Interesting complex number Q (1 Viewer)

[waxwing]

Attached as an image.

Comments: If you find yourself doing long drawn out trig algebra, you might get to the answer, but you're going the wrong way.
The geometrical interpretation of the answer is particularly interesting (well, to me anyway).

 

[vds700]

Attached as an image.

Comments: If you find yourself doing long drawn out trig algebra, you might get to the answer, but you're going the wrong way.
The geometrical interpretation of the answer is particularly interesting (well, to me anyway).

S(x) = e^-5ix + e^-4ix + ... + e^5ix.

which is a geometric series with a = e^-5ix , r = e^ix , n = 11

S(x) = a(rⁿ - 1)/(r - 1)
= e^-5ix (e^11ix - 1)/(e^ix -1)
=e^6ix - e^-5i/x/(e^ix - 1)
=e^5ix(e^ix - 1)/(e^ix - 1)
=e^5ix

im guessing to turn it into sines, you would use eulers formula, but we dont do that in extension 2.

Am i on the right track???

 

[waxwing]

S(x) = e^-5ix + e^-4ix + ... + e^5ix.

which is a geometric series with a = e^-5ix , r = e^ix , n = 11

S(x) = a(rⁿ - 1)/(r - 1)
= e^-5ix (e^11ix - 1)/(e^ix -1)
=e^6ix - e^-5i/x/(e^ix - 1)

Up to here, perfect!
But the next step is a mistake.
Actually that's the step that isn't so obvious. Look for a hidden symmetry.

About Euler, i.e. e^i@=cis@ : You're right! I teach British Maths A-level syllabus, we let that one slide in a lot earlier in our teaching, I just rather stupidly assumed you guys would know it too. Teach me to use you as guinea pigs
If anybody is interested in pursuing it, I should add this to the question:
"Given that e^ix=cisx, ..."

 

[vds700]

sorry about the mistake.

I can't work out what to do next. Maybe someone else can try...

 

[Affinity]

The correct answer is sin(5.5x)/sin(0.5x)

Spoiler

(e^6ix - e^-5ix) / (e^ix - 1) <- once you get here you multiply top and bottom by e^{-ix/2}

 

[waxwing]

Thanks Affinity, that's the symmetry I was talking about
I have a pretty picture of the answer, if anyone wants, I can post it later.

3unitz - interesting, I'm pondering your attempt - trying to work out your sine rule business there ...

 

[waxwing]

3unitz:
You seem to be finding the length of e^6ix + e^-5ix instead of e^6ix - e^-5ix. That's why your final answer is cosines (using sin 2@ formula) instead of sines.

 

[waxwing]

Here's my answer (with pictures )

3unitz, I like your answer, if anything I prefer it, at least for school level.

Even though you made an error, you can rejig it to find the other side of the triangle and it's right.

It's always nice to have both an algebraic and a geometric way to see an answer