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# Complex numbers question watch

1. When z is complex and |z|< 1, . Use this result to find the sum of the series , where theta and a are real and |a| < 1.

Not really sure where to start with this... I tried putting , then kind of assumed that a is equivalent to r (I'm not sure why ?), to get

... but I got stuck after that!

Any tips?
2. Can you prove that ? Hence what the sum equivalent to?
3. (Original post by GHOSH-5)
Can you prove that ? Hence what the sum equivalent to?
Ah yes, I remember proving that identity before (cba to do it again now )

So then we get

But ... and... erm... what about ? Am I on the right track?
4. (Original post by trm90)
Ah yes, I remember proving that identity before (cba to do it again now )

So then we get

But ... and... erm... what about ? Am I on the right track?
Yep, you're on the right track: - split up the sum:

Can you see how to proceed now?
Spoiler:
Show
and
5. (Original post by GHOSH-5)
Yep, you're on the right track: - split up the sum:

Can you see how to proceed now?
Spoiler:
Show
and
I can sort of see what to do... the first summation term is basically sum of zn which is just 1/(1-z), but I don't really know how what to do with ... is that the same thing as z-n? Or is that the same thing as (conjugate)? Regardless, not quite sure how I'd evaluate that sum, unless I'm missing something!

(Thanks by the way!)
6. They are almost the same sum, just the 'z' is different; in the first case, 'z' is ae^{i.theta}, and in the second case, 'z' is ae^{-i.theta}. Both of them sum to 1/(1-z), just 'z' is different...
7. (Original post by GHOSH-5)
They are almost the same sum, just the 'z' is different; in the first case, 'z' is ae^{i.theta}, and in the second case, 'z' is ae^{-i.theta}. Both of them sum to 1/(1-z), just 'z' is different...
Ah ok. So would my answer literally look something like:

Where z* = a.exp(-i.theta)?

I feel like I'm overcomplicating things here, though
8. (Original post by trm90)
Ah ok. So would my answer literally look something like:

Where z* = a.exp(-i.theta)?

I feel like I'm overcomplicating things here, though
Pretty much, yes. You can simplify your answer though. Leave z and z* in the exponential form, and rewrite everything over one fraction. Then recall that:

, and the analogous identity for sin(x) we used before.
9. There is another trick here .. (apologies - I've yet to learn LaTex or whatever)

Let S = the sum you seek
Let C = sum from n = 0 to inf (a^n cos(ntheta))

Consider C + iS = sum from n= 0 to inf (a^n . exp(intheta)) which is a GP I can do in one line

Equate real and imaginary components.
10. I'd go for the C + iS method described in ian.slater's post.
11. (Original post by ian.slater)
There is another trick here .. (apologies - I've yet to learn LaTex or whatever)

Let S = the sum you seek
Let C = sum from n = 0 to inf (a^n cos(ntheta))

Consider C + iS = sum from n= 0 to inf (a^n . exp(intheta)) which is a GP I can do in one line

Equate real and imaginary components.
Nice.

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