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MTH 5103 COMPLEX VARIABLES LECTURE NOTES, WEEK 2 2 Complex Functions In this Lecture we shall discuss the basic properties of functions of one complex variable. 2.1 Functions of a Complex Variable Let S ⊂ C. A function f on S is a rule which assigns to each z ∈ S a complex number f (z), called the value of f at z. Examples of (complex) functions include Polynomials e.g., f (z) = z 2 or f (z) = 1 + iz − z 2 + (1 + i)z 3 z . z2 + 2 These are defined everywhere except at roots z of the denominator polynomial. Rational Functions (One polynomial divided by another) e.g., f (z) = The Exponential Function e.g., f (z) = ez , which is defined to be the sum z e = ∞ X zn n=0 n! =1+ z z2 z3 + + + ··· , 1! 2! 3! (1) or equally well, by Euler’s Formula, as ez = ex (cos y+i sin y). We shall see later that the series above converges absolutely for all z ∈ C, therefore f (z) = ez can be defined for every z ∈ C (i.e., we can choose S = C). Trigonometric and Hyperbolic Functions For complex z, we define cos z = eiz + e−iz , 2 sin z = eiz − e−iz , 2i (2) which agrees with the definition for z real in Euler’s Formula. By analogy with the real case, we define cosh z = ez + e−z , 2 1 sinh z = ez − e−z . 2 (3) With these definitions, it is easy to check the following identity cos z + i sin z = eiz = cosh(iz) + sinh(iz). (4) Now, given any complex number may be expressed in the form x + iy, we may write a complex function f as f (z) = <(f (z)) + i=(f (z)) = u(x, y) + iv(x, y). (5) In this manner, we may think of f as a (real) mapping f : R2 −→ R2 (x, y) 7−→ (u(x, y), v(x, y)) . 1 Example 1. Rewrite the following complex functions as real mappings: z 2 , , ez . z 1. f (z) = z 2 can be written as f (x, y) = (x2 − y 2 , 2xy). 1 x −y 2. f (z) = can be written as f (x, y) = , . z x2 + y 2 x2 + y 2 3. f (z) = ez = ex+iy can be written as f (x, y) = (ex cos y, ex sin y). 2.2 Transformations of the Complex Plane A complex function is a map (transformation) C → C, but to draw a graph of such a mapping, we would need 2+2=4 real dimensions. So instead, we shall examine what happens to various “shapes” (e.g., lines, curves, etc.) under such a map. Example 2. Examine the mapping w = z 2 . In the previous example, we wrote down this mapping in (x, y), (u, v) coordinates as u + iv = x2 − y 2 + 2ixy, i.e., u = x2 − y 2 and v = 2xy. It is easier, however, to understand this map if we write z and w in polar coordinates. Then for z = reiθ , we have f (z) = w = r2 e2iθ . We first determine where the following curves are mapped under f : The real axis R is mapped to the positive real axis R+ : We can see this by representing the real axis in two pieces, the positive real axis by points of the form rei0 = r > 0 and the negative real axis by reiπ = −r < 0. These points are mapped under f to r2 ei0 = r2 ei2π = r2 , which lies on the positive real axis. 2 The imaginary axis iR is mapped to the negative real axis R− : Using a similar argument as above, we can represent the positive imaginary axis using points of the form reiπ/2 = ir and those on the negative imaginary axis as rei3π/2 = −ir (r > 0). These points are mapped under f to r2 eiπ = r2 ei3π = −r, which lies on the negative real axis. A quarter circle of radius r centred at 0 maps to a half circle of radius r2 centred at 0: The quarter circle is given byreiθ where 0 ≤ θ ≤ π/2, which is mapped to r2 e2iθ . Notice that as θ ranges from 0 to π/2, the angle 2θ ranges from 0 to π, giving us the half circle. A line `θ passing through the origin at angle θ maps to a ray `+ 2θ starting at 0 making an angle 2θ: In this case, the line is parametrised by reiθ , where θ is now fixed, but r may vary. The mapping sends r to r2 , resulting in a ray (and the angle is transformed from θ to 2θ). Example 3. What is the image of a vertical line x = c ∈ R under the mapping w = z 2 ? Recall that the function z 7→ z 2 can be expressed as (u, v) = (x2 − y 2 , 2xy). Setting x = c, we have the system u = c2 − y 2 , v = 2cy, from which we may eliminate the variable y to obtain the equation for a curve in v2 the u − v plane (equaivalently, the w-plane). Doing this gives us u = c2 − 2 , 4c which is a leftward-opening parabola with vertex (c2 , 0), intersecting the v axis at (0, ±2c2 ). Example 4. Using the results of the previous examples, determine which regions, if any, of the z-plane are mapped to the upper half of the w plane under the mapping w = z 2 . First, we identify the upper half of the w-plane to be {w ∈ C : =(w) ≥ 0}. An element z = reiθ maps to w = r2 e2iθ in the upper half plane if and only if the argument of w lies between 0 and π. That is, 0 + 2kπ ≤ 2θ ≤ π + 2kπ π kπ ≤ θ ≤ kπ + 2 (k ∈ Z) (k ∈ Z), which means that z must lie in the first or third quadrant in order to map to the upper half plane (plug in k = 0, 1, 2, 3, . . . to see this). 3 Alternatively, we could use the (x, y) and (u, v) coordinates to solve this problem as follows: w lies in the upper half plane if and only if v = 2xy ≥ 0. But this is true only when x and y are either both ≥ 0 or both ≤ 0, that is, when z lies in the first or third quadrant. Proposition 1. The following kinds of transformations all send straight lines to straight lines and circles to circles: 1. Translations w = z + c, for complex constant c 2. Rotations w = eiθ z for a real constant θ 3. Dilations w = rz for a real, nonzero constant r 4. Linear or Affine Maps w = λz + c for complex constants λ, c with λ 6= 0. 5. Complex Conjugation w = z̄ Proof. Translations, rotations and dilations all obviously satisfy that they map lines to lines and circles to circles (convince yourself geometrically!). To prove linear mappings also satisfy this property, we rewrite λ = reiθ and observe that λz +c can be written as a composition of a dilation followed by a rotation followed by a translation, reading left to right: z 7−→ rz 7−→ reiθ z 7−→ reiθ z + c. (6) Finally, note that complex conjugation z = (x, y) 7−→ z̄ = (x, −y) is simply reflection about the x-axis which also sends lines to lines and circles to circles. 1 Another function we often need to consider is z 7−→ w = . We may also regard z this as a composition (using the fact that z z̄ = |z|2 ): z 7−→ z̄ 1 z 7−→ 2 = . 2 |z| |z| z (7) The first mapping is “inversion” about the unit circle and the second is complex conjugation; it is an instructive exercise to draw a picture of each of these mappings and determine where various regions of the plane are mapped (e.g., in the first quadrant, the interior of unit circle maps to the exterior of unit circle in the fourth quadrant). 1 Consequently we can write down how z 7−→ w = maps the punctured plane C\{0} z to itself (or C ∪ {∞} to itself). 4 1 maps z 1. circles not through 0 to circles not through 0, Proposition 2. z 7−→ w = 2. circles through 0 to straight lines not through 0, 3. straight lines not through 0 to circles through 0, 4. straight lines through 0 to straight lines through 0. Remark 1. If we regard a straight line as a “circle passing through ∞” then the 1 above Proposition says the mapping z 7−→ sends circles to circles. z Proof. We outline the main ideas of the proof. Consider the equation a(x2 + y 2 ) + bx + cy + d = 0 (a, b, c, d ∈ R). (8) If a = 0, this is a straight line. If a 6= 0, then by completing the squares in x −b −c , and y separately, we see this is a circle centered at . In either case, it 2a 2a passes through 0 ∈ C (equivalently, (0, 0) ∈ R2 ) if and only if d = 0. 1 Now let w = so that for z = x + iy and w = u + iv, z 1 z = w v u −i 2 x + iy = 2 2 u +v u + v2 u −v ⇒x= 2 , y= 2 . 2 u +v u + v2 Substituting for x and y in (8) and clearing denominators, we obtain d(u2 + v 2 ) + bu − cv + a = 0. (9) But comparing this equation directly with (8), we see it must be a circle in the w-plane (equivalently, the (u, v)-plane) if and only if d 6= 0 and it is a line if and only if d = 0. Furthermore, it passes through the origin if and only if a = 0, which proves the Proposition. Example 5. Determine how the (horizontal) line y = k is transformed under 1 u −v z 7→ . In this case, we may substitute x = 2 ,y = 2 into the equa2 z u +v u + v2 tion y = k to find k(u2 + v 2 ) + v = 0. (10) 1 This is simply a circle in the w-plane centered at 0, −1 of radius 2k . 2k 5 2.3 Möbius Transformations Let a, b, c, d ∈ C be such that ad − bc 6= 0. A mapping of the form z 7−→ w = az + b cz + d (11) is called a fractional linear map or a Möbius transformation. Visibly, any Möbius transformation can be expressed as a composition of a linear map with an inversion, followed by another linear map: 1 1 7−→ α + β = w. (12) z 7−→ cz + d 7−→ cz + d cz + d Notice that if we require w = az + b , the solving for α, β gives us cz + d α + β(cz + d) = az + b ⇒ β= a ad and α = b − . c c Given the above calculation, we have immediately from Propositions 1 and 2 Proposition 3. Fractional linear transformations (Möbius transformations) send lines and circles to lines and circles. Möbius transformations are often represented by their corresponding!matrices, a b az + b e.g., for z 7→ , the associated matrix is given by M = . It is a cz + d c d straightforward calculation (do this!) to verify that the composition of two Möbius transformations is itself a Möbius transformation and that the matrix associated with the resulting transformation can be found by multiplying the corresponding matrices of the composition transformations. az + b Remark 2. z 7−→ can be extended to a map from C ∪ {∞} to C ∪ {∞} by cz + d a −d sending z = ∞ to w = and sending z = to w = ∞. It is a bijection from c c C ∪ {∞} to itself and has inverse given by w 7−→ w = dw − b . −cw + a (13) Example 6. Find the Möbius transformation sending −1 7→ i, 0 7→ 1, and 1 7→ −i. To do this, we require i= −a + b , −c + d b 1= , d 6 −i = a+b . c+d (14) From the second equation we have d = b so we may eliminate the other variables in the remaining equations to find a = −ib and c = ib. Hence the Möbius transformation is given by z 7−→ −ibz + b z+i = . ibz + b −z + i (15) Note that under this transformation, the real axis is mapped to the unit circle (R+ 7→ lower semicircle and R− 7→ upper semicircle), and the upper-half of the z-plane is mapped the exterior of this circle (first quadrant 7→ lower exterior and second quadrant 7→ upper exterior). It is a harder exercise to show that given any three distinct points z1 , z2 , z3 in the z-plane, and any three points w1 , w2 , w3 in the w-plane, there exists a Möbius transformation sending zj 7→ wj for each j = 1, 2, 3. This transformation is unique if we specify an orientation for the curve containing the three given points (or for the curve containing the prescribed points, but not both). 7