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Math Help - 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

    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
    (-\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
    (-\sqrt{3} + i)^\frac{1}{4} becoming 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

    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
    (-\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
    (-\sqrt{3} + i)^\frac{1}{4} becoming 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 a+ bi= r e^{i\theta}, then 128(a+ bi)= (128r)e^{i\theta)}! What ever "r" you got for -\sqrt{3}+ i, multiply it by 128.

    It happens that 128= 2^7 so that (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.

    Thanks
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  4. #4
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    Okay, so having done that step

    I end up doing

    z^4 = 2^8e^{i\theta}
    and then 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

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

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


    Tonio
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