Polar Form C.N

**taurus** · May 7th 2008, 08:35 PM

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

**Danshader** · May 7th 2008, 10:52 PM

given :

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

now
$ \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:

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

$\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} $
$ \left| \frac{z}{w} \right| = \frac{32}{11} = r $

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

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

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

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

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

hope i didn't mess things up.

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