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divide 

6√5 cis(11pi/6) ÷ 3√6 cis(pi/2)

 Apr 25, 2020
 #1
avatar+485 
+1

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:

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

.
 Apr 25, 2020
edited by jfan17  Apr 25, 2020
edited by jfan17  Apr 25, 2020
edited by jfan17  Apr 25, 2020
edited by jfan17  Apr 25, 2020
 #2
avatar+109527 
+1

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] \\~\\

 Apr 25, 2020
edited by Melody  Apr 25, 2020

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