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The power series of $z_{0} / (1 - z_{1})$ and the shape of its domain of convergence

Example

On the region ${(z_{0}, z_{1}) \in C^{2} : ∣ z_{1} ∣ < 1}$ one has

$\frac{z_{0}}{1 - z_{1}} = \sum_{j = 0}^{\infty} z_{0} z_{1}^{j} .$

Equivalently, in multi-index notation,

$\frac{z_{0}}{1 - z_{1}} = \sum_{α \in N^{2}} c_{α} z^{α},$

where $c_{(1, j)} = 1$ for every $j \in N$ and $c_{α} = 0$ otherwise. When $z_{0} \neq 0$ , the series converges absolutely exactly when $∣ z_{1} ∣ < 1$ ; when $z_{0} = 0$ , every term vanishes and the series converges absolutely for every $z_{1}$ . Thus its absolute-convergence set is $(C \times D (0, 1)) \cup ({0} \times C)$ , an unbounded set and not a bounded polydisc.

Facts & Assumptions

Given: The function $f (z) = z_{0} / (1 - z_{1})$ on the region $∣ z_{1} ∣ < 1$ .

[L1]

For complex $w$ , the geometric series $\sum_{j \geq 0} w^{j}$ converges absolutely exactly when $∣ w ∣ < 1$ : if $∣ w ∣ < 1$ , then $\sum_{j \geq 0} ∣ w ∣^{j}$ converges by For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges, so Every absolutely convergent complex series converges, and rearrangements preserve its sum applies, and the finite identity $(1 - w) \sum_{j < n} w^{j} = 1 - w^{n}$ together with $∣ w^{n} ∣ = ∣ w ∣^{n} \to 0$ gives the sum $1 / (1 - w)$ (Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive, For $∣ r ∣ < 1$ the sequence $r^{k}$ is null, and for $∣ r ∣ > 1$ the sequence $∣ r ∣^{k}$ diverges to $+ \infty$ , $C = R [x] / (x^{2} + 1)$ is a field, every element is uniquely $a + b i$ , and every nonzero element has inverse $(a - b i) / (a^{2} + b^{2})$ ). If $∣ w ∣ \geq 1$ , then the same real geometric-series criterion shows that $\sum_{j \geq 0} ∣ w ∣^{j}$ diverges, so for every nonzero complex constant $c$ the series $\sum_{j \geq 0} c w^{j}$ cannot be absolutely convergent.

Verification

technique · direct

1.1

If $∣ z_{1} ∣ < 1$ , then [L1] gives $(1 - z_{1})^{- 1} = \sum_{j \geq 0} z_{1}^{j}$ , so multiplying by $z_{0}$ yields $z_{0} / (1 - z_{1}) = \sum_{j \geq 0} z_{0} z_{1}^{j}$ .

givenL1

2.1

In multi-index form this is the stated coefficient rule: the only monomials that appear are $z_{0} z_{1}^{j}$ , so $c_{(1, j)} = 1$ and every other coefficient is $0$ .

step 1.1

2.2

The absolute-value series is $\sum_{j \geq 0} ∣ z_{0} ∣ ∣ z_{1} ∣^{j}$ . If $z_{0} \neq 0$ , division by the positive constant $∣ z_{0} ∣$ and [L1] show that it converges exactly when $∣ z_{1} ∣ < 1$ . If $z_{0} = 0$ , every term is $0$ , so it converges for every $z_{1}$ . Hence the absolute-convergence set is $(C \times D (0, 1)) \cup ({0} \times C)$ .

step 1.1L1

3.1

At points $(0, z_{1})$ with $∣ z_{1} ∣ \geq 1$ , every term of the series is $0$ , so the series still converges there to $0$ , although the quotient is undefined when $z_{1} = 1$ and this exceptional convergence set is not open.

step 2.2∎

Depends on

Used by

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Dependency tree · two levels

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Sources

J. Lebl, Tasty Bits of Several Complex Variables, v4.4, §1.2 (standard reference, not scraped)

The power series of z0/(1−z1)z0​/(1−z1​) and the shape of its domain of convergence