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Math Help - Complex numbers in this form: re^(a+bi) to polar form (r,thea)

  1. #1
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    Complex numbers, what are the values of r and theta?

    Write in the polar form ,.
    a)

    I know that I should use this: (e^a)(cosb+isinb)
    Which is: (e^4)(cos1+isin1)
    But then what? How do I get 7e^(a+i) into a+bi form?

    *To be more precise what are r=? and theta=? (without decimals)


    b)

    For question b r=

    But does Theta equal? I got but it's wrong.

    If there are other posts that could help please feel free to send me links.
    Last edited by SillyMath; July 27th 2012 at 02:59 PM.
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  2. #2
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    Re: Complex numbers in this form: re^(a+bi) to polar form (r,thea)

    e^{a+ bi}= e^a e^{bi}= e^a(cos(b+ i sin b) which is already a "polar form" with r= e^a and \theta= b.
    7e^{a+i}= 7e^a(e^{i})= 7e^a(cos(1)+ i sin(1)

    \frac{2i}{7e^{4+i}}= \frac{2}{7}(i)e^{-4- i}= \frac{2}{7}(e^{-\pi/4})(e^{-4}(e^{-i})= \frac{2}{7}e^{-4}e^{-(1+ \pi/4)i}
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    Re: Complex numbers in this form: re^(a+bi) to polar form (r,thea)

    Thank you but this looks like Chinese to me. I basically need the value of r and theta that's all.
    If you're saying that r= and theta=
    r is correct but theta is wrong
    Last edited by SillyMath; July 27th 2012 at 06:16 PM.
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  4. #4
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    Re: Complex numbers, what are the values of r and theta?

    Quote Originally Posted by SillyMath View Post
    b)

    For question b r=

    But does Theta equal? I got but it's wrong.
    In standard form, this complex number should be
    \frac{-\sqrt{3}-1}{2} + \frac{\sqrt{3}-1}{2}i

    Which means that theta should be
    \tan^{-1} \left( \frac{\sqrt{3} - 1}{-\sqrt{3} - 1}\right) + \pi

    Note: I normally use the
    \tan \theta = \frac{b}{a}
    formula to find theta instead of inverse sine.
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  5. #5
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    Re: Complex numbers, what are the values of r and theta?

    Thank you to all, this really helped me
    oh and the answer for a) theta=(pi/2)-1
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