Question

Use De Moivre's Theroem to find the square root of -5+12i.

Answer 1

You want to rewrite that in the form \[r(\cos x+i\sin x)\]

Answer 2

Yes. But I could not find the arg(-5+12i). Could you please show me how to?

Answer 3

\[r=\sqrt{(-5)^{2}+12^{2}}\]

Answer 4

Ok. So r=13, what next?

Answer 5

If we factor 13 out then we have \[13(\frac{ -5 }{ 13 }+\frac{ 12i }{13 })\]and so we want to find at what angle cosine equals -5/13 and sine equals 12/13.

Answer 6

Get it. How do we find that? :O

Answer 7

A calculator.
Enter\[\cos^{-1} (-5/13)\]

Answer 8

and\[\sin^{-1} (12/13)\]

Answer 9

112.62 and 67.38

Answer 10

Both should have the same sign, so we need to adjust the angles accordingly

Answer 11

The book gives an answer in this form. That was why I got confused. I don't get what this 0.98^c stands for? :O \[\pm \sqrt{5}\left\{ \cos(0.98^{c})+isin(0.98^{c}) \right\}\]

Answer 12

Sorry. That's \[\sqrt{13}\]

Answer 13

That does not make sense. I wonder if there is some sort of typing error.

Answer 14

The degrees should be approximately 112.62 for both sin and cos.

Answer 15

No. There are other questions like this and answers in this format. I'm just stuck. :-(

Answer 16

Before you divide by 2, that is.

Answer 17

Do they all have c's instead of degree signs?

Answer 18

Np. Some answers are in the form \[\pm \sqrt{12}(\cos \pi/12+isin \pi/12)\]

Answer 19

Okay. Your book is working with radians instead of degrees.

Answer 20

You should change the mode of your calculator to do radians.

Answer 21

Okay.

Answer 22

I got1.176 for sin^-1 (12/13) and 1.965 for cos^-1(-5/13).

Answer 23

That makes sense, but only one of those is correct.