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Math Help - Complex number to polarform

  1. #1
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    Oct 2008
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    Complex number to polarform

    Hello all,

    I was hoping for some help on turning complex number to polar form. I'm working on a nonhomogenous difference equation, and while solving I got the roots \frac{1.5\pm\sqrt{-1.75}}{2}.

    Ok, so far so good. The modulus, \rho=\sqrt{(\frac{3}{4})^2+(\frac{\sqrt{1,75}}{2})  ^2}, which gives me \rho=1. Then the argument, \cos \theta = \frac{3/4}{1}=\frac{3}{4}\mbox{, } \sin \theta = \frac{\sqrt{1,75}/2}{1}=\frac{\sqrt{1,75}}{2}. This gives me \cos^{-1} \theta =\sin^{-1} \theta = 0.7227342478. Is this correct?

    The root I got, \frac{1.5\pm\sqrt{-1.75}}{2}, when I want to find \sin \theta = \frac{b}{\rho}, what value of b do I use (because of the \pm)? The negative or positive value of \frac{\sqrt{-1.75}}{2}?

    Is there any way to make the argument a "pretty" \pi-fraction (if it exists)?

    Thank you for any help!
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  2. #2
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    st.catherine
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    complex number

    hi, dont think that make any sense something is not right.whats the original question.
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  3. #3
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    st.catherine
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    actually it does make sense sorry i was only glancing afeter i start thinking about it.anyways.let z=3/4 i√7/8 (rember that i=√-1 and the √-7/8 is the same as √-1 times √7 ok) polar form =r[cos(θ)+isin(θ)]

    [z]=r=(a^2+b^2)=85/64 (note that r is indeed the modulus of z)

    arg(z)=tan(θ)=6/7

    θ=tan^-1(6/7)
    θ=40.60
    therefore
    arg(z)=40.60 (you can convert to pi radians if you want ok)

    now we have everything to represent in polar form so now substitute.

    polar form of Z=85/64[cos(40.60) + isin(40.60)]. (the squared root of any -number does not exist thats its imaginary thats why i is used to replace √-n=i ok.) good luck hope this helps and talk to me if you need anything else god be with you.
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