1. ## 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

2. 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.