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Math Help - Complex equation

  1. #1
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    Complex equation

    First of all you'll have to excuse any miss-spelling an incorrect expressions, I am a swede and am not to good regarding the math grammar in english..

    Now for the problem at hand:

    Solve the equation and answer in polar form r(cosv + isinv)

    z^4 = -i


    Thanx!
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  2. #2
    MHF Contributor
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    Hello DenOnde

    Welcome to Math Help Forum!

    Your English is excellent. It's certainly much better than my Swedish!
    Quote Originally Posted by DenOnde View Post
    First of all you'll have to excuse any miss-spelling an incorrect expressions, I am a swede and am not to good regarding the math grammar in english..

    Now for the problem at hand:

    Solve the equation and answer in polar form r(cosv + isinv)

    z^4 = -i

    Thanx!
    If we write
    z = r(\cos\theta + i\sin\theta)
    then, using De Moivre's Theorem:
    z^4 = r^4(\cos\theta + i\sin\theta)^4
    =r^4(\cos4\theta+i\sin4\theta)
    We now express - i in polar form. On the Argand diagram it is represented by the point (0,-1), so has modulus 1 and argument -\pi/2. So:
    -i=1\Big(\cos(-\pi/2)+i\sin(-\pi/2)\Big)
    So:
    z^4=-i

    \Rightarrow r^4(\cos4\theta+i\sin4\theta)=\cos(-\pi/2)+i\sin(-\pi/2)


    \Rightarrow \left\{\begin{array}{l}<br />
r^4\cos4\theta=-\cos(-\pi/2)\\<br />
r^4\sin4\theta = -\sin(-\pi/2)<br />
\end{array}\right .


    \Rightarrow \left\{\begin{array}{l}<br />
r=1\\<br />
4\theta=-\pi/2+2n\pi, n \in \mathbb{Z}<br />
\end{array}\right .

    So the values of \theta are given by:
    4\theta = -\pi/2, -\pi/2\pm2\pi,-\pi/2\pm4\pi, ...
    and its principle values (those between -\pi and \pi) are given by:
    \theta = -5\pi/8, -\pi/8, 3\pi/8, 7\pi/8
    Grandad
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  3. #3
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    My hat off to you sir

    Excellent explaination!
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