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FP2 - Complex Exponents

Hello,
There are two series C = cosx + cos3x + cos5x +...+cos(2n-1)x
and S = sinx + sin3x + sin5x +...+sin(2n-1)x

I have summed the series and got C + iS = (cos2nx - 1 + isin2nx)/2isinx
(Not sure if that is right but looks like it?)

How would I find the modulus of C + iS? The answer is sin(nx)/sinx but I don't see how to get there which is why I'm unsure about getting the sum of the series correct. Also how would I find the argument of C + iS?

Thanks in advance
Original post by Quido
Hello,
There are two series C = cosx + cos3x + cos5x +...+cos(2n-1)x
and S = sinx + sin3x + sin5x +...+sin(2n-1)x

I have summed the series and got C + iS = (cos2nx - 1 + isin2nx)/2isinx
(Not sure if that is right but looks like it?)

How would I find the modulus of C + iS? The answer is sin(nx)/sinx but I don't see how to get there which is why I'm unsure about getting the sum of the series correct. Also how would I find the argument of C + iS?

Thanks in advance


I worked this through very quickly and got (1cos4nx)isin4nx2sinx\frac{(1-\cos 4nx)-i\sin 4nx}{2 \sin x} but I may have screwed up.

Also Is your quoted modulus correct? I got
Unparseable latex formula:

\frac{\sin 4nx}{\sin x}}



To find the modulus, write the complex number in the form z=a+ibz=a+ib, then z=a2+b2|z|=\sqrt{a^2+b^2} but your expression for C+iS won't give you the quoted result, so something is wrong somewhere.
Reply 2
Original post by atsruser
I worked this through very quickly and got (1cos4nx)isin4nx2sinx\frac{(1-\cos 4nx)-i\sin 4nx}{2 \sin x} but I may have screwed up.

Also Is your quoted modulus correct? I got
Unparseable latex formula:

\frac{\sin 4nx}{\sin x}}



To find the modulus, write the complex number in the form z=a+ibz=a+ib, then z=a2+b2|z|=\sqrt{a^2+b^2} but your expression for C+iS won't give you the quoted result, so something is wrong somewhere.


Yeah, I tried finding the modulus with mine and it doesn't work. By the way, what was first term and common ratio? Mine were e^ix and e^i2x respectively.
Original post by Quido
Hello,
There are two series C = cosx + cos3x + cos5x +...+cos(2n-1)x
and S = sinx + sin3x + sin5x +...+sin(2n-1)x

I have summed the series and got C + iS = (cos2nx - 1 + isin2nx)/2isinx
(Not sure if that is right but looks like it?)


Agreed.


How would I find the modulus of C + iS? The answer is sin(nx)/sinx but I don't see how to get there which is why I'm unsure about getting the sum of the series correct. Also how would I find the argument of C + iS?

Thanks in advance


The method is as atsruser suggested, and does work out to sinnxsinx\frac{\sin nx}{\sin x}

Post some working if you'd like someone to check.
Original post by Quido
Yeah, I tried finding the modulus with mine and it doesn't work. By the way, what was first term and common ratio? Mine were e^ix and e^i2x respectively.

Yes, that was what I worked with, but I'm short of time to check details now. I'll put up my working later/tomorrow if no one else has responded with a more informative answer.
Reply 5
Original post by ghostwalker
Agreed.



The method is as atsruser suggested, and does work out to sinnxsinx\frac{\sin nx}{\sin x}

Post some working if you'd like someone to check.


Working with the numerator only, I get C^2 = cos^2(2nx) - 2cos(2nx) + 1
and S^2 = -sin^2(2nx)

Adding those gives me 1 - cos(2nx)

After square rooting the denominator I am left with Root(1-cos(2nx))/(2isin(x))
Original post by Quido
Working with the numerator only, I get C^2 = cos^2(2nx) - 2cos(2nx) + 1
and S^2 = -sin^2(2nx)

Adding those gives me 1 - cos(2nx)

After square rooting the denominator I am left with Root(1-cos(2nx))/(2isin(x))


OK,

There are a few minor errors in there with the "i"s that don't really effect the outcome, but rather than correct, I'll redo.

If we work from C + iS = (cos2nx - 1 + isin2nx)/2isinx

Then C+iS=sin2nx2sinx+1cos2nx2sinxiC+iS= \frac{\sin 2nx}{2\sin x}+\frac{1-\cos 2nx}{2\sin x}i

Squaring and adding C and S, we get:

C2+S2=sin22nx4sin2x+(1cos2nx)24sin2xC^2+S^2= \frac{\sin^2 2nx}{4\sin^2 x}+\frac{(1-\cos 2nx)^2}{4\sin^2 x}

=1+12cos2nx4sin2x= \frac{1 +1-2\cos 2nx}{4\sin^2 x}

=1cos2nx2sin2x= \frac{1 -\cos 2nx}{2\sin^2 x}

Then double angle formula:

=1(12sin2nx)2sin2x= \frac{1 -(1-2\sin^2 nx)}{2\sin^2 x}

and I'm sure you can finish.
Reply 7
Original post by ghostwalker
OK,

There are a few minor errors in there with the "i"s that don't really effect the outcome, but rather than correct, I'll redo.

If we work from C + iS = (cos2nx - 1 + isin2nx)/2isinx

Then C+iS=sin2nx2sinx+1cos2nx2sinxiC+iS= \frac{\sin 2nx}{2\sin x}+\frac{1-\cos 2nx}{2\sin x}i

Squaring and adding C and S, we get:

C2+S2=sin22nx4sin2x+(1cos2nx)24sin2xC^2+S^2= \frac{\sin^2 2nx}{4\sin^2 x}+\frac{(1-\cos 2nx)^2}{4\sin^2 x}

=1+12cos2nx4sin2x= \frac{1 +1-2\cos 2nx}{4\sin^2 x}

=1cos2nx2sin2x= \frac{1 -\cos 2nx}{2\sin^2 x}

Then double angle formula:

=1(12sin2nx)2sin2x= \frac{1 -(1-2\sin^2 nx)}{2\sin^2 x}

and I'm sure you can finish.


Oh right, I was taking the real parts to be the cos and imaginary to be sin. Thanks for this!
Original post by Quido
Hello,
There are two series C = cosx + cos3x + cos5x +...+cos(2n-1)x
and S = sinx + sin3x + sin5x +...+sin(2n-1)x

I have summed the series and got C + iS = (cos2nx - 1 + isin2nx)/2isinx
(Not sure if that is right but looks like it?)

I now also agree - I'd managed to add in two different factors of 2, and i somewhere else. [Hmm - are you out by a factor of i though?]
(edited 7 years ago)

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