# Square root of complex

Discussion in 'Math' started by TheSpArK505, Dec 30, 2014.

1. ### TheSpArK505 Thread Starter Member

Sep 25, 2013
89
0
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

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
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
$\sqrt {A \angle {\pm \theta}} =\sqrt {A} \angle\pm \frac{\theta}{2}$

Last edited: Dec 31, 2014
3. ### TheSpArK505 Thread Starter Member

Sep 25, 2013
89
0
Thanks it's easier now

4. ### MrAl Well-Known Member

Jun 17, 2014
2,200
432
Hi,

It might also help sometimes to remember:

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

5. ### russ_hensel Well-Known Member

Jan 11, 2009
818
47

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.

6. ### amilton542 Active Member

Nov 13, 2010
491
64
Yeah it's called De Moivre's theorem.

Mar 2, 2015
69
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8. ### WBahn Moderator

Mar 31, 2012
17,446
4,699
Because some people want to actually learn the concepts behind the math instead of just letting a tool written by someone else do their thinking for them.