Question

how to prove this

Answer 1

@sirm3d

Answer 2

@niksva

Answer 3

\[1+r+r^2+\cdots+r^{n-1}\cdot\frac{1-r}{1-r}=\frac{1-r^n}{1-r}\]

because the series converges for \[|r|<1\]

\[\lim_{n\rightarrow \infty} r^n=0\]

\[\large \sum_{n=0}^\infty r^n=\lim_{n\rightarrow \infty} \frac{1-r^n}{1-r}=\frac{1}{1-r}\]

Answer 4

i have to go, @gerryliyana , i'm late for work.

Answer 5

ok thank you so much @sirm3d :)

Answer 6

can someone help me in my problem please

Answer 7

How to prove this

Answer 8

with \[z=pe^{ix}\]
\[\sum_{n=0}^\infty z^n =\frac{1}{1-z}=\frac{1}{1-(p\cos x +ip\sin x)}\\\qquad =\frac{1}{(1-p\cos x)-i(p\sin x)}\cdot \frac{(1-p\cos x)+i(p\sin x)}{(1-p\cos x)+i(p\sin x)}\\\qquad =\frac{1-p\cos x+i(p\sin x)}{(1-p\cos x)^2+p^2\sin^2 x}\\\qquad =\frac{(1-p\cos x)+i(p\sin x)}{1-2p\cos x+p^2} \]

Answer 9

\[z^n=p^ne^{inx}=p^n\cos nx + i(p^n\sin nx)\]
\[\sum z^n=\sum p^n\cos nx+i\sum p^n\sin nx\]
but \[\sum z^n=\frac{1-p\cos x}{1-2p\cos x+p^2}+i\frac{p\sin x}{1-2p\cos x+p^2}\]

Answer 10

aww thanks a lot :) great! What is the trick to solve this ??

Answer 11

I execute
\[\sum_{n=0}^{\infty} (pz)^{n} = \frac{ 1 }{ 1-p(\cos x + i \sin x ) } = \frac{ 1 }{ 1-p \cos x - i p \sin x } \times \frac{1+p \cos x + i p \sin x}{1+p \cos x + i p \sin x}\]
before

Answer 12

@sirm3d

Answer 13

to divide by a complex number \(a+bi\), multiply both sides by its conjugate, which is \(a-bi\)

Answer 14

your \(a+bi\) is \((1-p\cos x)+(-p\sin x)i\)

Answer 15

ok , i got it, thank you :)