Let $z=e^{i\theta}$ with $\theta \in [0,2\pi[$ . Consider the sum
$$
\sum_{n=1}^{N} (e^{i\theta})^n.
$$
How could this be equal to
$$
\frac{1-e^{iN+T\theta}}{1-e^{i\theta}} \quad ?
$$
I tried to apply the sum as if it were $z^n$ in place of $e^{i\theta}$, but I got a slightly different expression, that is
$$
\frac{1-e^{i\theta N}}{1-e^{i\theta}}.
$$
Any idea?
Complex analysis equality in a limited sum
0
$\begingroup$
calculus
complex-analysis
summation
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0Are you sure the "$+T$" belongs there at all? I don't see any $T$ in the original expression. – 2012-10-08
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0No, unfortunately I have no informations about $T$. I was looking for a summation that can be put in that form, but I am beginning to suspect that the equality we talk about is a mistake/typo. – 2012-10-08
1 Answers
1
Hint
Use geometric progression formula $1+q+q^2+\ldots+q^n=\dfrac{1-q^{n+1}}{1-q}$ for $\theta \notin \{0,\,2\pi \} $ and separately calculate sum $\sum\limits_{n=1}^{N} (e^{i\theta})^n$ for the case $e^{i \theta}=1.$