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Math Help - Two small things with Complex Numbers

  1. #1
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    Cool Two small things with Complex Numbers

    1)
    I got:
    z=2+2i and I have to find the polar representation of z. Them find z^4 and x+iy (when it's polar)...

    My bid:
    w=rCos+i rSin to find the polar representation... right? So it would be something like: w=r*cos(2)+r*sin(2i)?

    2a)
    w=x+iy find all values for w, when w^2=-2i

    2b)
    Find all z's when z^2+(1+i)z+i=0
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  2. #2
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    Quote Originally Posted by MissWonder View Post
    1)
    I got:
    z=2+2i and I have to find the polar representation of z. Them find z^4 and x+iy (when it's polar)...

    My bid:
    w=rCos+i rSin to find the polar representation... right? So it would be something like: w=r*cos(2)+r*sin(2i)?

    2a)
    w=x+iy find all values for w, when w^2=-2i

    2b)
    Find all z's when z^2+(1+i)z+i=0
    z = 2 + 2i.


    r = \sqrt{2^2 + 2^2}

     = \sqrt{8}

     = 2\sqrt{2}.


    Since z is in the first quadrant

    \theta = \arctan{\frac{2}{2}}

     = \arctan{1}

     = \frac{\pi}{4}.


    Thus

    z = 2 + 2i = 2\sqrt{2}\,\textrm{cis}\,\frac{\pi}{4}.


    To find z^4, use DeMoivre's Theorem.

    z^4 = (2\sqrt{2})^4\,\textrm{cis}\,\frac{4\pi}{4}

     = 64\,\textrm{cis}\,{\pi}

     = 64(\cos{\pi} + i\sin{\pi})

     = 64(-1 + 0i)

     = -64.
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  3. #3
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    Coool

    I did not know how to do that :S Strange! our teatcher did it in another way.. But your explaination made sense
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  4. #4
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    hint

    I maybe understand 2a and 2b now.. w^2*-2i=x+iy ??
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  5. #5
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    Quote Originally Posted by MissWonder View Post
    I maybe understand 2a and 2b now.. w^2*-2i=x+iy ??
    No, it would actually be

    w^2 = -2i

    w^2 + 2i = 0.


    You could solve this using the quadratic formula, but the solution would involve \sqrt{i}.


    Better to convert to polars and use DeMoivre's Theorem.

    w^2 = 2i = 2\,\textrm{cis}\,\frac{\pi}{2}.


    Supposing we knew what w was, if we squared it we would have

    w^2 = r^2\,\textrm{cis}\,(2\theta).


    Try to solve

    r^2\,\textrm{cis}\,(2\theta) = 2\,\textrm{cis}\,\frac{\pi}{2}.
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  6. #6
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    Try to solve

    .

    r=pi+cis ??
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  7. #7
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    Quote Originally Posted by MissWonder View Post
    Try to solve

    .

    r=pi+cis ??
    Wouldn't it make sense that r^2 = 2 and 2\theta = \frac{\pi}{2}?

    Solve for r and \theta.
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  8. #8
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    By the way,

    z = r\,\textrm{cis}\,\theta

    is shorthand for

    z = r(\cos{\theta} + i\sin{\theta}).
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  9. #9
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    r= 1.41421
    and the other... hmm pi?
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  10. #10
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    Hmm.. maybe rather cis=2/pi
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