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. :)

Spring Nov 23, 2020

#1**+3 **

Hint:

\(cis(\theta) = e^{i\theta}\)

Using this it is relatively easy to determine the answer.

As \(e^{i\theta}\div e^{i\alpha}=e^{i(\theta -\alpha)}\)

Melody Nov 24, 2020

#4**+3 **

\(cis(\theta) = e^{i\theta}\\ cis(7\pi/6) \div cis(\pi/3)= e^{i(7\pi/6)}\div e^{i(\pi/3)}=etc\)

Melody
Nov 24, 2020

#5**+2 **

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! :)

Spring
Nov 24, 2020

#8**+2 **

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?

Melody
Nov 24, 2020