Writing a Complex Number in Trigonometric Form

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June 9th 2007, 09:22 AM
alreadyinuse

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
June 9th 2007, 09:35 AM
CaptainBlack

Quote:

Originally Posted by alreadyinuse

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:

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

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

RonL
June 9th 2007, 09:45 AM
alreadyinuse

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.
April 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.
April 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 $\pi - \arctan \frac 47 \approx 2.62$
April 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.