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Thread: Polar Form C.N

  1. #1
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    Question Polar Form C.N

    If
    z = -8 sqrt(3) + 8 i
    AND
    w = (-(11/sqrt(2))/2) + ((11/sqrt(2))/2)i

    determine z / w in polar form:
    r cis(theta)

    Am stuck on the above, how do i solve it?
    Thanks
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  2. #2
    Member Danshader's Avatar
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    given :

    $\displaystyle
    z = -8\sqrt{3} +8i
    $
    and
    $\displaystyle
    w =
    \frac{-11}{2\sqrt{2}} + \frac{11}{2\sqrt{2}}i
    $

    now
    $\displaystyle
    \frac{z}{w} = \frac{-8\sqrt{3} +8i}{
    \frac{-11}{2\sqrt{2}} + \frac{11}{2\sqrt{2}}i}
    $

    multiplying the numerator and the denominator with the conjugate of w we get:

    $\displaystyle
    \frac{z}{w} = \frac{(-8\sqrt{3} +8i)( \frac{-11}{2\sqrt{2}} - \frac{11}{2\sqrt{2}}i)}{(
    \frac{-11}{2\sqrt{2}} + \frac{11}{2\sqrt{2}}i)( \frac{-11}{2\sqrt{2}} - \frac{11}{2\sqrt{2}}i)}
    $
    $\displaystyle
    \frac{z}{w} = \frac{\frac{88\sqrt{3}}{2\sqrt{2}} +\frac{88\sqrt{3}}{2\sqrt{2}}i - \frac{88}{2\sqrt{2}}i +\frac{88}{2\sqrt{2}}}{\frac{121}{8} +\frac{121}{8}}
    $
    $\displaystyle
    \frac{z}{w} = \frac{\frac{88}{2\sqrt{2}}(\sqrt{3}+1) +\frac{88}{2\sqrt{2}}(\sqrt{3}-1)i}{\frac{121}{4}}
    $
    $\displaystyle
    \frac{z}{w} = \frac{16}{11\sqrt{2}}(\sqrt{3}+1) +\frac{16}{11\sqrt{2}}(\sqrt{3}-1)i
    $

    to get the r value of the polar form is actually the modulus of $\displaystyle \frac{z}{w}$

    $\displaystyle \left| \frac{z}{w} \right| = \sqrt{ \left( \frac{16}{11\sqrt{2}}(\sqrt{3}+1)\right) ^2 +\left( \frac{16}{11\sqrt{2}}(\sqrt{3}-1)\right) ^2}
    $
    $\displaystyle
    \left| \frac{z}{w} \right| = \frac{32}{11} = r
    $

    to find the angle $\displaystyle \theta$ we use the fact that:

    if z = a +bi
    $\displaystyle
    tan (\theta) = \frac{b}{a}
    $

    therefore in our case it should be:
    $\displaystyle
    tan(\theta) = \frac{\frac{16}{11\sqrt{2}}(\sqrt{3}-1)}{\frac{16}{11\sqrt{2}}(\sqrt{3}+1)}
    $
    $\displaystyle
    \theta =15^o
    $

    the polar representation of a complex number is in the form of:
    $\displaystyle
    z = r(cos(\theta) + isin(\theta))
    $

    hence,
    $\displaystyle
    \left| \frac{z}{w} \right| = \frac{32}{11}(cos(15^o) + isin(15^o))
    $
    $\displaystyle
    \left| \frac{z}{w} \right|= \frac{32}{11}(cos(\frac{\pi}{12}) + isin(\frac{\pi}{12}))
    $
    $\displaystyle
    \left| \frac{z}{w} \right|= \frac{32}{11}cis(\frac{\pi}{12})
    $

    hope i didn't mess things up.
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