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Complex Exponentials

Need some help here with this question



I know the exponentials required, cos=1/2(e^ix+e^-ix) and sin=1/2i(e^ix-e^-ix) but Im not certain really what to do. I assume I have to multiiply the cos exponential by 4 and then cube it? But I cant seem to get the correct answer after taking just ther real parts.

Anyone able to help me?
compl3x.png13.8KB
Original post by Eremor
Need some help here with this question



I know the exponentials required, cos=1/2(e^ix+e^-ix) and sin=1/2i(e^ix-e^-ix) but Im not certain really what to do. I assume I have to multiiply the cos exponential by 4 and then cube it? But I cant seem to get the correct answer after taking just ther real parts.

Anyone able to help me?


Cube then multiply.

Post some working. I can imagine where you're going wrong but I'd like to see your working before commenting further.
What do you get when you work it:
4⋅(12(eix+e−ix))34 \cdot (\frac{1}{2}(e^{ix}+e^{-ix}))^3?
Reply 3
Avatar for ztibor
ztibor
10
Original post by Eremor
Need some help here with this question



I know the exponentials required, cos=1/2(e^ix+e^-ix) and sin=1/2i(e^ix-e^-ix) but
Anyone able to help me?


No. cos and sin do not equal anything, but
cos⁡x=eix+e−ix2\displaystyle \cos x=\frac{e^{ix}+e^{-ix}}{2}
and so for sin x
Reply 4
Avatar for TeeEm
TeeEm
19
Original post by Eremor
Need some help here with this question



I know the exponentials required, cos=1/2(e^ix+e^-ix) and sin=1/2i(e^ix-e^-ix) but Im not certain really what to do. I assume I have to multiiply the cos exponential by 4 and then cube it? But I cant seem to get the correct answer after taking just ther real parts.

Anyone able to help me?


quote me if help is still needed
Reply 5
Avatar for Eremor
Eremor
OP
9
Original post by TeeEm
quote me if help is still needed


Hey Know its been a while since I posted this but I'd spent time practicing for other exams and have just got back to This problem.

Here's what ive done 11251563_10155557813565414_1496897598_n.jpg

Im really close with what ive done but get the whole thing multiplied by 2 for some reason, any ideas?
Reply 6
Avatar for TeeEm
TeeEm
19
Original post by Eremor
Hey Know its been a while since I posted this but I'd spent time practicing for other exams and have just got back to This problem.

Here's what ive done 11251563_10155557813565414_1496897598_n.jpg

Im really close with what ive done but get the whole thing multiplied by 2 for some reason, any ideas?


you got me in a very busy spot

look at similar solved questions, 31, 33, 49, 57, 61, 67 ...
in this link
http://madasmaths.com/archive/maths_booklets/further_topics/various/complex_numbers_part_2_exam_questions.pdf
Original post by Eremor
Hey Know its been a while since I posted this but I'd spent time practicing for other exams and have just got back to This problem.

Here's what ive done 11251563_10155557813565414_1496897598_n.jpg

Im really close with what ive done but get the whole thing multiplied by 2 for some reason, any ideas?


Going from line5 to line6 your multiplying fraction should have changed from 1/8 to 1/4. Look back at your formula of cos⁡θ\cos\theta

Also note that your formula of sin⁡θ\sin\theta is incorrect, you're missing an "i".
Reply 8
Avatar for Eremor
Eremor
OP
9
Original post by ghostwalker
Going from line5 to line6 your multiplying fraction should have changed from 1/8 to 1/4. Look back at your formula of cos⁡θ\cos\theta

Also note that your formula of sin⁡θ\sin\theta is incorrect, you're missing an "i".


Sorry am sure im missing something obvious but why does it go from 1/8 to 1/4?
Original post by Eremor
Sorry am sure im missing something obvious but why does it go from 1/8 to 1/4?


2cos⁡θ=eiθ+e−iθ 2\cos\theta = e^{i\theta}+e^{-i\theta}
Original post by Eremor
Sorry am sure im missing something obvious but why does it go from 1/8 to 1/4?
Because e3θi+e−3θi=2cos(3θ)\displaystyle e^{3 \theta i}+e^{-3 \theta i}=2cos(3\theta) so you have 18×2cos(3θ)=14×cos(3θ)\displaystyle \frac{1}{8} \times 2cos(3\theta)=\frac{1}{4} \times cos(3\theta) This is because if you look for your formula for cos theta you will notice it has a 2 in the denominator which I'm guessing you have forgotten about.
Reply 11
Avatar for Eremor
Eremor
OP
9
Ah I see now. Thank you both!
Original post by Eremor
Ah I see now. Thank you both!


No problem. Good luck in your exams.

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