COMPLEX series question - please help asap (1 Viewer)

gamja · May 2, 2023

By expanding the left-hand side, show that $\left(1+2z+3z^{2}+4z^{3}+5z^{4}+...+nz^{n-1}\right)\left(z-1\right)=n$ where z is an nth root of unity, n≠1

I noticed that it looked like a differential, and if you integrate it you get $\left(1+z^{2}+z^{3}+z^{4}+...+z^{n}\right)=\int_{ }^{ }\frac{n}{z-1}dz$ . Obviously that's wrong, but I think I'm on the right track.

Thanks for your help!! (in advance - my test is tmrw)

[edit: by LHS I can infer it's from z^n-1=0]

011235 · May 2, 2023

$\text{LHS} = (1+2z+3z^2+\dots+nz^{n-1})(z-1) \\ = z + 2z^2 + 3z^3 + \dots + (n-1)z^{n-1} + nz^n - 1 - 2z - 3z^2 - \dots - nz^{n-1} \\ = -1 - z - z^2 - \dots - z^{n-1} + nz^n$
Now we know $z^n=1$ (roots of unity) and $-1-z-z^2-\dots -z^{n-1}$ is a sum of GP with $a=-1$ , $r=z$ and $n$ terms. Thus
$\text{LHS} = \frac{-(z^n-1)}{z-1} + n \\ = \frac{1-1}{z-1} + n \\ = n \\ = \text{RHS}$

does this help?

As a quick note. In Ext 2 it is incorrect to integrate or differentiate complex functions. To do this properly requires complex analysis (a uni branch of maths). Sometimes you will happen to get the right answer, but you haven't actually done the work required to be able to do that. (Unless you go to north sydney boys where their papers occasionally use this incorrect and very cursed technique).

COMPLEX series question - please help asap (1 Viewer)

gamja

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011235

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