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A disjoint two-circle cycle has indices $+ 1$ and $- 1$ in its two components

Example

Let $γ_{1} (t) = - 1 + \frac{1}{2} \exp (i t)$ and $γ_{2} (t) = 1 + \frac{1}{2} \exp (- i t)$ for $t \in [0, 2 π]$ , and let $Γ$ be the complex chain $((1, γ_{1}), (1, γ_{2}))$ . Then $Γ$ is a cycle whose trace is the disjoint union of the two circles ${∣ z + 1 ∣ = \frac{1}{2}}$ and ${∣ z - 1 ∣ = \frac{1}{2}}$ , and for $z$ off that trace

$n (Γ, z) = {\begin{cases} + 1, & ∣ z + 1 ∣ < \frac{1}{2}, \\ - 1, & ∣ z - 1 ∣ < \frac{1}{2}, \\ 0, & ∣ z + 1 ∣ > \frac{1}{2} and ∣ z - 1 ∣ > \frac{1}{2} . \end{cases}$

Facts & Assumptions

Given: The contours $γ_{1}, γ_{2}$ above and the chain $Γ = ((1, γ_{1}), (1, γ_{2}))$ .

[L1]

The trace of a sum of chains is the union of their traces; a sum of cycles is a cycle; and for $p$ off the traces involved, $n (Γ_{1} + Γ_{2}, p) = n (Γ_{1}, p) + n (Γ_{2}, p)$ (Chain integration and the index are additive in the chain, and reverse with it).

[L2]

For $a \in C$ , $r > 0$ and $k \in Z$ , the contour $γ_{k} (t) = a + r \exp (i k t)$ on $[0, 2 π]$ is a closed complex contour with $n (γ_{k}, z) = k$ for $∣ z - a ∣ < r$ and $n (γ_{k}, z) = 0$ for $∣ z - a ∣ > r$ ; for $k \neq 0$ its trace is ${z : ∣ z - a ∣ = r}$ (A circle traversed $k$ times has winding number $k$ inside and $0$ outside).

[L3]

A complex chain is a finite list of pairs $(m_{k}, γ_{k})$ ; a list of closed contours is a cycle; and a single closed contour with coefficient $1$ is a cycle whose trace is the trace of that contour (Complex chains, their traces, and cycles).

[L4]

$n (Γ, p) = (2 π i)^{- 1} \int_{Γ} d z / (z - p)$ , and for a single closed contour with coefficient $1$ this is the winding number of that contour (Integration over a complex chain and the index of a chain).

[L5]

$∣ z + w ∣ \leq ∣ z ∣ + ∣ w ∣$ for complex $z, w$ (Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

Verification

technique · direct

1.1givenL2

By [L2] with $a = - 1$ , $r = \frac{1}{2}$ , $k = 1$ the contour $γ_{1}$ is closed with trace ${∣ z + 1 ∣ = \frac{1}{2}}$ , $n (γ_{1}, z) = 1$ for $∣ z + 1 ∣ < \frac{1}{2}$ and $n (γ_{1}, z) = 0$ for $∣ z + 1 ∣ > \frac{1}{2}$ ; by [L2] with $a = 1$ , $r = \frac{1}{2}$ , $k = - 1$ the contour $γ_{2}$ is closed with trace ${∣ z - 1 ∣ = \frac{1}{2}}$ , $n (γ_{2}, z) = - 1$ for $∣ z - 1 ∣ < \frac{1}{2}$ and $n (γ_{2}, z) = 0$ for $∣ z - 1 ∣ > \frac{1}{2}$ .

1.2L5

The two circles are disjoint: if $∣ z + 1 ∣ \leq \frac{1}{2}$ and $∣ z - 1 ∣ \leq \frac{1}{2}$ then $2 = ∣ (z + 1) - (z - 1) ∣ \leq ∣ z + 1 ∣ + ∣ z - 1 ∣ \leq 1$ by [L5], which is false.

2.1step 1.1step 1.2L1L3

Both contours are closed, so $Γ$ is a cycle by [L3], and by [L1] and [L3] its trace is the union of the two circles, which is disjoint by step 1.2.

3.1step 1.1step 2.1L1L4∎

For $z$ off that trace, [L1] and [L4] give $n (Γ, z) = n (γ_{1}, z) + n (γ_{2}, z)$ ; with step 1.1 this is $1 + 0 = 1$ when $∣ z + 1 ∣ < \frac{1}{2}$ , which forces $∣ z - 1 ∣ > \frac{1}{2}$ by step 1.2, and $0 + (- 1) = - 1$ when $∣ z - 1 ∣ < \frac{1}{2}$ , and $0 + 0 = 0$ when both moduli exceed $\frac{1}{2}$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

52 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

L. V. Ahlfors, Complex Analysis, 3rd ed., Ch. 4 §2.1 (standard reference, not scraped)

A disjoint two-circle cycle has indices +1+1 and −1−1 in its two components