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Thread: Two small things with Complex Numbers

  1. #1
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    Cool Two small things with Complex Numbers

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

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

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

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

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

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

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


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

    $\displaystyle = \sqrt{8}$

    $\displaystyle = 2\sqrt{2}$.


    Since $\displaystyle z$ is in the first quadrant

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

    $\displaystyle = \arctan{1}$

    $\displaystyle = \frac{\pi}{4}$.


    Thus

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


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

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

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

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

    $\displaystyle = 64(-1 + 0i)$

    $\displaystyle = -64$.
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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.. $\displaystyle w^2*-2i=x+iy$ ??
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  5. #5
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    Quote Originally Posted by MissWonder View Post
    I maybe understand 2a and 2b now.. $\displaystyle w^2*-2i=x+iy$ ??
    No, it would actually be

    $\displaystyle w^2 = -2i$

    $\displaystyle w^2 + 2i = 0$.


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


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

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


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

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


    Try to solve

    $\displaystyle r^2\,\textrm{cis}\,(2\theta) = 2\,\textrm{cis}\,\frac{\pi}{2}$.
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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 $\displaystyle r^2 = 2$ and $\displaystyle 2\theta = \frac{\pi}{2}$?

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

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

    is shorthand for

    $\displaystyle z = r(\cos{\theta} + i\sin{\theta})$.
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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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