+499 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:
\((6\sqrt{90} + 6i\sqrt{30})/36i = (3\sqrt{10} + i\sqrt{30})/6i\)
+118667 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] \\~\\
