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Suppose that I have the following sum: $\sum_{m=0}^{\infty}(e^{it}(1-p))^{m}$, where $i^2 = -1$.

This is a geometric series, but involving the complex number $i$. Can I just apply the geometric series formula and conclude that the sum if $\frac{1}{1-e^{it}(1-p)}$?

Thank you,

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    $\begingroup$ Yes, as long as $|1-p|<1$. $\endgroup$ – Angina Seng Oct 14 '20 at 19:58
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The geometric series formula you are referring to holds for real numbers as long as the common ratio is less than 1. Analogously, the formula holds for complex numbers when the common ratio has modulus less than 1.

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To prove that if $|z|<1$ then $\sum_{m\ge0}z^m=\tfrac{1}{1-z}$, note that $\tfrac{1}{1-z}-\sum_{m=0}^{n-1}z^m=\frac{z^n}{1-z}$ has $n\to\infty$ limit $0$. (By contrast, if $|z|\ge1$ then this $n\to\infty$ behaviour of the error term is lost, so the series diverges.) Note nothing about this reasoning cares whether $z$ is real or complex.

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