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Thread: nth roots of complex number fun!

  1. #1
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    Talking nth roots of complex number fun!

    I have been asked to find all the solutios for the following equation

    $\displaystyle z^4 = 128(-\sqrt{3} + i)$

    and leave my answers in exponential form and then plot them on an argand diagram.

    I think I could probably do it if the 128 wasn't there. If it isn't there I put
    $\displaystyle (-\sqrt{3} + i)$ on an argand diagram and find that the argument or radius from the centre is 2. I think I can then change that to
    $\displaystyle (-\sqrt{3} + i)^\frac{1}{4}$ becoming $\displaystyle 2e^{i\theta}$ or something along those lines.

    Can anybody tell me how to put the 128 in and if I got the rest of it sort of correct? or if it is a trainwreck?
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  2. #2
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    Quote Originally Posted by Beard View Post
    I have been asked to find all the solutios for the following equation

    $\displaystyle z^4 = 128(-\sqrt{3} + i)$

    and leave my answers in exponential form and then plot them on an argand diagram.

    I think I could probably do it if the 128 wasn't there. If it isn't there I put
    $\displaystyle (-\sqrt{3} + i)$ on an argand diagram and find that the argument or radius from the centre is 2. I think I can then change that to
    $\displaystyle (-\sqrt{3} + i)^\frac{1}{4}$ becoming $\displaystyle 2e^{i\theta}$ or something along those lines.

    Can anybody tell me how to put the 128 in and if I got the rest of it sort of correct? or if it is a trainwreck?
    If $\displaystyle a+ bi= r e^{i\theta}$, then $\displaystyle 128(a+ bi)= (128r)e^{i\theta)}$! What ever "r" you got for $\displaystyle -\sqrt{3}+ i$, multiply it by 128.

    It happens that $\displaystyle 128= 2^7$ so that $\displaystyle (128r)^{1/4}= 2^{7/4}r^{1/4}= (2)(2^{3/4})(r^{1/4})$.
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  3. #3
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    Ahhhhhh, yes that makes sense, so, because my r = 2 I multiply that by 2^7 to get 2^8 and then the fourth root of that is 2^2 which is 4.

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  4. #4
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    Okay, so having done that step

    I end up doing

    $\displaystyle z^4 = 2^8e^{i\theta}$
    and then $\displaystyle z = 4e^{\frac{i\theta}{4}}$ and then from there I just...
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  5. #5
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    Quote Originally Posted by Beard View Post
    Okay, so having done that step

    I end up doing

    $\displaystyle z^4 = 2^8e^{i\theta}$
    and then $\displaystyle z = 4e^{\frac{i\theta}{4}}$ and then from there I just...

    You have to find the argument $\displaystyle \theta$ , of course! $\displaystyle \theta=\arg(-\sqrt{3}+i)=\arctan\left(-\frac{1}{\sqrt{3}}\right)=\frac{5\pi}{6}$ , since $\displaystyle \cos\theta<0\,,\,\,\sin\theta>0$


    Tonio
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