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Math Help - Polar Form/Finding all solutions

  1. #1
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    Post Polar Form/Finding all solutions

    How do I find all solutions of (2z + 6)^5= 1 + i

    Do I have to find the polar form and determine what angle it has and then use
    z=r(cosx + isinx)? Or is that not helpful in this situation?

    Thanks.
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by JollyJolly15 View Post
    How do I find all solutions of (2z + 6)^5= 1 + i

    Do I have to find the polar form and determine what angle it has and then use
    z=r(cosx + isinx)? Or is that not helpful in this situation?

    Thanks.
    Oh, yes, yes: the polar form is very helpful in this case. So you have

    (2z+6)^5=1+i=\sqrt{2}\cdot e^{\displaystyle i\tfrac{\pi}{4}+i2n\pi}, n\in\mathbb{Z}
    (I am using z=|z|\cdot e^{i\varphi} here as the "polar form" of z, instead of z=|z|\cdot\big(\cos\varphi+i\sin\varphi\big), because it is easier to type.)

    Next you can take the fifth root by dividing the exponents on both sides by 5, and you get

    2z+6=\sqrt[10]{2}\cdot e^{\displaystyle i\tfrac{(8n+1)\pi}{20}}, n\in \mathbb{Z}

    Finally subtract 6 and divide by 2: done. - Except that you still need to figure out how many different solutions you have now got...
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