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The complex geometric power series has radius $1$ and sums to $1/(1-z)$ for $|z|<1$

Example

For $|z|<1$ , the complex geometric series $\sum_{n\ge0}z^n$ has radius $1$ and sum $1/(1-z)$ . The conventions and prerequisite facts used below are recorded in Cauchy-Hadamard for complex power series, including zero and infinite radius, The complex numbers form a field, and every nonzero $x+iy$ has inverse $(x-iy)/(x^2+y^2)$ , For $|r| < 1$ , $\sum_{k \ge 0} r^k = 1/(1-r)$ , and for $|r| \ge 1$ the series diverges, For $|r| < 1$ the sequence $r^k$ is null, and for $|r| > 1$ the sequence $|r|^k$ diverges to $+\infty$ , Every absolutely convergent complex series converges, and rearrangements preserve its sum, Conjugation laws, $z\overline z=|z|^2$ , multiplicativity of modulus, and the triangle inequality, A vector-valued function has a limit, or is continuous, if and only if each of its components does; with the algebra of continuous vector-valued functions.

Facts & Assumptions

Given: A complex $z$ with $|z|<1$ .

[L1]

For $|r| < 1$ , $\sum_{k \ge 0} r^k = 1/(1-r)$ , and for $|r| \ge 1$ the series diverges says that the real series $\sum_{n\ge0}r^n$ converges for $|r|<1$ .

[L2]

Conjugation laws, $z\overline z=|z|^2$ , multiplicativity of modulus, and the triangle inequality gives $|z^n|=|z|^n$ .

[L3]

Every absolutely convergent complex series converges, and rearrangements preserve its sum says that an absolutely convergent complex series converges.

[L4]

For $|r| < 1$ the sequence $r^k$ is null, and for $|r| > 1$ the sequence $|r|^k$ diverges to $+\infty$ gives $|z|^n\to0$ when $|z|<1$ .

[L5]

Cauchy-Hadamard for complex power series, including zero and infinite radius defines the shifted coefficient limsup and its radius cases.

Verification

1.1

The shifted coefficient roots in [L5] are all $1$ , so it gives radius $1$ .

2.1

By [L2], the modulus series is the real geometric series $\sum|z|^n$ , which converges by [L1]; therefore the complex series converges by [L3], say to $S$ . The finite identity $(1-z)\sum_{k<n}z^k=1-z^n$ holds in the complex field. Since $|z^n|=|z|^n\to0$ by [L2] and [L4], passing to the limit gives $(1-z)S=1$ ; since $z\ne1$ , $S=1/(1-z)$ .

L1L2L3L4∎

Depends on

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Sources

J. Lebl, Basic Analysis I: Complex Numbers and the Complex Exponential (standard reference, not scraped)

The complex geometric power series has radius 111 and sums to 1/(1−z)1/(1-z)1/(1−z) for ∣z∣<1|z|<1∣z∣<1