Precalc question

cis(x) = cos(x) + isin(x)

cis(11pi/6) = cos(11pi/6) + isin(11pi/6)

cos(11pi/6) = $\sqrt3/2$

sin(11pi/6) = 1/2

So the top then becomes:

$6\sqrt5 *(\sqrt{3}+i)/2$

This expands to:

$(6\sqrt{15} + 6i\sqrt{5})/2$

Now we do the same with the denominator.

cis(pi/2) = cos(pi/2) + isin(pi/2)

cos(pi/2) = 0

sin(pi/2) = 1

cis(pi/2) = 0 + i

Our denominator then becomes:

$3\sqrt6 * 0 + 3i\sqrt6$

Our answer is then:

$((6\sqrt{15} + 6i\sqrt{5})/2)/3i\sqrt{6}$ which ends up simplifying to:

$(6\sqrt{15} + 6i\sqrt5)/6i\sqrt6$

Rationalizing our denominator, we get:

$(6\sqrt{90} + 6i\sqrt{30})/36i = (3\sqrt{10} + i\sqrt{30})/6i$

6√5 cis(11pi/6) ÷ 3√6 cis(pi/2)

$6\sqrt5\;e^{(11\pi i/6)}\div 3\sqrt6 \;e^{(\pi i/2)}\\~\\ =2\sqrt5\;e^{[(11\pi i/6)-(\pi i/2)]}\div \sqrt6 \\~\\ =2\sqrt5\;e^{[(11\pi i/6)-(3i\pi/6)]}\div \sqrt6 \\~\\ =2\sqrt5\;e^{[(4\pi i/3)]}\div \sqrt6 \\~\\ =2\sqrt{30}[\;cos(4\pi i/3)+isin(4\pi i/3)]\div 6 \\~\\ =\frac{\sqrt{30}}{3}[\;-cos(\pi /3)-isin(\pi /3)] \\~\\ =\frac{\sqrt{30}}{3}[\;-\frac{1}{2}-i(\frac{\sqrt3}{2})] \\~\\ =\frac{\sqrt{30}}{6}[\;-1-\sqrt3i] \\~\\ $

I think this is the same as jfan's answer,.

jfan, you need to finish by making your denominator real. 

I will multiply mine out as well, it might look better.....

$=\frac{\sqrt{30}}{6}[\;-1-\sqrt3i] \\~\\ =\frac{-1}{6}[\;\sqrt{30}+3\sqrt{10}i] \\~\\ =\frac{-\sqrt{10}}{6}[\;\sqrt{3}+3i] \\~\\$

Does that look better?  Not sure.

LaTex:

Latex:

6\sqrt5\;e^{(11\pi i/6)}\div 3\sqrt6 \;e^{(\pi i/2)}\\~\\

=2\sqrt5\;e^{[(11\pi i/6)-(\pi i/2)]}\div \sqrt6 \\~\\

=2\sqrt5\;e^{[(11\pi i/6)-(3i\pi/6)]}\div \sqrt6 \\~\\

=2\sqrt5\;e^{[(4\pi i/3)]}\div \sqrt6 \\~\\

=2\sqrt{30}[\;cos(4\pi i/3)+isin(4\pi i/3)]\div 6 \\~\\

=\frac{\sqrt{30}}{3}[\;-cos(\pi /3)-isin(\pi /3)] \\~\\

=\frac{\sqrt{30}}{3}[\;-\frac{1}{2}-i(\frac{\sqrt3}{2})] \\~\\

=\frac{\sqrt{30}}{6}[\;-1-\sqrt3i] \\~\\