)

#### russ_hensel

Joined Jan 11, 2009
825
It's not that difficult a computation - if that's your main concern.
Starting with the original complex ratio value as a single value polar form, the square root is readily determined - as you presumably know. If your calculator can do the complex division and display the result in polar form the last step is relatively easy.
The polar form of the square root
$\sqrt {A \angle {\pm \theta}} =\sqrt {A} \angle\pm \frac{\theta}{2}$

In the square root you can add 2pi to theta ( one or more times ) before dividing the angle to get other square roots ( here just 2 in total)> More for cube roots, 4thf roots, etc.

#### amilton542

Joined Nov 13, 2010
496
Yeah it's called De Moivre's theorem.