Complex-Variable Theory

5.1 Analytic Functions

A complex-valued function f (z) of a complex variable z is differentiable at z with derivative f⁰(z) if the limit

f⁰(z) = lim

z⁰→z

f (z⁰) − f (z)

z⁰− z (5.1)

exists as z⁰ approaches z in the complex plane. The interesting point is that the limit must exist no matter how or from what direction the variable z⁰ approaches z.

If the function f (z) is differentiable in a small region, e.g., a small disk, around a point z₀, then f (z) is said to be analytic at z₀.

Example, Polynomials: Suppose f (z) = zⁿ is an integral power of z.

Then for tiny dz and z⁰= dz + z, the difference f (z⁰) − f (z) is

f (z⁰) − f (z) = (z + dz)ⁿ− zⁿ≈ nzⁿ⁻¹dz (5.2) and so the limit

zlim⁰→z

f (z⁰) − f (z)

z⁰− z = lim

dz→0

nzⁿ⁻¹dz

dz = nzⁿ⁻¹ (5.3)

exists independently of the direction from which z⁰ approaches z. Thus the function zⁿ is analytic for all z with derivative

dzⁿ

dz = nzⁿ⁻¹. (5.4)

A function that is everywhere analytic is entire. All polynomials P (z) =

N

X

n=0

cnzⁿ (5.5)

5.2 Cauchy’s Integral Theorem 173 are entire.

5.2 Cauchy’s Integral Theorem Suppose f (z) is analytic at z0. Then to first order in z − z0

f (z) = f (z₀) + f⁰(z₀) (z − z₀) (5.6) near z₀. Let’s compute the contour integral of f (z) along a small circle of radius  and center z0. The points on the contour are

z = z0+  e^iθ (5.7)

for θ ∈ [0, 2π]. So dz = i e^iθdθ, and the contour integral is I

f (z) dz = Z 2π

0

f(z0) + f⁰(z0) (z − z0) i e^iθdθ. (5.8) Since z − z₀ =  e^iθ, the contour integral breaks into two pieces

I

f (z) dz = f (z0) Z 2π

0

i e^iθdθ + f⁰(z0) Z 2π

0

 e^iθi e^iθdθ (5.9) which vanish because the θ-integrals are zero. So the contour integral of the analytic function f (z)

I

f (z) dz = 0 (5.10)

is zero around the tiny circle — at least to order ².

What about the contour integral of an analytic function f (z) around a tiny square of size ? Again, we use the analyticity of f (z) at z = z₀ to expand it as

f (z) = f (z0) + f⁰(z0) (z − z0) (5.11) on the tiny square. Now the contour consists of the four complex segments dz_nare , i , −, and −i , so that dz_n=  iⁿ⁻¹for n = 1, 2, 3, 4. The centers of these segments are displaced from z₀ by −i /2, /2, i /2, and −/2, so z_n− z₀ = iⁿ⁺²/2. The contour integral of f (z) around the square then is

I

f (z) dz =

4

X

n=1

f (z_n) dz_n=

4

X

n=1

f(z0) + (z_n− z₀) f⁰(z₀) dzn. (5.12) Again the contour integral splits into two pieces, one involving f (z0), and

the other involving the derivative f⁰(z0) I

f (z) dz = f (z0) I1+ f⁰(z0) I2. (5.13) The four complex segments dz_n add up to a path that goes around the square and ends where it started, so the first piece f (z₀)I₁ is zero

f (z₀) I₁= f (z₀) [ + i  + (−) + (−i )] = 0. (5.14) And so is the second one f⁰(z0)I2

f⁰(z₀) I₂ = f⁰(z₀)

4

X

n=1

(z_n− z₀) dz_n

= f⁰(z0) [(−i /2)  + (/2) i  + (i /2) (−) + (−/2) (−i )]

= f⁰(z₀) (²/2) [(−i ) + i + (−i ) + i] = 0. (5.15) So the contour integral of an analytic function f (z) around a tiny square of side  is zero to order ². Thus, the integral around such a square can be at most of order ³. This is very important. We’ll use it to prove Cauchy’s integral theorem.

Let’s now consider a function f (z) that is analytic on a square of side L, as pictured in Fig. 5.1. The contour integral of f (z) around the square can be expressed as the sum of L²/² contour integrals around tiny squares of side

. All interior integrals cancel, leaving the integral around the perimeter.

Each contour integral around its tiny square is at most of order ³. So the sum of the L²/² tiny contour integrals is at most (L²/²) ³ =  L², which vanishes as  → 0. Thus the contour integral of a function f (z) along the perimeter of a square of side L vanishes if f (z) is analytic on the perimeter and inside the square. This is an example of Cauchy’s integral theorem.

Suppose a function f (z) is analytic in a region R and that I is a contour integral along a straight line within that region from z1 to z2. The contour integral of f (z) around any square inside the region R of analyticity is zero.

So by successively adding contour integrals around small squares to the straight-line contour integral, one may deform the straight-line contour into an arbitrary contour from z1 to z2 without changing its value.

So if a function f (z) is analytic in a region R, then its integral I along a contour C from z1 to z2 remains invariant as we continuously deform its contour C to C⁰ as long as these contours and all the intermediate contours lie entirely within the region R and the end points z1 and z2 are kept fixed

5.2 Cauchy’s Integral Theorem 175

Figure 5.1 A contour integral around a big L × L square is equal to the sum of the contour integrals around the L²/² tiny  ×  squares that tile the big square.

and lie within R:

I = Z z2

z1C

dz f (z) = Z z2

z1C⁰

dz f (z). (5.16)

Thus a contour integral depends upon its end points and upon the function f (z) but not upon the actual contour as long as the deformations of the contour do not push it outside of the region R of analyticity.

If the end points z1 and z2 are the same, then the contour C is closed, and

Figure 5.2 As long as the four contours are within the domain of analyt-icity and have the same end-points, the value of the contour integral is unchanged.

we write the integral as

I = I z1

z1C

dz f (z) ≡ I

C

dz f (z) (5.17)

with a little circle to denote that the contour is a closed loop. The value of that integral is independent of the contour as long as our deformations of the contour keep it within the domain of analyticity of the function and as long as the contour starts and ends at z1 = z2. Now suppose that the function f (z) is analytic at all points within the contour. Then we can shrink the contour, staying within the domain of analyticity of the function, until the area enclosed is zero and the contour is of zero length — all this without changing the value of the integral. But the value of the integral along this null contour of zero length is zero. Thus the value of the original contour integral also must be zero

I z1

z1C

dz f (z) = 0. (5.18)

Thus we arrive at Cauchy’s integral theorem: The contour integral of a function f (z) around a closed contour C lying entirely within the domain R

5.2 Cauchy’s Integral Theorem 177 of analyticity of the function vanishes

I

C

dz f (z) = 0 (5.19)

as long as the function f (z) is analytic at all points within the contour.

A region in the complex plane is simply connected if we can shrink every loop in the region to a point while keeping the loop in the region. A slice of American cheese is simply connected, but a slice of Swiss cheese is not. A dime is simply connected, but a washer isn’t. The surface of a sphere is, but the surface of a bagel isn’t.

With this definition, we can restate the integral theorem of Cauchy (1789–

1857): The contour integral of a function f (z) around a closed contour C vanishes

I

C

dz f (z) = 0 (5.20)

if the contour lies within a simply connected domain of analyticity of the function.

If a region R is simply connected, then we may deform any contour C from z1 to z2 in R into any other contour C⁰ from z1 to z2 in R while keeping the moving contour in the region R. So another way of understanding the Cauchy integral theorem is to ask, What is the value of the contour integral

IM = Z z1

z2C⁰

dz f (z)? (5.21)

This integral is the same as the integral along C⁰ from z1 to z2, except for the sign of the dz’s and the order in which the terms are added. Thus

I_M = Z z1

z2C⁰

dz f (z) = − Z z2

z1C⁰

dz f (z). (5.22)

Now consider a closed contour running along the contour C from z₁ to z₂ and backwards along C⁰ from z₂ to z₁ all within a simply connected region R of analyticity. Since I_M = −I, the integral of f (z) along this closed contour vanishes:

I

dz f (z) = I + I⁰ = I − I = 0 (5.23) and we have again derived Cauchy’s integral theorem.

Example: Every polynomial P (z) =

N

X

n=0

cnzⁿ (5.24)

is analytic everywhere, and so the integral of any polynomial along any closed contour C is zero

I

P (z) dz = 0. (5.25)

5.3 Cauchy’s Integral Formula

Let’s now consider a simply connected region R in which a function f (z) is analytic. Pick a point z₀ well inside the region R . We will compute the value of a closed contour integral around the point z₀ . We take the contour to be a tiny counterclockwise circle of radius  with center at the point z₀. Then the points on the contour are

z = z₀+  e^iθ (5.26)

for θ ∈ [0, 2π], and so dz = i e^iθdθ. As our function in this contour integral I0, we’ll pick not f (z) but f (z)/(z − z0). So since z − z0 =  e^iθ, the contour integral I0 in the limit  → 0 is

I0=

I f (z) z − z0

dz = Z 2π

0

[f (z0) + f⁰(z0) (z − z0)]

z − z0

i e^iθdθ

= Z 2π

0

f(z0) + f⁰(z0)  e^iθ

 e^iθ i e^iθdθ

= Z 2π

0

h

f (z0) + f⁰(z0)  e^iθi

idθ = 2πi f (z0) (5.27) since the θ-integral involving f⁰(z0) vanishes. This result

f (z0) = 1 2πi

I



f (z) z − z0

dz (5.28)

is a miniature version of Cauchy’s integral formula, in which the subscript

 means that the contour is around a tiny circle.

Now consider a big counterclockwise contour C⁰ within the simply con-nected region R in which a function f (z) is analytic as in Fig. 5.3. The whole contour lies within a simply connected part of R in which the ratio f (z)/(z − z₀) is analytic. So the integral along the contour C⁰ vanishes

0 = 1 2πi

I

C⁰

f (z)

z − z₀ dz. (5.29)

We let the two straight-line segments approach each other so that they

5.3 Cauchy’s Integral Formula 179

z

Figure 5.3 The full contour is the sum of a big ccw contour and a small cw contour, both around z0.

cancel. Now split the contour C⁰ into a contour C that is a big counter-clockwise contour around z₀ and a tiny clockwise circle of radius  around z₀. That tiny clockwise-circle integral is the negative of the corresponding counterclockwise-circle integral, so Using the miniature result (5.28), we find

f (z0) = 1

which is Cauchy’s integral formula.

Let’s now use this formula to compute the first derivative f⁰(z) of f (z):

f⁰(z) = f (z + dz) − f (z)

which is what we’d have gotten had we simply differentiated behind the integral sign.

If we continue in this way, we find that the second derivative f⁽²⁾(z) of f (z) is And the nth derivative f⁽ⁿ⁾(z) is

f⁽ⁿ⁾(z) = n!

2πi I

dz⁰ f (z⁰)

(z⁰− z)ⁿ⁺¹, (5.35) where the contour runs counter-clockwise about the point z and all the points inside the contour are within the domain R in which f (z) is analytic. Thus, if a function f (z) is analytic in a region R, then it is infinitely differentiable there.

Example 5.1(Bessel Functions of the First Kind) A simple application of Cauchy’s integral formula is the counter-clockwise integral around the unit circle z = e^iθ of the ratio z^m/zⁿ in which both m and n are integers integral is 2π/2π = 1. Thus, the original integral is the Kronecker delta

1 2πi

I

dzz^m

zⁿ = δm+1,n. (5.38)

The generating function for Bessel functions J_m of the first kind is e^t(z−1/z)/2=

∞

X

m=−∞

z^mJm(t). (5.39)

Applying our integral formula (5.38) to it, we find 1

5.4 The Cauchy-Riemann Conditions 181 Thus letting z = e^iθ, we have

Jn(t) = 1 2π

Z 2π 0

dθ exp

"

t e^iθ− e^−iθ

2 − inθ

#

(5.41)

or more simply (problem 1) Jn(t) = 1

2π Z 2π

0

dθ ei(t sin θ−nθ) = 1 π

Z π 0

dθ cos(t sin θ − nθ). (5.42)

5.4 The Cauchy-Riemann Conditions

Why do the real and imaginary parts u and v of an analytic function f (z) = f (x + iy) = u(x, y) + iv(x, y) (5.43) satisfy the Cauchy-Riemann conditions

∂u

∂x = ∂v

∂y and ∂u

∂y = −∂v

∂x ? (5.44)

The reason is that the derivative f⁰ is independent of the direction from which dz → 0. If we use subscripts for partial differentiation, ux = ∂u/∂x and uy = ∂u/∂y, etc., then the derivative in the x-direction is

df

dx = (u_x+ iv_x) (5.45)

while that in the iy-direction is df

diy = 1

i (u_y + iv_y) = −i (u_y+ iv_y) . (5.46) Equating the two directional derivatives, we get the Cauchy-Riemann con-ditions

u_x = v_y and u_y = −v_x. (5.47) The directions in the x-y plane in which the real u(x, y) and imaginary v(x, y) parts of an analytic function increase most rapidly are given by the vectors (ux, uy) and (vx, vy). The Cauchy-Riemann conditions (5.47) imply that these directions must be perpendicular

(u_x, u_y) · (v_x, v_y) = u_xv_x+ u_yv_y = v_yv_x− v_xv_y = 0. (5.48) The Cauchy-Riemann conditions (5.47) and a two-dimensional version of

Stokes’s theoremprovide an alternative proof of Cauchy’s integral theo-rem (5.19). We recall that Stokes’s theotheo-rem says that the line integral of a vector field V(r) along a closed contour C is equal to the surface integral of the curl of V(r) over any surface bounded by the contour

Z

The direction of the normal vector da is given by the right-hand rule: If one curls the index finger of one’s right hand parallel to a representative fraction of the contour, then one’s right thumb will point more or less in the direction of da.

Let us try to express the line integral of a not necessarily analytic function f (x, y) = u(x, y) + iv(x, y) along a closed ccw contour C as an integral over the surface enclosed by the contour. The contour integral is

I Now since the contour C is counterclockwise, the differential dx is negative at the top of the curve with coordinates (x, y+(x) and positive at the bottom (x, y−(x). So the first line integral is the surface integral

I

in which da = |dxdy| is a positive element of area. Similarly, we find i

The dy integrals are then:

−

Combining (5.50–5.54), we find I

5.4 The Cauchy-Riemann Conditions 183 This formula holds whether or not the function f (x, y) is analytic. But if f (x, y) is analytic on and within the contour C, then it satisfies the Cauchy-Riemann conditions (5.47) within the contour, and so both surface integrals vanish. The contour integral then is zero, which is Cauchy’s integral theorem (5.19).

The extent to which the contour integral of the function f (x, y) = u(x, y)+

iv(x, y) differs from zero, its value if f (x, y) is analytic in z = x+iy, is related to surface integrals of u_y+ v_x and u_x− v_y which vanish when f = u + iv satisfies the Cauchy-Riemann conditions (5.47).

Example: The Integral of Nonanalytic Function. The integral formula (5.55) can be useful for evaluating contour integrals of functions that are not analytic. The function

f (x, y) = 1 x + iy + i

1

1 + x²+ y² (5.57)

is the product of an analytic function 1/z and a nonanalytic real one r(x, y) = 1/(z^∗z) . The real and imaginary parts of f are in which 1/z = u + iv. Let’s use Eq.(5.55) to compute the contour integral I of f along the real axis from −∞ to ∞ and then around the upper half plane along z = x + iy = Re^iθ as θ runs from 0 to π and R → ∞ The contour integral around the UHP is an example of the ghost contours discussed in Sec.(5.13). Because u and v satisfy the Cauchy-Riemann con-ditions (5.47), the terms in the area integral simplify to

−Uy− Vx= −ury− vrx (5.61) and

−Vy+ Ux= −vry+ urx. (5.62)

So the integral I is I =

Z ∞

−∞

dx Z ∞

0

dy [−ury− vrx+ i(−vry+ urx)] (5.63) or explicitly

I = Z ∞

−∞

dx Z ∞

0

dy −2x − 2i(x²+ y²+ y)

[x²+ (y + )²] (1 + x²+ y²)². (5.64) We let  → 0 and find

I = −2i Z ∞

−∞

dx Z ∞

0

dy 1

(1 + x²+ y²)². (5.65) Changing variables to ρ²= x²+ y², we have

I = −4πi Z ∞

0

dρ ρ

(1 + ρ²)² = 2πi Z ∞

0

dρ d dρ

1

1 + ρ² = −2πi. (5.66)

5.5 Harmonic Functions

Let’s use the Cauchy-Riemann conditions (5.47) to compute the laplacian of the real part u of an analytic function f . First, the second x-derivative uxx is

uxx = vyx= vxy = −uyy. (5.67) So the real part u of an analytic function f is a harmonic function

u_xx+ u_yy= 0 (5.68)

or equivalently one with a vanishing laplacian. Similarly,

vxx= −uyx= −vyy (5.69)

so the imaginary part of an analytic function f also is a harmonic function

vxx+ vyy = 0. (5.70)

A harmonic function h(x, y) can have saddle points, but not local minima or maxima because at a local minimum of both hxx> 0 and hyy > 0, while at a local maximum both hxx < 0 and hyy < 0. So in its domain of analyticity, the real and imaginary parts of an analytic function f have neither minima nor maxima.

For static fields, the electrostatic potential φ(x, y, z) is a harmonic function

5.5 Harmonic Functions 185 of the three spatial variables x, y , and z in regions that are free of charge because the electric field is

E= −∇φ (5.71)

and its divergence vanishes

∇ · E = 0 (5.72)

where the charge density is zero. Thus the laplacian of the electrostatic potential φ(x, y, z) vanishes

∇ · ∇φ = φ_xx+ φ_yy+ φ_zz = 0 (5.73) and so φ(x, y, z) is harmonic. Note that in the presence of point charges, the location of each positive charge is a local maximum of the electrostatic potential φ(x, y, z) and the location of each negative charge is a local mini-mum of φ(x, y, z). But in the absence of charges, the electrostatic potential has neither local maxima nor local minima. For this reason, one cannot trap charged particles with an electrostatic potential, a result known as Earnshaw’s theorem.

We have seen that the real and imaginary parts of an analytic function are harmonic functions with two-dimensional gradients that are mutually perpendicular. And we know that the electrostatic potential is a harmonic function. Thus the real part u(x, y) (or the imaginary part v(x, y)) of any analytic function f (z) = u(x, y) + iv(x, y) describes the electrostatic poten-tial φ(x, y) for some electrostatic problem that does not involve the third spatial coordinate z. The surfaces of constant u(x, y) are the equipotential surfaces, and since the two gradients are orthogonal, the surfaces of constant v(x, y) are the electric field lines.

Examples of Two-dimensional Potentials: The function

f (z) = Ez = Ex + iEy (5.74)

has u = Ex and v = Ey. It may represent a potential V (x, y, z) = Ex for which the electric-field lines E = −E ˆx are lines of constant y. Equiva-lently, it may represent a potential V (x, y, z) = Ey in which E points in the negative y-direction, which is to say along lines of constant x.

Another simple example is the function

f (z) = z²= x²− y²+ 2ixy (5.75) for which u = x² − y² and v = 2xy. This function gives us a potential V (x, y, z) whose equipotentials are the hyperbolas

u = x²− y² = c² (5.76)

and whose electric-field lines are the perpendicular hyperbolas

v = 2xy = d². (5.77)

Equivalently, we may take these last hyperbolas (5.77) to be the equipoten-tials and the other ones (5.77) to be the lines of the electric field.

As a third example, the function of z = re^iθ = exp(ln r + iθ) of charge per unit length λ = q/L. The electric-field lines are the lines of constant v, that is, of constant θ

E = λ

2π₀

(x, y, 0)

x²+ y². (5.79)

5.6 Taylor Series for Analytic Functions

Let’s consider the contour integral of the function f (z⁰)/(z⁰ − z) along a circle C inside the region R in which f (z) is analytic. For any point z inside the circle, Cauchy’s integral formula (5.31) tells us that

f (z) = 1

Now, we rewrite the denominator z⁰ − z in terms of the center z0 of the circle: Next, we factor the denominator

f (z) = 1 than unity, so the power series

 converges absolutely and uniformly on the circle by (4.23–4.26). We are

5.6 Taylor Series for Analytic Functions 187

z

0

z

z

Figure 5.4 Contour integral for Taylor series.

therefore allowed to integrate the series f (z) = 1

2πi I

C

f (z⁰) dz⁰ z⁰− z0

∞

X

n=0

 z − z0

z⁰− z0

n

(5.84)

term by term

f (z) =

∞

X

n=0

(z − z₀)ⁿ 1 2πi

I

C

f (z⁰) dz⁰

(z⁰− z0)ⁿ⁺¹. (5.85) By Eq.(5.35), the integral is just the nth derivative f⁽ⁿ⁾(z) divided by

n-factorial. Thus the function f (z) possesses the Taylor series f (z) =

∞

X

n=0

(z − z0)ⁿ

n! f⁽ⁿ⁾(z₀) (5.86)

which converges as long as the point z is inside a circle centered at z₀ that lies within a simply connected region R in which f (z) is analytic.

5.7 Cauchy’s Inequality

Suppose a function f (z) is analytic in a region that includes the disk with perimeter

z = R e^iθ (5.87)

and that f (z) is bounded by

|f (z)| ≤ M (5.88)

on that circle. Then by using Cauchy’s integral formula (5.35), we may bound the nth derivative f⁽ⁿ⁾(0) of f (z) at z = 0 by

|f⁽ⁿ⁾(0)| ≤ n!

2π

I |f (z)||dz|

|z|ⁿ⁺¹

≤ n!M 2π

Z 2π 0

R dθ

Rⁿ⁺¹ = n!M

Rⁿ (5.89)

which is called Cauchy’s inequality.

5.8 Liouville’s Theorem

Suppose now that f (z) is analytic everywhere (such functions are called entire) and is bounded by

|f (z)| ≤ M for all |z| ≥ R₀. (5.90) Then by applying Cauchy’s inequality (5.89) at successively larger values of R, we have

|f⁽ⁿ⁾(0)| ≤ lim

R→∞

n!M

Rⁿ = 0 for n ≥ 1 (5.91)

which shows that every derivative of f (z) vanishes

f⁽ⁿ⁾(0) = 0 for n ≥ 1 (5.92)

5.9 The Fundamental Theorem of Algebra 189 at z = 0. But then the Taylor series (4.53) about z = 0 for the function f (z) consists of only a single term, and f (z) is a constant

f (z) =

∞

X

n=0

zⁿ

n! f⁽ⁿ⁾(0) = f⁽⁰⁾(0) = f (0). (5.93) In particular, any function that is everywhere bounded and analytic is a constant, which is Liouville’s theorem.

5.9 The Fundamental Theorem of Algebra Gauss applied Liouville’s theorem to the function

f (z) = 1 PN(z) =





N

X

j=0

c_jz^j





−1

(5.94) which is the inverse of an arbitrary polynomial of order N . Suppose that the polynomial PN(z) had no zero, that is, no root anywhere in the complex plane. Then f (z) would be analytic everywhere. Moreover, for sufficiently large |z|, the polynomial PN(z) is approximately PN(z) ≈ cNz^N, and so f (z) is bounded by something like

|f (z)| ≤ 1

|c_N|R^N₀ ≡ M for all |z| ≥ R₀. (5.95) So if P_N(z) had no root, then the function f (z) would be analytic every-where and would satisfy condition (5.90); hence f (z) would be a constant, Eq.(5.93). But of course, f (z) = 1/P_N(z) is not a constant unless N = 0.

Thus, the polynomial PN(z) must have a root, a pole of f (z), so that f (z) is not entire; this is the only exit from the contradiction.

If the root of PN(z) is at z = z1, then PN(z) = (z − z1) PN −1(z), in which PN −1(z) is a polynomial of order N − 1, and we may repeat the argument for its reciprocal f1(z) = 1/PN −1(z). In this way, one arrives at the fundamental theorem of algebra: Every polynomial PN(z) =PN

j=0cjz^j has N roots somewhere in the complex plane

PN(z) = cN N

Y

j=1

(z − zj). (5.96)

5.10 Laurent Series

Consider a function f (z) that is analytic between an outer circle C1of radius R1 and an inner circle C2 of radius R2, as in Fig. 5.5. We will integrate f (z)

z0

z

*

*

*

*

*

*

Figure 5.5 The contour consisting of two concentric circles with center at z0encircles the point z in a counter-clockwise sense. The asterisks are poles or other singularities of the function f (z).

along a contour C₁₂that encircles the point z in a counter-clockwise fashion by following C₁counter-clockwise and C₂clockwise and a line joining them in both directions. By Cauchy’s integral formula (5.31), this contour integral yields f (z)

f (z) = 1 2πi

I

C12

f (z⁰)dz⁰

z⁰− z . (5.97)

The integrations in opposite directions along the line joining C₁ and C₂ can-cel, and we are left with a counter-clockwise integral around the outer circle C₁ and a clockwise one around C₂ or minus a counter-clockwise integral around C₂

f (z) = 1 2πi

I

C1

f (z⁰)dz⁰ z⁰− z − 1

2πi I

C2

f (z⁰⁰)dz⁰⁰

z⁰⁰− z . (5.98) Now from the figure (5.5), the center z0 of the two concentric circles is

5.10 Laurent Series 191 closer to the points z⁰⁰ on the inner circle C2 than it is to z and also closer to z than to the points z⁰ on C1

To use these inequalities, as we did in Eq.(5.83), we add and subtract z₀ from each of the denominators and absorb the minus sign before the second integral into its denominator

f (z) = 1 After factoring the two denominators

f (z) = 1 we expand them, as in Eq.(5.83), in power series that converge absolutely and uniformly on the two contours

f (z) = Having removed the point z from the two integrals, we now we apply

f (z) = Having removed the point z from the two integrals, we now we apply

In document Cahill-Physical Mathematics What Physicists and Engineers Need to Know (Page 180-200)