We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
57
3
avatar

Find the exponential form of the complex number

 

[e^{\pi i/4} + e^{5 \pi i/12} + e^{7 \pi i/12} + e^{3\pi i/4} + e^{11\pi i/12}

 

\(e^{\pi i/4} + e^{5 \pi i/12} + e^{7 \pi i/12} + e^{3\pi i/4} + e^{11\pi i/12}\\ \)


with proof.

 

I've graphed each of the points on the complex plane to find that they are separated by an angle measure of pi/6. 2 pairs of 2 points have the same x-coordinate, but negative, but I don't know where to go from there. I know I'm supposed to use the magnitude of the sum of some of the points as well as the symmetry of the diagram, but I'm not sure how.

 

 

Thanks!

 Jun 26, 2019
edited by Melody  Jun 28, 2019
edited by Melody  Jun 28, 2019
edited by Melody  Jun 28, 2019

Best Answer 

 #2
avatar
+1

Use the identity

 \(\displaystyle re^{\imath \theta}\equiv r(\cos \theta + \imath \sin \theta)\)

to switch each term to trig and then use the identities

\(\displaystyle \cos A + \cos B = 2\cos((A +B)/2)\cos((A-B)/2) \\\sin A +\sin B = 2\sin((A+B)/2)\cos((A-B)/2)\)

to group terms, (1st with 5th, 2nd with 4th).

Switch back to exponential form at the end.

 Jun 27, 2019
 #1
avatar+625 
+4

I have no idea what the answer is but for convenience's sake:

 

THE QUESTION:

 

Find the exponential form of the complex number

 

\( [e^{\pi i/4} + e^{5 \pi i/12} + e^{7 \pi i/12} + e^{3\pi i/4} + e^{11\pi i/12} \)


with proof.

 Jun 27, 2019
 #2
avatar
+1
Best Answer

Use the identity

 \(\displaystyle re^{\imath \theta}\equiv r(\cos \theta + \imath \sin \theta)\)

to switch each term to trig and then use the identities

\(\displaystyle \cos A + \cos B = 2\cos((A +B)/2)\cos((A-B)/2) \\\sin A +\sin B = 2\sin((A+B)/2)\cos((A-B)/2)\)

to group terms, (1st with 5th, 2nd with 4th).

Switch back to exponential form at the end.

Guest Jun 27, 2019
 #3
avatar+102447 
0

Thanks guest, I was wondering about this one too   laugh

Melody  Jun 28, 2019

4 Online Users