In this chapter we consider selected functions of a real variable that were
introduced in ... f (z) = na0zn−1 +(n−1)a1zn−2 +···+an−1 which also exists for all
values.
59
Chapter 3 Elementary Functions In this chapter we consider selected functions of a real variable that were introduced in calculus. We shall learn how to represent these functions when complex variables replace the real variables. Each function which is single-valued can be viewed as mapping a point z = x + iy of the z -plane to an image point ω = u + iv of the ω-plane. We investigate and define selected elementary functions and in later chapters we illustrate how these functions can be employed to analyze applied problems.
Polynomial Functions A polynomial function of degree n has the form ω = f(z) = a0z n + a1z n−1 + · · · + an−1 z + an ,
(3.1)
where a0, a1, . . . , an are complex constants. These type of functions exist for all values z and = f 0 (z) = na0z n−1 +(n−1)a1z n−2 +· · ·+an−1 which also exists for all values have the derivative dω dz of z . Consequently, polynomial functions are analytic in the entire finite z-plane and so are called entire functions. Note that a polynomial of degree n ≥ 1 can be expressed a product of linear factors f(z) = a0 (z − z1 )(z − z2 ) · · · (z − zn ). The polynomial equation f(z) = 0 thus has n roots z1 , z2, . . . , zn. The fact that every polynomial of degree n has n roots is sometimes referred to as the fundamental theorem of algebra.
Example 3-1. Let f(z) = a0 z n + a1 z n−1 + · · · + an−1z + an denote a polynomial with real coefficients a0 , a1, . . ., an and assume that β = β1 + iβ2 is a complex root of this equation. We would then have f(β) = a0β n + a1 β n−1 + · · · + an−1β + an = 0
Now take the conjugate of this equation to obtain a0β n + a1 β n−1 + · · · + an−1β + an = 0
The conjugate of a sum is the sum of the conjugates of each term and the conjugate of a product is the product of the conjugate terms and because the coefficients are real, there results the equation n
f(β) = a0 β + a1 β
n−1
+ · · · + an−1β + an = 0
60 This shows that if β is a complex root of the polynomial equation f(z) = 0, with real coefficients, then the conjugate β is also a root of this equation. This shows the complex roots of polynomial equations with real coefficients must occur in conjugate pairs.
Rational algebraic functions Rational algebraic functions can be expressed as the quotient of polynomials. That is, one can write ω = f(z) =
P (z) , Q(z)
(3.2)
where both P (z) and Q(z) are polynomial functions. These functions have the derivative Q(z)P 0 (z)−P (z)Q0 (z) dω 0 and consequently rational algebraic functions are analytic at dz = f (z) = Q2 (z) all points in the z-plane except at those points where the denominator Q(z) is zero. If z0 is a point where Q(z) is zero, then z0 is called a singular point of the function f(z) or one says that a singularity exists at the point z0 . If z0 is a zero of Q(z), then the function ω = f(z) = P (z)/Q(z) is not defined at z0 , and hence it can not be analytic at z0 . The linear fractional transformation ω = az+b cz+d , where a, b, c, d are constants such that ad − bc 6= 0, is an example of a rational algebraic function.
Linear fractional transformation The linear fractional transformation (sometimes referred to as a bilinear or M¨obius1 transformation) has the form ω = f(z) =
az + b , cz + d
(3.3)
where a, b, c, d are constants and ad − bc 6= 0.
Example 3-2. The linear fractional transformation
az+b cz+d
, where a, b, c, d are constants such that ad − bc = 6 0, has the property that if z1 , z2, z3 , z4 are distinct point in the z -plane, then the quantity ω=
(z4 − z1 )(z2 − z3 ) , (z4 − z3 )(z2 − z1 )
called a cross ratio, is an invariant under the linear fractional transformation. To prove this statement calculate the cross ratio associated with the image points ω1 =
az1 + b , cz1 + d
ω2 =
az2 + b , cz2 + d
ω3 =
az3 + b , cz3 + d
ω4 =
az4 + b cz4 + d
We form the cross ratio az2 + b az3 + b az4 + b az1 + b − − (ω4 − ω1)(ω2 − ω3) cz4 + d cz1 + d cz + d cz3 + d 2 = az4 + b az3 + b az2 + b az1 + b (ω4 − ω3)(ω2 − ω1) − − cz4 + d cz3 + d cz2 + d cz1 + d
1
August Ferdinand M¨ obius (1790-1868) German mathematician.
(3.4)
61 and simplify the right-hand side of this equation to obtain the result. (ω4 − ω1)(ω2 − ω3) (z4 − z1)(z2 − z3 ) = (ω4 − ω3)(ω2 − ω1) (z4 − z3)(z2 − z1 )
(3.5)
which demonstrates that the cross ratio is an invariant of the linear fractional transformation. The cross ratio property of the linear fractional transformation can be used to map any three distinct points from the z -plane to three distinct points of the ω-plane. This is accomplished by writing the cross ratio in the form (ω − ω1)(ω2 − ω3) (z − z1)(z2 − z3 ) = (ω − ω3)(ω2 − ω1) (z − z3)(z2 − z1 )
(3.6)
The linear fractional transformation can be described by using a sequence or succession of elementary mappings involving translation, rotation and inversion with respect to the unit circle. To show this assume that c 6= 0 and then write equation (3.3) in the form a ω= c
z+
b a
+
z+
d c d c
−
d c
!
a = c
1+
b a
−
z+
d c d c
!
which simplifies to ω=
a K + , c cz + d
where K =
bc − ad . c
(3.7)
This form of the linear fractional transformation shows that it can be obtained from the following sequence of elementary transformations: ω1 = cz
a rotation of z with contraction or expansion
ω2 = ω1 + d a translation of ω1 , ω2 = cz + d 1 1 ω3 = an inversion of ω2, ω3 = ω2 cz + d ω4 = Kω3 ω=
a + ω4 c
a rotation of ω3 , with contraction or expansion, a translation of ω4 , ω =
Figure 3-1.
ω4 =
K cz + d
a K + c cz + d
An example of sequence of elementary mappings.
62 By representing the linear fractional transformation as a sequence of transformation we can utilize four intermediate complex planes ω1 , ω2, ω3 and ω4 between the z and ω planes to represent the mapping. An example is given in the figure 3-1. Note that each intermediate complex plane is a mapping constructed using one of the previous elementary transformations of translation, rotation (with expansion or contraction) and inversion with respect to the unit circle. Our previous analysis of the equation a(x2 + y2 ) + bx + cy + d = 0 can now be applied to the above sequence of elementary mappings. One finds that the linear fractional transformation transforms lines into circles or lines into lines. Also circles map to circles or circles map to straight lines.
Example 3-3.
Figure 3-2.
Mapping of unit disk to upper half plane.
A mapping of special interest in certain applications of complex variable theory is the linear fractional transformation which maps the unit disk |z| ≤ 1 to the upper half plane Im {ω} ≥ 0 The mapping constructed is such that the points z1 = i, z2 = −1, z3 = −i map onto the points ω1 = −1, ω2 = 0, ω3 = 1 as illustrated in the figure 3-2. Solution: We use the cross ratio property of the linear fractional transformation and substitute the values z1 , z2 , z3, ω1, ω2, ω3 into the cross ratio equation (3.6) to obtain [ω − (−1)](0 − 1) (z − i)[−1 − (−i)] = . (ω − 1)[0 − (−1)] [z − (−i)](−1 − i) z+1
Solving for ω we find ω = −i . This is the transformation which maps the unit disk to the z−1 upper half plane. One can verify the mapping of one or two specific points to check that the mapping is the one desired. For example, if z = re i θ with 0 ≤ r ≤ 1 one can verify the relation ω = −i
re i θ + 1 1 − r2 −2r sin θ + i . = re i θ − 1 1 − 2r cos θ + r2 1 − 2r cos θ + r2
Observe that the denominator of this relation is positive. By completing the square one can show 1 − 2r cos θ + r2 = 1 − 2r cos θ + r2 cos2 θ − r2 cos2 θ + r2 = (1 − r cos θ)2 + r2 sin2 θ which is always
63 positive. In addition, the condition that 0 < r < 1, implies that Im {ω} is positive which implies that a point inside the unit disk |z| ≤ 1 maps to the upper half plane. For r = 1, the point z = e i θ is on the unit circle and its image point is given by
ω=
− sin θ θ = − cot 1 − cos θ 2
Therefore, ω is real and for 0 < θ < π we have −∞ < ω < 0 andfor π < θ < 2π we have 0
