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Tetration for Complex Bases

William Paulsen

Abstract

In this paper we will consider the tetration, deﬁned by the equation F(z+ 1) = bF(z)in the complex

plane with F(0) = 1, for the case where bis complex. A previous paper determined conditions for a unique

solution the case where bis real and b>e1/e. In this paper we extend these results to ﬁnd conditions

which determine a unique solution for complex bases. We also develop iteration methods for numerically

approximating the function F(z), both for bases inside and outside the Shell-Thron region.

2010 Mathematics Subject Classiﬁcation: 26A18, 30D05, 39B12

Keywords: Tetration, Abel’s functional equation, iteration.

1 Background

Much research has already been done on the tetration function, whose name comes from tetra- (four) and

iteration. Addition by a positive integer ncan be thought of as repeated incrementing, multiplication by n

is done by repeated addition, and exponentiation by nis repeated multiplication. So a fourth level operation

would be repeated exponentiation using the initial base b:

bb·

·b

n

Although there are many diﬀerent notations for this operation, we will use the iterated exponential notation

expn

b(x) = bb·

·bx

with n b’s.

The tetration function for base bcan then be expressed as F(n) = expn

b(1). Thus we see that F(0) = 1,

F(1) = b,F(2) = bb, and so on. When bis complex, we must deﬁne a branch cut so that bzis uniquely

deﬁned. Once this branch cut is established, we can recursively deﬁne F(x) by F(x+ 1) = bF(x). Working

backwards from this relation, we see that F(−1) = 0 and F(−2) must be undeﬁned, even for complex b. In

[1], Kneser showed that if bis real and b > e1/e, we can extend F(z) to the whole complex plane, minus a

branch cut at z≤ −2. This solution will be referred to as κb(z).

Kneser’s solution begins by ﬁnding the ﬁxed points of the exponential function bz, that is, points Lfor

which bL=L. To ﬁnd the principle ﬁxed points, we ﬁrst deﬁne the Wright ωfunction to be the solution to

the initial value problem

ω′(z) = ω(z)

1 + ω(z), ω(1) = 1,

1

where the branch cuts are chosen to start at −1±πi, run parallel to the real axis in the negative direction

[2]. The Wright ωfunction can in fact be deﬁned in terms of the nth branch of the Lambert’s Wfunction.

ω(z) = Wn(ez),where n=ℑ(z)−π

2π.

Note that with this deﬁnition, ω(a) = ω(a−) whenever ais on either of the branch cuts. (In this context,

a−means to approach afrom below the branch cut rather than from the left. For example, ln(−1−) = −πi,

and ln(−1+) = πi.) We prefer a function which is continuous from above the branch cuts, so we deﬁne

ω2(z) = ω(z).

Then ω2will be identical to ωexcept at the branch cuts, where now ω2(a) = ω2(a+) at both branch cuts.

The principle ﬁxed point of the exponential function bzcan now be expressed by

L1=ω2(ln(ln(b)) −πi)

−ln(b).

A second ﬁxed point can be found with a similar formula,

L2=ω2(ln(ln(b)) + πi)

−ln(b).

In both formulas, we use the principle branch of the logarithm. If b > e1/e, or if bis not real, then we can

see that ℑ(L1)>0 and ℑ(L2)<0.

Although we have two ﬁxed points for most bases b, there is a question as to whether or not these are

attractive ﬁxed points of the function bz. We say that the base bis in the Shell-Thron region if the sequence

of values

{b, bb, bbb, bbbb

, . . .}(1)

converge to a ﬁnite ﬁxed point. Note that this can only happen if |ln L|<1 for one of the ﬁxed points.

The Shell-Thron region is depicted in ﬁgure 1. For b̸= 1 within this region, the sequence converges to L1if

ℑ(b)≥0, and converges to L2if ℑ(b)<0.

We begin by solving the Schr¨oder equation σ(bz) = sσ(z), where sis the derivative of bzat one of the

ﬁxed points. At the ﬁxed point L1,s1= ln(L1) = L1ln(b), and at the ﬁxed point L2,s2= ln(L2) = L2ln(b).

For each ﬁxed point, there is a unique solution to Schr¨oder’s equation which is analytic near the ﬁxed point,

and for which the derivative at the ﬁxed point is 1. By equating coeﬃcients of the power series, we ﬁnd that

σ1(z) = (z−L1) + s1

2L1(1 −s1)(z−L1)2+s2

1(1 + 2s1)

6L2

1(1 −s1)(1 −s2

1)(z−L1)3

+s3

1(1 + 6s1+ 5s2

1+ 6s3

1)

24L3

1(1 −s1)(1 −s2

1)(1 −s3

1)(z−L1)4+s4

1(1 + 14s1+ 24s2

1+ 45s3

1+ 46s4

1+ 26s5

1+ 24s6

1)

120L4

1(1 −s1)(1 −s2

1)(1 −s3

1)(1 −s4

1)(z−L1)5

+s5

1(1 + 30s1+ 89s2

1+ 214s3

1+ 374s4

1+ 416s5

1+ 511s6

1+ 461s7

1+ 330s8

1+ 154s9

1+ 120s10

1)

720L5

1(1 −s1)(1 −s2

1)(1 −s3

1)(1 −s4

1)(1 −s5

1)(z−L1)6

+···.

σ2(z) = (z−L2) + s2

2L2(1 −s2)(z−L2)2+s2

2(1 + 2s2)

6L2

2(1 −s2)(1 −s2

2)(z−L2)3

2

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....................................................................................................................................................................................................................................................................................................................................................................................................................................................

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..................................................................................................................................................

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........

........

........

........

e1/e

e−e

−1 1 2 3

−2

−1

1

2

Figure 1: The Shell-Thron region.

+s3

2(1 + 6s2+ 5s2

2+ 6s3

2)

24L3

2(1 −s2)(1 −s2

2)(1 −s3

2)(z−L2)4+s4

2(1 + 14s2+ 24s2

2+ 45s3

2+ 46s4

2+ 26s5

2+ 24s6

2)

120L4

2(1 −s2)(1 −s2

2)(1 −s3

2)(1 −s4

2)(z−L2)5

+s5

2(1 + 30s2+ 89s2

2+ 214s3

2+ 374s4

2+ 416s5

2+ 511s6

2+ 461s7

2+ 330s8

2+ 154s9

2+ 120s10

2)

720L5

2(1 −s2)(1 −s2

2)(1 −s3

2)(1 −s4

2)(1 −s5

2)(z−L2)6

+···.

Unfortunately, when bis on the boundary of the Shell-Thron region, one of the series will probably have

a zero radius of convergence, requiring us to treat these as asymptotic series. Even for other basis, it is

unclear what the radius of convergence will be. However, we can easily extend the region of convergence. In

the case where bis within the Shell-Thron region, and the series in Eq. 1 converges to L, then if zis in the

basin of attraction for bz, so that

{z, bz, bbz, bbbz

, . . .}

also converges to L, then

σ(z) = lim

n→∞ s−n(expn

b(z)−L),

where expn

b(z) represents applying bza total of ntimes. If Lis one of the repulsive ﬁxed points of bz, and

if zis in the basin of attraction for logb(z), so that repetitively applying the principal value of logbto zwill

converge to L, then

σ(z) = lim

n→∞ sn(logn

b(z)−L).

3

Next, we let ψ(z) = ln(σ(z))/ln(s), so that ψ(z) will solve Abel’s equation ψ(bz) = ψ(z) + 1. We can

compute, for j= 1 or 2,

ψj(z) = ln(z−Lj)

ln(sj)+sj

2Ljln(sj)(1 −sj)(z−Lj) + s2

j(1 + 5sj)

24L2

jln(sj)(1 −sj)(1 −s2

j)(z−Lj)2(2)

+s4

j(2 + sj+ 3s2

j)

24L3

jln(sj)(1 −sj)(1 −s2

j)(1 −s3

j)(z−Lj)3

+s4

j(61sj−1 + 71s2

j+ 290s3

j+ 299s4

j+ 109s5

j+ 251s6

j)

2880L4

jln(sj)(1 −sj)(1 −s2

j)(1 −s3

j)(1 −s4

j)(z−Lj)4

+s6

j(6 + 15sj+ 69s2

j+ 143s3

j+ 115s4

j+ 212s5

j+ 221s6

j+ 155s7

j+ 49s8

j+ 95s9

j)

1440L5

jln(sj)(1 −sj)(1 −s2

j)(1 −s3

j)(1 −s4

j)(1 −s5

j)(z−Lj)5

+···.

As long as bis not too close to the boundary of the Shell-Thron region, we can compute ψj(z) very quickly

to high levels of precision. If |sj|>1, and zis in the basin of attraction for logb(z) converging to Lj, then

ψj(z) = lim

n→∞

ln(sn

jTm(logn

b(z)))

ln(sj),

where Tm(z) is the m-th degree Taylor polynomial for σb(z). The larger mis, the faster the convergence.

Likewise, if |sj|<1, and zis in the basin of attraction for bzconverging to Lj, then

ψj(z) = lim

n→∞

ln(s−n

jTm(expn

b(z)))

ln(sj).

Note that these two are not mutually exclusive—zmight be in the basin of attraction for bzconverging to

L1and also in the basin of attraction for logb(z) converging to L2.

Next, we ﬁnd the inverse function ψ−1

j(z) = σ−1

j(ezln sj). This function will satisfy the functional equation

ψ−1

j(z+ 1) = bψ−1

j(z).

By reversing the series for σ(z), we ﬁnd that

ψ−1

j(z) = Lj+ezln sj−sj

2Lj(1 −sj)e2zln sj+s2

j(2 + sj)

6L2

j(1 −sj)(1 −s2

j)e3zln sj(3)

−s3

j(6 + 6sj+ 5s2

j+s3

j)

24L3

j(1 −sj)(1 −s2

j)(1 −s3

j)e4zln sj+s4

j(24 + 36sj+ 46s2

j+ 40s3

j+ 24s4

j+ 9s5

j+s6

j)

120L4

j(1 −sj)(1 −s2

j)(1 −s3

j)(1 −s4

j)e5zln sj

−s5

j(120 + 240sj+ 390s2

j+ 480s3

j+ 514s4

j+ 416s5

j+ 301s6

j+ 160s7

j+ 64s8

j+ 14s9

j+s10

j)

720L5

j(1 −sj)(1 −s2

j)(1 −s3

j)(1 −s4

j)(1 −s5

j)e6zln sj

+···.

Again, if zis in the basin of attraction of logb(z) converging to Lj,

ψ−1

j(z) = lim

n→∞ expn

b(Lj+e(z−n) ln sj),

4

and if zis in the basin of attraction of bzconverging to Lj,

ψ−1

j(z) = lim

n→∞ logn

b(Lj+e(z+n) ln sj).

In fact, we can speed up the convergence by considering the limit

ψ−1

j(z) = lim

n→∞ expn

b(Sm(e(z−n) ln sj)) if |sj|>1,

logn

b(Sm(e(z+n) ln sj)) if |sj|<1,

where Sm(z) is the m-th degree Taylor polynomial for σ−1

j(z). This function will be entire if |sj|>1, but

will have branch cuts if |sj|<1.

In either case, it should be pointed out that as long as bis not a real number ≤e1/e, then ℑ(ln(s1)) >0,

since ℑ(L1)>0, and likewise ℑ(ln(s2)) <0. Hence, for any x,

lim

y→∞ ψ−1

1(x+iy) = L1and lim

y→−∞ ψ−1

2(x+iy) = L2.

2 The Uniqueness of a complex tetration

Even though ψ−1

1(z) and ψ−1

2(z) satisfy the tetration problem F(z+ 1) = bF(z)for at least some portion

of the complex plane, they are far from unique. We can combine each of these with any periodic function

to produce another solution. If we let p(z) be a periodic function of period 1, and let ρj(z) = p(z) + z,

then ψ−1

j(ρj(z)) will also solve the tetration problem. Since p(z) is periodic, we could express ρj(z) with a

complex Fourier series

ρj(z) = z+∞

k=−∞

cke2πikz .

We would like for the new tetration ψ−1

1(ρ1(z)) to also have the property that limy→∞ F(x+iy) = L1.

This means that ρ1(z) must approach ∞ias zapproaches ∞i. In order for this to happen in the Fourier

series, we see that ck= 0 for all k < 0. Thus,

ρ1(z) = z+∞

k=0

cke2πikz =z+c0+c1e2πiz +c2e4π iz +c3e6πiz +···.(4)

Likewise, for the tetration ψ−1

2(ρ2(z)) to also have the property that limy→−∞ F(x+iy) = L2, the Fourier

series for L2must be in the form

ρ2(z) = z+∞

k=0

dke−2πikz =z+d0+d1e−2πiz +d2e−4π iz +d3e−6πiz +···.(5)

The question is whether we can ﬁnd ρ1(z) and ρ2(z) so that the two iterations ψ−1

1(ρ1(z)) and ψ−1

2(ρ2(z))

meet in the middle, that is, one is an analytic continuation of the other. The ﬁrst step is to show that if this

is possible, it is unique.

Proposition 1:

Suppose that F(z) is an analytic function deﬁned for ℜ(z)>−2 for which F(z+ 1) = bF(z), and that

for all x > −2,

lim

y→+∞F(x+iy) = L1and lim

y→−∞ F(x+iy) = L2.

5

Then there is some Msuch that for ℑ(z)> M,ψ1(F(z)) can be expressed as

ψ1(F(z)) = z+∞

k=0

fke2πikz ,

and for ℑ(z)<−M,ψ2(F(z)) can be expressed as

ψ2(F(z)) = z+∞

k=0

gke−2πikz .

Proof:

Since ψ1(F(z+ 1)) = ψ1(bF(z)) = ψ1(F(z)) + 1, we see that ψ1(F(z)) −zis periodic with period 1. By

the limit property, we have that F(z) is in the region where ψ1(z) is deﬁned whenever 0 ≤ ℜ(z)≤1 and

ℑ(z)> M for suﬃciently large M, and by periodicity, we can deﬁne ψ1(F(z)) whenever ℑ(z)> M . By the

Fourier series,

ψ1(F(z)) = z+∞

k=−∞

fke2πikz .

Thus,

F(z) = ψ−1

bz+∞

k=−∞

fke2πikz .

If fk̸= 0 for some negative k, then as ℑ(z)→ ∞,

fke2πikz → ∞ as ℑ(z)→ ∞.

If there are many such terms, they will have diﬀerent exponential growth rates asymptotically, so such terms

cannot cancel out asymptotically. Thus

∞

k=−∞

fke2πikz → ∞ as ℑ(z)→ ∞.

Also note that the argument of this sum will depend on the real part of z. Thus, there is a path Cgoing

towards +∞ifor which ψ1(F(z)) is real and positive along this path. That is, for every y > 0 there is an x

such that with z=x+iy,

arg z+∞

k=−∞

fke2πikz = 0.

But then along this path,

lim

CF(z) = lim

Cψ−1

1z+∞

k=−∞

fke2πikz = lim

z→+∞ψ−1

b(z),

and limz→+∞ψ−1

1(z) does not exist, since the real part is unbounded, and the imaginary part is negative.

We can use a similar argument for ψ2(z), and pick the larger of the two M’s.

6

Proposition 2:

Suppose that F1(z) and F2(z) are analytic functions deﬁned for ℜ(z)>−2 for which F(z+ 1) = bF(z),

F(0) = 1, and that for all x > −2,

lim

y→+∞F1(x+iy) = lim

y→+∞F2(x+iy) = L1,and

lim

y→−∞ F1(x+iy) = lim

y→−∞ F2(x+iy) = L2.

Then F1(z) = F2(z) for all ℜ(z)>−2.

Proof:

By proposition 1,

ψ1(F1(z)) = z+∞

k=0

fke2πikz ,for ℑ(z)> M,

ψ1(F2(z)) = z+∞

k=0

gke2πikz ,for ℑ(z)> M,

ψ2(F1(z)) = z+∞

k=0

hke−2πikz ,for ℑ(z)<−M,

ψ2(F2(z)) = z+∞

k=0

jke−2πikz ,for ℑ(z)<−M.

Manipulating the series, we can ﬁnd

F−1

1(F2(z)) = z+∞

k=0

ℓke2πikz ,for ℑ(z)> M,

F−1

1(F2(z)) = z+∞

k=0

mke−2πikz ,for ℑ(z)<−M.

We know that F−1

1(F2(z)) −zis periodic, and approaches one constant as ℑ(z)→ ∞, and another as

ℑ(z)→ −∞. Even though the series may not converge in the same region, the Fourier coeﬃcients are

unique, so F−1

1(F2(z)) = z+Cfor some constant C. Since F1(0) = F2(0) = 1, we have C= 0, and so

F1(z) = F2(z).

Note that this proof does not guarantee existence of the function F(z). In fact, for some bases b, the

complex tetration with the required properties is impossible. Nonetheless, if the complex tetration does

exist for a base b, then we can deﬁne ρ1(z) = ψ1(F(z)) and ρ2(z) = ψ2(F(z)), for which both ρ1(z)−zand

ρ2(z)−zwill be periodic. In fact, ρ1(z) and ρ2(z) must have series expansions of the form of Eqs. 4 and 5.

We can even ﬁnd the domains of ρ1(z) and ρ2(z).

Lemma 1:

If bis not on the boundary of the Shell-Thron region, and there is a F(z) satisfying the conditions of

proposition 2, then ρ1(z) is analytic for ℑ(z)>0, and ρ2(z) is analytic for ℑ(z)<0.

7

Proof:

If ℑ(z)>0, then F(z−n) is deﬁned for all integer n. If |s1|>1 so that L1is a repulsive ﬁxed point

of bz, then limn→∞ F(z−n) = L1, so F(z) will be in the basin of attraction for logb(z) converging to

L1, hence in the domain of ψ1(z), so ρ1(z) is deﬁned. If |s1|<1, then L1is an attractive ﬁxed point of

bz, and limn→∞ F(z+n) = L1, so F(z) is in the basin of attraction for bzconverging to L1, so again,

ρ1(z) = ψ1(F(z)) will be deﬁned. To show that ρ1(z) is analytic, note that the series in Eq. 4 can be written

as ρ1(z) = z+g(e2πiz ), where g(z) has the Maclaurin series

g(z) = ∞

k=0

ckzk.

Since ρ1(z) is deﬁned for ℑ(z)>0, g(z) has no singularities within the unit disk, so g(z) will have a radius of

convergence of 1, making ρ1(z) analytic for ℑ(z)>0 whenever bis not on the boundary of the Shell-Thron

region. Likewise, ρ2(z) is analytic for ℑ(z)<0.

Even though we have proven uniqueness, we have yet to prove that a tetration exists for complex b.

Unfortunately, Kneser’s existence proof for κb(z) in [1] breaks down if bis complex. Nonetheless, we can

extend Kneser’s result to the base bin the complex plane. The plan is to form an analytic continuation of

Kneser’s solution.

Proposition 3:

There is an open connected set Sin the complex plane containing all real numbers greater than e1/e,

such that for b∈S, then there will be a κb(z) which satisﬁes the conditions of Propositions 1 and 2, that is,

κb(z+ 1) = bκb(z),κb(0) = 1, and that for x > −2,

lim

y→+∞κb(x+iy) = L1and lim

y→−∞ κb(x+iy) = L2.

Proof:

By the way that Kneser deﬁned κb(z), we see that for each z0with ℜ(z)>−2, κb(z0) is a real analytic

function of bfor b>e1/e. Hence, we can analytically extend κb(z0) to the complex b-plane, at least to an

open region containing (e1/e,∞). Since

κb(z0+ 1) −bκb(z0)= 0

for all real b, this must be true for complex bas well. Letting z0= 0 shows that κb(0) = 1 for complex b.

Let b0be a point in this open set, and let Cbe a path from b0to ein the open set that avoids the point

e1/e. Note that for a ﬁxed x0and breal, limy→∞ κb(x+iy) = L1(b) uniformly on every closed subset of

(e1/e,∞). Since

L1=ω2(ln(ln(b)) −πi)

−ln(b)

is obviously an analytic function of bexcept for a branch cut at e1/e, we can extend this to say that

limy→∞ κb(x0+iy) exists for each b, and converges uniformly on the path C. So limy→∞ κb(x0+iy) = L1(b)

for all points on C, and in particular b0. Likewise,

lim

y→−∞ κb0(x+iy) = L2.

So the analytic continuation κb0(z) will satisfy the conditions of propositions 1 and 2, hence will be the

unique tetration which satisﬁes these conditions.

Of course the main question is determining the open set Sfor which we can extend the complex tetration.

Before we can answer this, we will need a way of computing the κb(z) to a high degree of accuracy.

8

3 First approximation

One of the goals of this paper is to produce an iterative method for calculating the tetration for a given base

b. However, this iterative method requires having a ﬁrst order approximation to begin to process. In [3], the

ﬁrst order approximation was created by forcing a high degree polynomial to satisfy Julia’s equation, which

in turn produces a solution to Abel’s equation. This seems to work well if bis real, but when bis complex,

it produces a function with a highly periodic component. So we will resort to another method.

The goal is to ﬁnd a function with several key properties:

f(−1) = 0, f(0) = 1, f (1) = b, lim

z→i∞f(z) = L1,and lim

z→−i∞f(z) = L2.

We also want the function to be nearly periodic for large imaginary component. From [4], we expect the

upper half plane to approach a periodic function with period 2πi/ ln(s1) as ℑ(z)→ ∞, and the lower half

plane to approach a periodic function with period 2πi/ ln(s2), since these are the periods of ψ1(z) and ψ2(z).

Thus, we want

f(z)∼L1+K1eln(s1)zas x→i∞, f(z)∼L2+K2eln(s2)zas x→ −i∞.

In [4], such a function was found using a piecewise deﬁned function, but we can ﬁnd a single function having

all of these properties by assuming the function is of the form

f(z) = Ae−ir1z+B+C eir2z

De−ir1z+E+F eir2z,(6)

with ℜ(r1)>0 and ℜ(r2)>0. It is not too hard to establish that r1must be −iln(s1), and r2must be

iln(s2). As long as bis complex, or b > e1/e , then indeed ℜ(r1) and ℜ(r2) are both positive. Then we need

to ﬁnd A,B,C,D,E, and Fso that

As1+B+Cs2= 0,

A+B+C=D+E+F,

bD

s1

+bE +bF

s2

=A

s1

+B+C

s2

,

A=DL1,and C=F L2.

This gives us 5 equations with 6 unknowns, but we can multiply all 6 variables by a constant to eliminate

denominators. We can algebraically solve for the constants to produce

A=L2

1(L2ln b−1)(b−L2+ (b−1)L2

2ln b),

B= (L1−L2) ln b(−L1L2(L1+L2) + b(L2

1+L1L2+L2

2) + bL1L2(L1L2−L1−L2) ln b),

C=−L2

2(L1ln b−1)(b−L1+ (b−1)L2

1ln b),

D=L1(L2ln b−1)(b−L2+ (b−1)L2

2ln b),

E= (L1−L2)(L1L2−b(L1+L2−1) + (−L1L2(L1+L2) + b(L2

1+L1L2+L2

2)) ln b

+L1L2(L1L2−L1−L2)(ln b)2),

F=−L2(L1ln b−1)(b−L1+ (b−1)L2

1ln b).

This function may not seem to be that accurate of an initial guess, since the graphs of f(z+ 1) and bf(z)

are visibly diﬀerent. However, we do not need an extremely accurate ﬁrst guess, just one close enough for

9

the double dagger track method to converge. This is akin to the initial guess for Newton’s method, for which

even a crude ﬁrst estimate is good enough for the method to ﬁnd a root.

Coincidentally, when bis close to 1, then f(z+ 1) and bf(z)do become very close together, with the

diﬀerences going to 0 as bapproaches 1. Thus, f(z) is a very close approximation to κb(z) for bclose to 1.

This is interesting because, as we will see later, b= 1 is a singular point for κb(z), since one of the two ﬁxed

points goes to ∞as b→1.

4 The Double dagger track Method

The ﬁrst step in ﬁnding the analytic continuation of κb(z) is to consider bto be outside the Shell-Thron

region. Technically b= 0 is outside this region, and will be a singular point of κb(z), so to avoid this we will

only consider bto be in the connected portion of the exterior of the Shell-Thron region with ℜ(b)≥0. As

long as bis not too close to the boundary, we can modify the cross-track method of [3] to apply to complex

bases. The key diﬀerence is that we cannot use the symmetry Fb(z) = Fb(z), so we will have to use both

ψ1(z) and ψ2(z). We also will modify the contour Ω to one which optimizes computations for ℑ(b)≥0,

shown in ﬁgure 2, involving the parameter A, which will typically have a value of 1.

A: integrate along the line x= 1 from t= 1 −Ai to t= 1 + Ai.

B: integrate along the upper half of the circle x2+ (y−A)2= 1 counterclockwise.

C: integrate along the line x=−1 from t=−1 + Ai to t=−1−Ai.

D: integrate along the line x+ 2y=−1−2Afrom t=−1−Ai to t=−(A+ 1/2)i.

E: integrate along the line x−2y= 1 + 2Afrom t=−(A+ 1/2)ito t=−1 + Ai.

Then by the Cauchy contour integral, we have for zwithin Ω,

κb(z) = 1

2πi Ω

κb(t)

t−zdt.

Note that along the integral A, we can let t= 1 + iy, and use the fact that κb(z+ 1) = bκb(z)to simplify.

Thus, we see that

1

2πi