# Writing a Complex Number in Trigonometric Form

• Jun 9th 2007, 09:22 AM
Writing a Complex Number in Trigonometric Form
I feel ridiculous asking this question because I know how simple it is, yet I can't seem to get the answer the book is getting.

Write this in a trigonometric form: -7 + 4i
• Jun 9th 2007, 09:35 AM
CaptainBlack
Quote:

I feel ridiculous asking this question because I know how simple it is, yet I can't seem to get the answer the book is getting.

Write this in a trigonometric form: -7 + 4i

Write:

$
-7+4i=A \left( \cos(\theta)+i \sin(\theta) \right)
$

Then $A=\sqrt{7^2+4^2}$, and $\theta=\arctan(-7/4)$

RonL
• Jun 9th 2007, 09:45 AM
I knew how to do that, but somehow I can't get the answer:

square root 65(cos2.62 + isin2.62)

Edit: Ack. Just saw what I did wrong.

Thanks.
• Apr 13th 2009, 04:58 PM
Struggle_Through
Can someone explain how this the answer becomes 2.62. I have this same problem and really don't understand how they get it. It would be the arctan of 4/7 but how do they get 2.62
Thanks.
• Apr 13th 2009, 05:39 PM
Jhevon
Quote:

Originally Posted by Struggle_Through
Can someone explain how this the answer becomes 2.62. I have this same problem and really don't understand how they get it. It would be the arctan of 4/7 but how do they get 2.62
Thanks.

it is just a matter of knowing that this is the reference angle of the one you want. you want the positive angle measured from the positive real-axis. so you need the angle which arctan(4/7) is a reference angle for, which is $\pi - \arctan \frac 47 \approx 2.62$
• Apr 13th 2009, 06:02 PM
Struggle_Through
Quote:

Originally Posted by Jhevon
it is just a matter of knowing that this is the reference angle of the one you want. you want the positive angle measured from the positive real-axis. so you need the angle which arctan(4/7) is a reference angle for, which is $\pi - \arctan \frac 47 \approx 2.62$

Alright, I see how they are getting the answer. I really appreciate the help.
Thanks.