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Math Help - Writing a Complex Number in Trigonometric Form

  1. #1
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    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
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  2. #2
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    Quote Originally Posted by alreadyinuse View Post
    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
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  3. #3
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    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.
    Last edited by mr fantastic; April 13th 2009 at 08:55 PM. Reason: Merged posts
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  4. #4
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    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.
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Struggle_Through View Post
    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
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  6. #6
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    Quote Originally Posted by Jhevon View Post
    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.
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