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# Need urgently! Help and check!

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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. :) Nov 23, 2020

#1
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Hint:

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

Using this it is relatively easy to determine the answer.

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

Nov 24, 2020
#2
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I got 7π/6. Is that correct?

Spring  Nov 24, 2020
#3
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No it is not.

What is theta?

What is alpha?

What is theta-alpha ?

Melody  Nov 24, 2020
#4
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$$cis(\theta) = e^{i\theta}\\ cis(7\pi/6) \div cis(\pi/3)= e^{i(7\pi/6)}\div e^{i(\pi/3)}=etc$$

Melody  Nov 24, 2020
#5
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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
#6
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Yes the number out the front is correct.

How do you do       x^7 divided by   x^2   ?

Melody  Nov 24, 2020
edited by Melody  Nov 24, 2020
#7
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Okay, thank you.

You subtract 7-2 to get 5. The answer is x^5.

Spring  Nov 24, 2020
#8
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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
#9
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Next, you subtract 7π/6 & π/3. When you subtract you get 5π/6.

Nov 24, 2020
#10
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yes so it will be

e^(i(5pi/6)  = cis 5pi/6

Melody  Nov 24, 2020
#11
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Wow! Thank you so much Melody. I greatly appreciate it! 😊

Spring  Nov 24, 2020
#12
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So long as I have helped you to learn, I am pleased. Melody  Nov 24, 2020
#13
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Thanks, Melody  !!!!   Nov 24, 2020