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Math Help - Complex Number Help

  1. #1
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    Complex Number Help

    Hi guys, I am totally lost in my class and am not sure about this tutorial. Help please...

    Solve the equation : giving your answers in the form of a + jb. Show that there can be only be 8 distinct roots.
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  2. #2
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    Re: Complex Number Help

    Hello, timeforchg!

    Where did this problem come from?
    Do you have the answers? .Any hints?

    Is there a typo?
    If the right side is a constant, there are only six roots.


    \text{Solve the equation: }\:z^6-1 \;=\;\left[\frac{64j(1-j)}{1+2j}\right]^6

    . . \text{giving your answers in the form }a+ib.

    \text{Show that there can be only be 8 distinct roots.}

    Consider the fraction: . \frac{64j(1-j)}{1+2j} \;=\;\frac{64(j - j^2)}{1+2j} \;=\; \frac{64(1 + j)}{1+2j}

    Rationalize the denominator: . \frac{64(1+j)}{1+2i}\cdot\frac{1-2j}{1-2j} \;=\; \tfrac{64}{5}(3-j)

    The equation becomes: . z^6-1 \;=\; \left[\tfrac{64}{5}(3-j)\right]^6 \quad\Rightarrow\quad z^6 \;=\;\left[\tfrac{64}{5}(3-j)\right]^6 + 1

    . . . . . . . . . . . . . . . . . . . . . z \;=\; \sqrt[6]{\left[\tfrac{64}{5}(3-j)\right]^6 + 1}
    Ugly!
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  3. #3
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    Re: Complex Number Help

    Quote Originally Posted by timeforchg View Post
    Hi guys, I am totally lost in my class and am not sure about this tutorial. Help please...

    Solve the equation : giving your answers in the form of a + jb. Show that there can be only be 8 distinct roots.
    It's ugly, but not unmanageable...

    \displaystyle \begin{align*} z^6 - 1 &= \left[\frac{64j(1-j)}{1+2j}\right]^6 \\ z^6 - 1 &= \left(\frac{64+64j}{1 + 2j}\right)^6 \\ z^6 - 1 &= \left[\frac{(64 + 64j)(1 - 2j)}{(1 + 2j)(1 - 2j)}\right]^6 \\ z^6 - 1 &= \left( \frac{192}{5} - \frac{64}{5}j \right)^6 \\ z^6 - 1 &= \left[\frac{64\sqrt{10}}{5}\,e^{j\left( \pi - \arctan{\frac{1}{3}}\right)}\right]^6 \\ z^6 - 1 &= \frac{ 2^{36}\cdot 10^3 }{5^6}\,e^{6j\left( \pi - \arctan{\frac{1}{3}}\right)} \\ z^6 - 1 &= \frac{ 2^{39} }{5^3}\,e^{6\pi j}\,e^{j\left(-6\arctan{\frac{1}{3}}\right)} \\ z^6 - 1 &= \frac{2^{39}}{5^3} \, e^{j\left(-6\arctan{\frac{1}{3}}\right)} \\ z^6 - 1 &= \frac{2^{39}}{5^3}\left(-\frac{44}{125} - \frac{117}{125}j \right) \\ z^6 - \frac{5^6}{5^6} &= -\frac{2^{41} \cdot 11}{5^6} - \frac{2^{39}\cdot 3^2 \cdot 13}{5^6}j \\ z^6 &= \frac{5^6 - 2^{41} \cdot 11}{5^6} - \frac{2^{39} \cdot 3^2 \cdot 13}{5^6}j \end{align*}

    Now by converting the RHS to the most general polar form and taking both sides to the power of 1/6 will yield the desired answer.
    Thanks from timeforchg
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    Re: Complex Number Help

    Hi Guys,

    Sorry there was a typo. Yes it is 6 distinct roots. Thanks a lot for the head start.
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  5. #5
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    Lightbulb Re: Complex Number Help

    {z6}=1+\left(\frac{64 i (1-i)}{1+2 i}\right)^6

    z6 = ((64 i (1 - i))/(1 + 2 i))^6 + 1

    {zz}({k}){=}\sqrt[6]{|{z6}|} e^{i \arg \left(\frac{2 \pi k}{6}+\frac{\text{z6}}{6}\right)}

    k = {0, 1, 2, 3, 4, 5}

    z1, z2, z3, z4, z5, z6=
    -\sqrt[3]{\frac{3}{5}} \sqrt[12]{149250101163100633121} e^{i \tan ^{-1}\left(\frac{7146825580544}{2687695088383}\right)  }
    -\sqrt[3]{\frac{3}{5}} \sqrt[12]{149250101163100633121} e^{i \tan ^{-1}\left(\frac{64321430224896}{24189255795447-31250 \pi }\right)}
    -\sqrt[3]{\frac{3}{5}} \sqrt[12]{149250101163100633121} e^{i \tan ^{-1}\left(\frac{64321430224896}{24189255795447-62500 \pi }\right)}
    -\sqrt[3]{\frac{3}{5}} \sqrt[12]{149250101163100633121} e^{i \tan ^{-1}\left(\frac{21440476741632}{8063085265149-31250 \pi }\right)}
    -\sqrt[3]{\frac{3}{5}} \sqrt[12]{149250101163100633121} e^{i \tan ^{-1}\left(\frac{64321430224896}{24189255795447-125000 \pi }\right)}
    -\sqrt[3]{\frac{3}{5}} \sqrt[12]{149250101163100633121} e^{i \tan ^{-1}\left(\frac{64321430224896}{24189255795447-156250 \pi }\right)}
    Thanks from manoj9585
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  6. #6
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    Re: Complex Number Help

    Thanks for the help guys.

    Hi Proveit,

    A quick question. how do you derive from equation A to B. Pls See attached. Complex Number Help-complex2.png
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  7. #7
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    Re: Complex Number Help

    Expanded the brackets in the power, then applied \displaystyle \begin{align*} a^{m+n} = a^ma^n \end{align*}.
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  8. #8
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    Re: Complex Number Help

    Quote Originally Posted by Prove It View Post
    Expanded the brackets in the power, then applied \displaystyle \begin{align*} a^{m+n} = a^ma^n \end{align*}.
    I see. Correct me if I am wrong. Do we need the same base in order to apply that theorem?
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  9. #9
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    Re: Complex Number Help

    Yes, and the base in question is e.
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    Re: Complex Number Help

    hi jasper i read the question i dont really get what u mean on the part zz(K)=
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  11. #11
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    Re: Complex Number Help

    Quote Originally Posted by hiroyoki View Post
    hi jasper i read the question i dont really get what u mean on the part zz(K)=
    hmmm... someone interested in this question uh.. from SG some more. haha
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    Re: Complex Number Help

    hmm
    Last edited by xotinalx; September 18th 2012 at 05:13 PM. Reason: delete
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  13. #13
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    Re: Complex Number Help

    SO timeforchg u understand that part?
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  14. #14
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    Re: Complex Number Help

    SOLVED!! Thanks a lot guys!! Appreciate it.
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