NUMBER SYSTEMS AND ANALYSIS

Complex Variables

Contents

Complex Functions
Differentiation
Differentiation of Algebraic Expressions
Exponential Function
Logarithm
Trigonometric Functions
Integration
Cauchy's Integral Formula
Applications of Complex Variables

Complex Functions

A function of a complex variable is a mapping f which gives for each complex number z = x + i.y another complex number w = u + i.v. We write:
w = f(z),   u + i.v = f(x + i.y).

EXAMPLES

Differentiation

Let w = f(z), or u + i.v = f(x + i.y). Then dw/dz = d(u + i.v)/d(x + i.y) = (du + i.dv)/(dx + i.dy) = [(du.dx + dv.dy) + i.(dv.dx - du.dy)] / (dx2 + dy2).

Let dy = 0. Then dz = dx, and we have

(dw/dz)dy=0 = (du/dx)dy=0 + i.(dv/dx)dy=0.
Let dx = 0. Then dz = i.dy, and we have
(dw/dz)dx=0 = (dv/i.dy)dx=0 - i.(du/dy)dx=0.

A complex function w = f(z) is differentiable when dw/dz exists and is the same for all directions defined by the ratio dx:dy. Therefore, if f is differentiable, we have

(dw/dz)dx=0 = (dw/dz)dy=0.

This is true if and only if

(du/dx)dy=0 = (dv/dy)dx=0 and (dv/dx)dy=0 = -(du/dy)dx=0.
These two equations are called the Cauchy-Riemann equations.

We may also write the Cauchy-Riemann equations in the following notation, which represents partial differentiation:

ux = vy and vx = -uy.
In polar coordinates the Cauchy-Riemann equations are:
r.ur = vtheta and r.vr = -utheta.

EXAMPLES

The following functions satisfy the Cauchy-Riemann equations, and are differentiable: The following functions do not satisfy the Cauchy-Riemann equations, and are therefore not differentiable:

Differentiation of Algebraic Expressions

Since the laws of algebra for complex numbers are the same as the laws of algebra for real numbers, it follows that the rules for differentiating complex algebraic expressions are the same as the rules for differentiating real algebraic expressions:

Let w1 = f(z) and w2 = g(z). Then:

We may use these laws to differentiate polynomials. If pn(z) is a complex polynomial
pn(z) = zn + a1.zn-1 + ... + an-1.z + an,
then the derivative is
pn'(z) = n.zn-1 + (n - 1).a1.zn-2 + ... +an-1.
We may also use these laws to differentiate rational functions
pm(z) / qn(z),
where pm(z) and qn(z) are polynomials. (The general formula for the derivative of a rational function is long, but not difficult.)

EXAMPLES

Exponential Function

The complex exponential function
exp(x + i.y) = u + i.v
should have the following properties:
  1. exp(x + i.y) = exp(x).exp(iy).
  2. exp(x + i.y) = ex on the real axis (y = 0) in the complex plane.
  3. exp(x + i.y) satisfies the Cauchy-Riemann equations.
We shall find formulas for u and v which give these three properties.

By property (1) we have

u = Re[exp(x + i.y)] = Re[exp(x).exp(iy)] = ex.Re(exp i.y),
v = Im[exp(x + i.y)] = Im[exp(x).exp(iy)] = ex.Im(exp i.y).
From these equations and property (3) we have
Re(exp i.y) = d Im(exp i.y)/dy, . . . (4)
Im(exp i.y) = -d Re(exp i.y)/dy.
Eliminating first Im(exp i.y), and second Re(exp i.y) gives
Re(exp i.y) = -d2Re(exp i.y)/dy2,
Im(exp i.y) = -d2Im(exp i.y)/dy2.
The solutions of these differential equations are
Re(exp i.y) =A cos y + B sin y,
Im(exp i.y) = C cos y + D sin y,
where A, B, C, and D are constants to be found. Using property (2) we get A = 1, C = 0. Using (4) we get B = 0, and D = 1. Therefore u = ex.cos y, and v = i.ex.sin y. In other words:
exp(x + i.y) = ex(cos y + i.sin y).
To calculate the derivative of the exponential function, let dy = 0. Then dz = dx, and we have:
d(exp z)/dz = [d(excos y)/dx]dy=0 + i.[d(exsin y)/dx]dy=0 = excos y + i.exsin y.
But excos y + i.exsin y = exp z. Therefore:
d(exp z)/dz = exp z.

Logarithm

The complex logarithm is the inverse of the complex exponential function. It should have the following properties:
  1. log(z1.z2) = log z1 + log z2.
  2. log(exp z) = z.
Let z = r.(cos theta + i.sin theta). Then z = r.exp(i.theta), and log z = log[r.exp(i.theta)].

Therefore, by property (1), we have log z = log r + log[exp(i.theta)], and by property (2), we have:

log z = log r + i.theta,
where theta may be replaced by any of the values theta + 2.n.pi, n = ±1, ±2, ...

Note that log z has infinitely many values. The principal logarithm (which is single-valued) is:

Log z = log r + i.theta,   where -pi < theta < +pi.
To calculate the derivative of the logarithm, let d(theta) = 0. Then:
d(log z) = d(log r + i.theta) = d(log r) = (1/r)dr, and
dz = d(r.cos theta + i.r.sin theta)= dr(cos theta + i.sin theta) = dr(z/r).
It follows that:
d(log z)/dz = 1/z.

Trigonometric Functions

Since exp(i.y) = cos y + i.sin y, we have:
cos y = (½).(ei.y + e-i.y),   sin y = -i.(½).(ei.y - e-i.y).
Therefore, more generally, we require:
cos(x + i.y) = (½).[exp i.(x + i.y) + exp -i.(x + i.y)],
sin(x + i.y) = -i.(½).[exp i.(x + i.y) - exp -i.(x + i.y)].
These formulas give:
cos(x + i.y) = cos x.cosh y - i.sin x.sinh y,
sin (x + i.y) = sin x.cosh y + i.cos x.sinh y.
To calculate the derivatives of the trigonometric functions, let dy = 0. Then we have dz = dx, and it follows that:
d(cos z)/dz = -sin z,
d(sin z)/dz = cos z.

Integration

The definite integral of a complex function f(z) from an initial point z1 to a final point z2 in the complex plane can be written:
Integral (z1, z2): f(z).dz
= Integral (x1 + i.y1, x2 + i.x2): (u + i.v).(dx + i.dy)
= Integral (x1 + i.y1, x2 + i.x2): [(u.dx - v.dy) + i.(v.dx + u.dy)].
The path of integration in the complex plane from z1 to z2 (see Fig.1) may be given by x = g(t), dx = g'(t).dt, y = h(t), dy = h'(t).dt, from t1 to t2.

Fig. 1.

Fig. 1. Path from initial point z1 to final point z2.

Then the integral becomes:

Integral (t1, t2): [u.g'(t) - v.h'(t)].dt + i.Integral (t1, t2): [v.g'(t) + u.h'(t)].dt.

EXAMPLE

Let f(z) = z2, z1 = 1, z2 = i. Let the path of integration be the straight line from z1 to z2. On this path we may let x = 1 - t, y = t, and integrate from t = 0 to t = 1. Then u = 1 - 2t, v = 2(1 - t)t, g'(t) = -1, h'(t) = 1, and the integral becomes:
Integral (0, 1): [(1 - 2t).(-1) - 2(1 - t)t.(1)].dt + i.Integral (0, 1): [2(1 - t)t.(-1) + (1 - 2t).(1)].dt
= -(1/3) - i.(1/3).

Suppose the complex function f(z) is differentiable on and within a closed curve C in the complex plane, and let R be the region inside the closed curve C. The integral of f(z) around C is:

Integral (C): f(z).dz = Integral (C): [(u.dx - v.dy) + i.(v.dx + u.dy)].
By Green's theorem, we have:
Integral (C): (u.dx - v.dy) = Integral (R): (-vx - uy).dx.dy,
Integral (C): (v.dx + u.dy) = Integral (R): (ux - vy).dx.dy,
and, by the Cauchy-Riemann equations, the double integrals over R are zero. Therefore:
Integral (C): f(z).dz = 0, where f is differentiable on and inside C.
It follows that:
In a region R of the complex plane where f(z) is everywhere differentiable the integral of f(z) from a given initial point z1 to a given final point z2 has the same value for every path between the two points.

EXAMPLE

The function f(z) = z2 is differentiable everywhere in the complex plane. Therefore the integral of z2 from z1 to z2 is the same for all paths. Let the path of integration from z1 = 1 to z2 = i be the unit circle from theta = 0 to theta = pi/2. Then
Integral (1, i): z2.dz
= Integral (0, pi/2): [cos(2.theta) + i.sin(2.theta)].[-sin theta + i.cos theta].d(theta)
= -(1/3) - i.(1/3).
This is the same as the value of the integral along the straight path from 1 to i.

EXAMPLE

The function f(z) = 1/z is differentiable everywhere in the complex plane except at the point z = 0. The value of
Integral (1, -1): (1/z).dz
over a path where y > 0 (see Fig. 2) is i.pi. But over a path where y < 0 the value of the integral is -i.pi.

Fig. 2.

Fig. 2. Alternative paths for the integration of 1/z from +1 to -1.

The two values of the integral are not equal because 1/z is not differentiable at z = 0, and the point 0 is in the region enclosed by the two paths.

Cauchy's Integral Formula

Let f(z) be differentiable on and inside a closed curve C, and let zo be any point inside C. Then
Integral (C): f(z).dz / (z - zo) = 2.pi.i.f(zo).
This is called Cauchy's integral formula.

PROOF: Let Co be a small circle with center zo and radius ro inside the curve C. (See Fig. 3)

Fig. 3.

Fig. 3. Diagram for the proof of Cauchy's integral formula.

Since f(z)/(z - zo) is differentiable on and inside the complete path shown by the arrows in Fig. 3, it follows that

Integral : f(z).dz / (z - zo)
integrated around the complete path is zero. But the integrals in opposite directions along AB and BA cancel. Therefore the integral around C equals the integral around Co (against the arrows). Then
Integral (C): f(z).dz / (z - zo)
= Integral (0, 2.pi): f(zo + ro.exp i.theta).d(zo + ro.exp i.theta) / (ro.exp i.theta).
But d(zo + ro.exp i.theta) = i.(ro.exp i.theta).d(theta). Therefore
Integral (C): f(z).dz / (z - zo) = Integral (0, 2.pi): f(zo + ro.exp i.theta).i.d(theta).
Let ro go to zero. Then we get
Integral (C): f(z).dz / (z - zo) = 2.pi.i.f(zo),
because f(z) is continuous.

Differentiating Cauchy's integral formula n times with respect to zo we get:

Integral (C): f(z).dz / (z - zo)n+1 = 2.pi.i.f(n)(zo) / n!.

Applications of Complex Variables

There are many applications of complex variables. Some of them are listed below.

In mathematics:

In engineering and physics:

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By R. H. B. Exell, 1998, revised 2003. King Mongkut's University of Technology Thonburi.