+133 I have posted a question before with the same problem, but it was not answered. I solved half the problem. But I am confused about the rest. Can someone check my answer and help me figure out the answer? For some reason, I keep getting 7/2 for the second half, which is incorrect and not even an option. Thank you ahead of time for your help! Greatly appreciate it. :)
+118609
Hint:
\(cis(\theta) = e^{i\theta}\)
Using this it is relatively easy to determine the answer.
As \(e^{i\theta}\div e^{i\alpha}=e^{i(\theta -\alpha)}\)
+118609 \(cis(\theta) = e^{i\theta}\\ cis(7\pi/6) \div cis(\pi/3)= e^{i(7\pi/6)}\div e^{i(\pi/3)}=etc\)
+133 Well, to be quite honest, I am not 100% sure. I didn't learn it in class with the equation you provided me.
So solving it to my ability with your equation, correct me if I am incorrect, I got 5π/6.
Also, is the 2√15/5 correct?
Thank you Melody for your time and patience in helping me figure this out. I am not the best at Pre-Calc/Trig! :)
+118609 right!
x^7 divided by x^2 = x^(7-2)= x^5
When you do not understand what to do always convert to a simply but similar example and see what happens.
\(cis(\theta) = e^{i\theta}\\~\\ cis(7\pi/6) \div cis(\pi/3)\\= e^{i(7\pi/6)}\div e^{i(\pi/3)}\\ =e^{[(i(7\pi/6)-i(\pi/3)]}\\ =e^{i[(7\pi/6)-(\pi/3)]}\)
What is next?
