MATH F112 (MATHEMATICS-II) Gaurav Dwivedi Department of Mathematics BITS Pilani, Pilani Campus. Module-9(Functions of Complex Variable) Function of a Complex Variable Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) March 14, 2020 3 / 22 Function of a Complex Variable Function of a complex variable Let S be a set of complex numbers. Then function f defined on S is a rule that assigns to each z ∈ S, a complex number w, and we write f (z) = w. The set S is called domain of definition of f . Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) March 14, 2020 4 / 22 Suppose that w = u + iv is the value of the function f at z = x + iy, so that u + iv = f (x + iy). Thus each of real number u and v depends on the real variables x and y i.e., f (z) = u(x, y) + iv(x, y). Similarly, in the polar form f (z) = u(r, θ) + iv(r, θ). Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) March 14, 2020 5 / 22 Examples f (z) = z 2 = (x2 − y 2 ) + i(2xy). Here u(x, y) = x2 − y 2 and v(x, y) = 2xy. f (z) = ez = ex (cos y + i sin y). Here u(x, y) = ex cos y and v(x, y) = ex sin y. In polar form f (z) = z 2 gives f (z) = (reiθ )2 = r2 e2iθ = r2 (cos 2θ + i sin 2θ). Here u(r, θ) = r2 cos 2θ and v(r, θ) = r2 sin 2θ. Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) March 14, 2020 6 / 22 Polynomial function: f (z) = a0 + a1 z + a2 z 2 + · · · + an z n , where n is zero or a positive integer and a0 , a1 , . . . , an are complex constants and an 6= 0. The domain of definition is the entire z-plane. Rational function: the quotients P (z)/Q(z) of polynomials. The domain of definition is {z ∈ C : Q(z) 6= 0}. Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) March 14, 2020 7 / 22 Limit Limit Let f be a function defined at all points of z in some deleted nbd of z0 , then lim f (z) = w0 , z→z0 if given ε > 0, there exists a δ > 0 such that |f (z) − w0 | < whenever 0 < |z − z0 | < δ. Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) March 14, 2020 8 / 22 That means the point w = f (z) can be made arbitrarily close to w0 if we choose the point z close enough to z0 but distinct from it. Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) March 14, 2020 9 / 22 Exercise Show that if f (z) = |z| < 1, then iz̄ in the open disk 2 i lim f (z) = . z→1 2 2 Exercise Show that lim zz does not exist. z→0 Sol. Use two paths test. Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) March 14, 2020 10 / 22 Some Theorems on Limits Theorem 1 Let f (z) = u(x, y) + iv(x, y), z0 = x0 + iy0 , (z = x + iy), and w0 = u0 + iv0 , then lim f (z) = w0 , z→z0 if and only if lim (x,y)→(x0 ,y0 ) u(x, y) = u0 Gaurav Dwivedi (BITS Pilani) and lim (x,y)→(x0 ,y0 ) MATH F112 (MATHEMATICS-II) v(x, y) = v0 . March 14, 2020 11 / 22 Some Theorems on Limits Theorem 2 If lim f (z) and lim g(z) both exist, then z→z0 z→z0 lim [f (z) ± g(z)] = lim f (z) ± lim g(z). z→z0 z→z0 z→z0 lim [f (z)g(z)] = lim f (z) lim g(z). z→z0 lim z→z0 z→z0 h f (z) g(z) i Gaurav Dwivedi (BITS Pilani) z→z0 lim f (z) = z→z0 lim g(z) , provided lim g(z) 6= 0. z→z0 MATH F112 (MATHEMATICS-II) z→z0 March 14, 2020 12 / 22 Stereographic Projection and the Riemann Sphere Let us think of the complex plane as passing through the equator of the unit sphere S centred at origin. To each point z in the plane there corresponds exactly one point P on the surface of the sphere. The point P is the point where the line through z and the north pole N intersects the sphere. In like manner, to each point P on the surface of the sphere, other than the north pole N, there corresponds exactly one point z in the plane. By letting the point N of the sphere correspond to the point at infinity, we obtain one to one correspondence between S and the extended complex plane. Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) March 14, 2020 13 / 22 Stereographic Projection and the Riemann Sphere Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) March 14, 2020 14 / 22 Point at Infinity Point at Infinity The point at infinity is denoted by ∞, and the complex plane together with ∞ is called the extended complex plane. For each small > 0, the set 1 , S = z : |z| > is called an neighborhood of ∞. Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) March 14, 2020 15 / 22 Definition of lim f (z) = f0 z→∞ Let f (z) be a complex function of the complex variable z, and f0 be a complex constant. If for every real number ε, there exists a real number r such that |f (z) − f0 | < ε for every |z| > r, then we say that lim f (z) = f0 . z→∞ Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) March 14, 2020 16 / 22 Definition of lim f (z) = ∞ z→z0 Let f (z) be a complex function of the complex variable z, and z0 be a complex constant. If for every real number ε, there exists a δ > 0 such that |f (z)| > ε for every 0 < |z − z0 | < δ, then we say that lim f (z) = ∞. z→z0 Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) March 14, 2020 17 / 22 Theorem Related to ∞ 1 f (z) z→z0 lim f (z) = ∞ ⇔ lim z→z0 1 z lim f (z) = w0 ⇔ lim f z→∞ z→0 lim f (z) = ∞ ⇔ lim f z→∞ z→0 1 ( z1 ) = 0. = w0 . = 0. Proof. Proofs are exercise. Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) March 14, 2020 18 / 22 Continuity Continuous Function A function f (z) is said to be continuous at a point z0 if f (z0 ) is defined. lim f (z) exists. z→z0 lim f (z) = f (z0 ). z→z0 Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) March 14, 2020 19 / 22 Continuity Continuous Function In terms of ε, δ notation, a function f (z) is said to be continuous at a point z0 if for a given > 0, there is a δ > 0, such that |f (z) − f (z0 )| < whenever |z − z0 | < δ. A function f (z) is said to be continuous in a region R if it is continuous at all the points of the region R. Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) March 14, 2020 20 / 22 Some Results on Continuity Theorem 1. A composition of continuous functions is itself continuous. Theorem 2. f (z) = u(x, y) + iv(x, y) is continuous iff both u(x, y) and v(x, y) are continuous. Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) March 14, 2020 21 / 22 Some Results on Continuity Theorem 3. If f (z) and g(z) are continuous, then f (z) ± g(z), f (z)g(z), and f (z) g(z) , g(z) 6= 0, all are continuous. Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) March 14, 2020 22 / 22 Some Results on Continuity Theorem 4. Let f (z) is continuous at z0 and f (z0 ) 6= 0. Then f (z) 6= 0 throughout in some nbd of z0 . Proof. Since f (z) is continuous at z0 , therefore for each > 0, there is a δ > 0 such that |f (z) − f (z0 )| < whenever |z − z0 | < δ. (1) Equation (??) is valid for all , so in particular it holds for = |f (z2 0 )| . Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) March 14, 2020 23 / 22 Some Results on Continuity Then there exists δ > 0 such that when |z − z0 | < δ |f (z) − f (z0 )| < |f (z0 )| . 2 If f (z) = 0 in some nbd of z0 , then |f (z0 )| < |f (z2 0 )| , which is a contradiction. Therefore, we must have f (z) 6= 0 for all z in Nδ (z0 ). Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) March 14, 2020 24 / 22 Some Results on Continuity Theorem 5. Every continuous function in a closed and bounded region R is bounded i.e., |f (z)| ≤ M for all z ∈ R. Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) March 14, 2020 25 / 22 Examples Discuss the continuity of f (z) = Sol. We have f (0) = 0. Now lim f (z) = z→0 Re z 1+|z| at z = 0. x p = 0 = f (0). (x,y)→(0,0) 1 + x2 + y 2 lim Therefore f (z) is continuous at 0. Discuss the continuity of f (z) = Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) Re z z at z = 0. March 14, 2020 26 / 22 Derivatives Differentiable Function Let f be a function defined on S containing Nρ (z0 ). If f (z) − f (z0 ) , z→z0 z − z0 lim or equivalently f (z0 + ∆z) − f (z0 ) , ∆z→0 ∆z lim exists then we say f is differentiable at z0 and the value of the limit denoted by f 0 (z0 ) is the derivative of f at z0 . Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) March 14, 2020 27 / 22 Thus f (z) − f (z0 ) , z→z0 z − z0 f 0 (z0 ) = lim or f (z0 + ∆z) − f (z0 ) . ∆z→0 ∆z f 0 (z0 ) = lim Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) March 14, 2020 28 / 22 Example If f (z) = z 2 + 2, then show that f 0 (z0 ) = 2z0 . Sol. We have f (z) − f (z0 ) (z 2 + 2) − (z02 + 2) = lim z→z0 z→z0 z − z0 z − z0 = lim (z + z0 ) lim z→z0 = 2z0 . Thus f 0 (z0 ) = 2z0 . Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) March 14, 2020 29 / 22 Differentiability implies continuity If f is differentiable at z0 then f is continuous at z0 . Proof. f (z) − f (z0 ) .(z − z0 ) lim [f (z) − f (z0 )] = lim z→z0 z→z0 z − z0 f (z) − f (z0 ) = lim . lim (z − z0 ) z→z0 z→z0 z − z0 = f 0 (z0 ).0 = 0. Thus limz→z0 f (z) = f (z0 )(Why?). Hence f is continuous at z0 . Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) March 14, 2020 30 / 22 Remark The converse of above result is not true i.e., a continuous function may not be differentiable. Example The function f (z) = |z|2 is continuous everywhere but differentiable only at 0. Sol. We have f (z) = |z|2 = x2 + y 2 = u(x, y) + iv(x, y). Since both u(x, y) = x2 + y 2 and v(x, y) = 0 are continuous everywhere, therefore f (z) is continuous everywhere. Now, for z 6= z0 , we have Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) March 14, 2020 31 / 22 Example f (z) − f (z0 ) |z|2 − |z0 |2 = z − z0 z − z0 z z̄ − z0 z0 = z − z0 z z̄ − z̄z0 + z̄z0 − z0 z0 = z − z0 z̄(z − z0 ) + z0 (z̄ − z0 ) = z − z0 ∆z = z̄ + z0 , ∆z = z − z0 ∆z Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) March 14, 2020 32 / 22 f (z) − f (z0 ) ∆x − i∆y = z + z0 z − z0 ∆x + i∆y ( z + z0 when ∆z = (∆x, 0) = z − z0 when ∆z = (0, ∆y) f (z) − f (z0 ) exists, then, z + z0 = z − z0 , z→z0 z − z0 (z0 ) i.e., z0 = 0. Thus, if z0 6= 0, limz→z0 f (z)−f does not z−z0 exist and so f is not differentiable at z0 6= 0. Thus if lim Remark: Note that, this does not prove that f (z) is differentiable at z0 = 0. Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) March 14, 2020 33 / 22 Now for z0 = 0: |z|2 f (z) − f (0) = lim z→0 z z→0 z−0 = lim z̄ lim z→0 = 0. Thus f is differentiable at 0 and f 0 (0) = 0. Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) March 14, 2020 34 / 22 Differentiation Rules d dz (c) = 0 for a constant c. d n dz (z ) = nz n−1 . [f (z) ± g(z)]0 = f 0 (z) ± g 0 (z). [f (z)g(z)]0 = f (z)g 0 (z) + f 0 (z)g(z). h f (z) g(z) i0 = g(z)f 0 (z)−f (z)g 0 (z) , (g(z))2 Gaurav Dwivedi (BITS Pilani) g(z) 6= 0. MATH F112 (MATHEMATICS-II) March 14, 2020 35 / 22 Chain Rule for Differentiation Let F (z) = g(f (z)), and assume that f is differentiable at z0 and g is differentiable at f (z0 ), then F (z) is differentiable at z0 and F 0 (z0 ) = g 0 (f (z0 ))f 0 (z0 ). Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) March 14, 2020 36 / 22 Q. 9, Page - 63 Let f (z) be a function defined by ( 2 (z̄) z , if z 6= 0, f (z) = 0, if z = 0. Show that f 0 (0) does not exist. Sol. We have 2 (z̄) −0 f (z) − f (0) = lim z lim z→0 z→0 z − 0 z−0 2 z̄ 2 x − iy = lim = lim . z→0 z (x,y)→(0,0) x + iy Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) March 14, 2020 37 / 22 Now, along x-axis lim (x,y)→(0,0) x − iy x + iy 2 = 1, while along the line y = x 2 x − iy lim = −1. (x,y)→(0,0) x + iy (0) Hence limz→0 f (z)−f does not exist and hence f 0 (0) z−0 does not exist. Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) March 14, 2020 38 / 22 Homework If f (z) = z̄, show that f 0 (z) does not exist at any point z. Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) March 14, 2020 39 / 22 Cauchy Riemann Equations (C-R Equations) Cauchy Riemann Equations (C-R Equations) Let f (z) = u(x, y) + iv(x, y), then the equations ux = vy and uy = −vx , are called Cauchy Riemann Equations (C-R Equations). Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) March 14, 2020 40 / 22 Necessary Condition for Differentiability Necessary Condition for Differentiability Suppose f (z) = u(x, y) + iv(x, y) is differentiable at a point z0 = x0 + iy0 . Then the first order partial derivatives of u and v exist and satisfy C-R equations at (x0 , y0 ). Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) March 14, 2020 41 / 22 Proof Since f (z) = u(x, y) + iv(x, y) is differentiable at z0 = x0 + iy0 , therefore f (z0 + 4z) − f (z0 ) ∆z→0 4z u(x0 + 4x, y0 + 4y) − u(x0 , y0 ) = lim 4x + i4y (∆x,∆y)→(0,0) v(x0 + 4x, y0 + 4y) − v(x0 , y0 ) +i . 4x + i4y f 0 (z0 ) = lim Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) March 14, 2020 42 / 22 u(x0 + 4x, y0 + 4y) − u(x0 , y0 ) 4x + i4y (∆x,∆y)→(0,0) v(x0 + 4x, y0 + 4y) − v(x0 , y0 ) . +i lim 4x + i4y (∆x,∆y)→(0,0) = lim Now along the path where ∆y = 0, it gives f 0 (z0 ) = ux (x0 , y0 ) + ivx (x0 , y0 ), and along the path where ∆x = 0, it gives f 0 (z0 ) = −iuy (x0 , y0 ) + vy (x0 , y0 ). Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) March 14, 2020 43 / 22 This shows the existence of the first order partial derivatives of u and v. Now since f is differentiable at z0 therefore f 0 (z0 ) is unique (limits along every path unique) and so, on comparing: ux (x0 , y0 ) = vy (x0 , y0 ), Gaurav Dwivedi (BITS Pilani) uy (x0 , y0 ) = −vx (x0 , y0 ). MATH F112 (MATHEMATICS-II) March 14, 2020 44 / 22 Sufficient Condition Sufficient Condition for Differentiability Let f (z) = u(x, y) + iv(x, y) be any function defined throughout in N (z0 ) such that 1 The first order partial derivatives of u and v (ux , uy , vx , vy ) exist in N (z0 ), The first order partial derivatives of u and v are continuous and satisfy C-R equations at (x0 , y0 ). Then f is differentiable at z0 and 2 f 0 (z0 ) = ux (x0 , y0 ) + ivx (x0 , y0 ). Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) March 14, 2020 45 / 22