Chapter II. Complex Variables

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1 hapter II. omplex Variables Dates: October 2, 4, 7, These three lectures will cover the following sections of the text book by Keener omplex valued functions and branch cuts; Differentiation and analytic functions, auchy-riemann conditions; Integration; auchy integral formula; Taylor series and expansion Application: Inviscid potential flow omplex valued functions. 1. omplex numbers. We introduce the imaginary number i, whose square is 1: i 2 = 1. omplex numbers are in the form a + ib where a and b are real numbers. omplex numbers can be represented in the Argand diagram by the vector (a, b): (Figure to be provided later). Addition and subtraction of two complex numbers are simple: (a + bi) ± (c + di) =(a ± c)+(b ± d)i. Multiplication and division are as follows: (a + bi)(c + di) = (ac bd)+i(ad + bc) a+bi c+di = (a+bi)(c di) (1) c 2 +d 2 provided that c 2 + d 2 0 for the division. From these one can calculate the power (a + bi) n when n is an integer. 2. Functions. Let z = a + bi. We call z a complex variable when we use z as a variable. In general we let z = x+iy to be consistent with our habit of real variables. onsider f(z) =z 2. It is called a complex valued function. Other complex valued functions are z 3, g(z) = z +1 z 1 ; h(z) =az + b cz + d

2 where a, b, c, d are complex numbers. An important function is f(z) = z = x iy where the bar is called complex conjugate. We introduce the exponential function e z z n = n! n=0 for all complex z. Note that this definition is consistent with the real exponential. We see this opens a new world. First we see e iθ = (iθ) n n=0 n! = (iθ) 2n n=0 (2n)! + (iθ) 2n+1 n=0 (2n+1)! = n=0 ( 1) n θ2n (2n)! + i n=0 ( 1) n θ2n+1 (2n+1)! =cosθ + i sin θ. (2) Multiplying it with any real number r, we find re iθ = r cos θ + ir sin θ. If r and θ is the polar representation of the point (a, b), then we find the polar representation of complex numbers: a + bi = re iθ. Here there is no restriction on θ. From this we have a special case: e iπ +1=0 which is called Euler s identity. This identity is fun to watch since it involves the simplest symbols of mathematics: 0, 1, +, =, iand two transcendental numbers e and π. Anotherversionis e iπ +1=0 which involves additionally the minus sign. In polar representation, multiplications of complex numbers is extremely easy: (a + bi)(c + di) =(r 1 e iθ 1 )(r 2 e iθ 2 )=(r 1 r 2 )e i(θ 1+θ 2 ). 2

3 We have more examples: e z = e x+iy = e x e iy = e x (cos y + i sin y) e z 1+z 2 = e z 1 e z 2. (3) 3. Inverse functions of z 2 and e z We know that 5isanumber and satisfies the equation x 2 =5. Another solution to this equation is 5. We can verify that both z 1 = 2 + i 2, z 2 = 2 i 2 satisfy z 2 = i. So there are two answers z 1,z 2 to the square root equation z 2 = i. Thus in general there are multiple solutions to the square root. onsider the equation w 2 = z. We calculate a solution w 1 = z = re iθ = re iθ/2. We can verify that this satisfies w 2 = z. When we restrict θ to be in [0, 2π), this root is called the principal branch. Wecanseeanothersolutionis w 2 = z = re i(θ+2π) = re i(θ/2+π) = re iθ/2. If you try all possible representations θ +2nπ where n is an integer, you only find the previous two roots and no more. For the exponential function w = e z we define the inverse w =lnz as One solution is ln z = w iff z = e w. ln z =ln(re iθ )=lnr +lne iθ =lnr + iθ for z = re iθ. Or simply ln z =lnr + iθ for z = re iθ. We note that if a ln z works, then ln z +2nπi all work for all integer n: e ln z+2nπi = e ln z e 2nπi = e ln z = z. 3

4 So there are multiple inverses. The branch with the restriction π < Im (ln z) π is called the principal branch. The ray θ = π is called a branch cut. The origin is called a branch point. Any continuous interval of 2π length is called a branch. The principal branch for a function may vary from discipline to discipline. ==End of Lecture 15=== (Excluding lecture on memorizing formula/knowledge ) 4

5 2.2. alculus of omplex Functions Differentiation. We define the derivative f (z) of a complex valued function f(z) like the derivative of a real function: f f(ξ) f(z) (z) = lim ξ z ξ z where the limit is over all possible ways of approaching z. If the limit exists, the function f is called differentiable and f (z) is the derivative. Definition. If f (z) is continuous, then f is called analytic. ontinuity is defined like that for real functions of two variables. Theorem 2.1 (auchy-riemann conditions) The function f(z) = u(x, y) + iv(x, y) for z = x + iy is analytic in some region Ω if and only if u x, u y, v x, v y exist, are continuous, and satisfy the equations u x = v y, v x = u y. Proof. Let f be continuously differentiable. Then take the special path along x-axis: f(z+ x) f(z) x Then along the path y-axis: = u(x+ x,y)+iv(x+ x,y) u(x,y) iv(x,y) x = u x + i v x u x + i v x. (4) f(x+iy+i y) f(z) i y 1 i f y = i f y = i u y + v y. (5) The two limits have to be the same by definition, so we have obtained the auchy- Riemann equations u x = v y, v x = u y. onversely, suppose the auchy-riemann conditions hold; i.e., the existence and continuity of the partial derivatives and the equations of auchy-riemann all hold. Let z 0 = x 0 + iy 0. From theory of real variables we have the expansion u(x, y) v(x, y) = u(x 0,y 0 )+ u x (x 0,y 0 ) x + u y (x 0,y 0 ) y + R 1 ( x, y), = v(x 0,y 0 )+ v x (x 0,y 0 ) x + v y (x 0,y 0 ) y + R 2 ( x, y), where x = x x 0, y = y y 0,and (6) lim x, y 0 R i ( x) 2 +( y) 2 =0. 5

6 Now we have f(z 0 + z) f(z 0 ) = u x (x 0,y 0 ) x + u y (x 0,y 0 ) y + R 1 +i[ v x (x 0,y 0 ) x + v y (x 0,y 0 ) y + R 2 ] = u x ( x + i y)+i v x ( x + i y)+r 1 + R 2 i. (7) So f(z 0 + z) f(z 0 ) z This completes the proof. We list some practical rules of differentiation: = u x + i v x + R 1+R 2 i z u x + i v x. (8) f(z) =z 2 f (z) =2z f(z) =z k f (z) =kz k 1 (k integer ) (e z ) = e z (f(z)g(z)) = f (z)g(z)+f(z)g (z) [F (g(z))] = F (g(z))g (z) ( 1 f ) = 1 f f. 2 (9) Integration. Integration in the complex plane is defined in terms of real line integrals of the complex function f = u + iv. If is any (geometric) curve in the complex plane we define the line integral f(z)dz = (u + iv)(dx + idy) = u(x, y)dx v(x, y)dy + i vdx + udy. Example 1. Find zdz, where = {(x, y) x =0, 0 <y<1} oriented upward. Solution. Use parametrization x =0,y = s, s (0, 1). Then zdz = 1 0 (0 + iy)(dx + idy) = 1 Theorem 2.2. If f(z) is analytic in a domain Ω, then f(z)dz =0 for any closed curve whose interior lies entirely in Ω. 0 iyidy = 1/2. 6

7 Note that a curve whose interior lies entirely in Ω is a stronger requirement than a curve which lies entirely in Ω. The stronger requirement rules out the situation that the relevant part of Ω is not simply connected. Proof. Recall Green s Theorem Ω φdx + ψdy = Ω ( ψ x φ y )dxdy for a simply connected domain Ω. We apply this formula to our complex integral to obtain f(z)dz = (u + iv)(dx + idy) = u(x, y)dx v(x, y)dy + i vdx + udy = c ( x ( v) y u)dxdy + i c x u y v)dxdy =0 where we use c to denote the interior of the contour. This completes the proof. Examples. 2. We have z n dz =0 for any integer n and any contour that does not enclose the origin. This follows from Theorem We can calculate z =1 z 1 dz = 2π 4. We leave as an exercise the claim 0 z =1 1 1 e iθ 1e iθ idθ =2πi. z n dz =0 for all integer n 1. We note that the notation z = 1 means all points of the unit circle x 2 + y 2 =1. The default direction of the circle is counterclockwise. (10) 7

8 auchy integral formula We have found that contour integrals of analytic functions are always zero. Only a few integrands with singularities result in nonzero values. The following auchy integral formula describes contour integrals extremely well. Theorem 2.3. (auchy integral formula) Let be a simple noninteracting closed curve traversed counterclockwise. Suppose f(z) is analytic everywhere inside. For any point z inside, there holds f(ξ) dξ =2πif(z). (11) ξ z Proof. For any ɛ>0fixed, we deform the curve to where is the circle ξ z = ɛ such that f(z) f(ξ) <ɛfor all points ξ inside. Note that the integrand in (11) f(z)/(ξ z) is analytic in the region between and, we conclude that the integral in (11) is equal to the same integral over. (This can be achieved by the previous Theorem and a double-sided cut (or bridge) connecting and.) Now on we have f(ξ) ξ z dξ = f(ξ) ξ z dξ = f(z) 1 ξ z dξ + f(ξ) f(z) ξ z dξ =2πif(z)+i 2π 0 [f(z + ɛeiθ ) f(z)]dθ =2πif(z)+iI where the integral I is such that I 2πɛ. Let ɛ 0. We recover the auchy integral formula. This completes the proof of the theorem. orollary 2.4. Under the same assumptions of theorem 2.3, there hold f(ξ) (ξ z) 2 dξ =2πif (z) (13) and f(ξ) n! (ξ z) n+1 dξ =2πif(n) (z) (14) for all n-th (n a positive integer) order derivatives. And thus analyticity implies that f(z) is infinitely differentiable. orollary 2.5. (Poisson formula) A solution to the boundary value problem of the Laplacian u 2 u + 2 u =0, in x 2 + y 2 1 x 2 y 2 (15) u(r, θ) =u 0 (θ) on the boundary r = 1 (12) 8

9 where (r, θ) is the polar coordinate and u 0 (θ) is a given continuous function, is given by the formula u(r, θ) = 1 2π 1 r 2 u 0 (θ) 2π 0 1 2r cos(θ φ)+r 2 dθ. Proof. onsider an analytic function f(z) = u(x, y)+ iv(x, y) inr < 1. We have the auchy-riemann equations: u x = v y, v x = u y. So we have 2 u x 2 = 2 v y x = y ( u y ), thus 2 u x u y 2 =0. That is, the real part of an analytic function is a harmonic function (satisfying the Laplace equation). Now we use the auchy integral formula f(z) = 1 2π f(ξ)ξ 2π 0 ξ z dθ (letting ξ = eiθ ) for z inside the unit circle and the same formula 0= 1 2π f(ξ)ξ dθ 2π 0 ξ ( z) 1 applied at the point ( z) 1 which is outside of the unit circle ( 1 z > 1if z < 1.) Noting that ξ =( ξ) 1 on the unit circle, we can add the previous formulas f(z) = 1 2π 2π 0 ξ f(ξ)[ ξ z 1/ ξ 1/ ξ 1/ z ]dθ. Or f(z) = 1 2π f(ξ) 1 z 2 2π 0 ξ z 2 dθ. Taking the real part of the formula, we obtain Poisson formula. This completes the proof Taylor series. 9

10 We would like to expand an analytic function in a powerful series: f(z) = a n (z z 0 ) n, n=0 where a n = f (n) (z 0 ) n! and z 0 is a convenient point for an application. Using auchy integral formula for derivatives (orollary 2.4), we find that a n = 1 2πi f(ξ) dξ (ξ z 0 ) n+1 for any simple contour that contains z 0. We can use other simple ways to find Taylor series: For example we have 1 1 z =1+z + z2 + z 3 + valid for all z < 1. Using these lecture notes together with the text book is recommended Application: Inviscid incompressible steady potential flow. Text: pp Take a two-dimensional inviscid fluid of density ρ and velocity vector u =(u, v). Suppose there is no source and sink, then ρu n dl =0 Ω for any domain Ω in the fluid, where n is the unit exterior normal to the boundary and dl represents length element. Using the divergence theorem we find ρu n dl = (ρu) dxdy =0. Ω Ω Since the domain is arbitrary, we conclude that (ρu) = 0. Assuming that the fluid is incompressible; i.e., ρ = constant, then u =0. If the velocity u is a gradient of a scalar function φ(x, y): u = φ, 10

11 then the flow is called a potential flow. A potential flow is als ocalled irrotational or curl free flow since one can readily check that the curl of u is zero. The scalar function φ(x, y) is called a potential function of the flow. For incompressible and potential flow we have u = 2 φ 2 x + 2 φ 2 y =0. Thus the potential (function) φ of a potential flow satisfies the Laplace equation for which we have already a solution formula from last lecture. If we let f(z) =φ(x, y) +iψ(x, y) be an analytic function, then both φ and ψ satisfy the Laplace equation. Also, if we have a φ that satisfies the Laplace equation, we can always find a ψ, for example by ψ(x, y) = x φdy y φdx such that the f(z) =φ(x, y) +iψ(x, y) isanalytic. Here can be any path from the origin to the point (x, y). We can use Green s formula and auchy-riemann conditions to verify those. The physical interpretation of ψ is very interesting. We can verify that φ ψ =0 since φ x = ψ y,ψ x = φ y. Thus the level curves of φ and ψ are orthogonal. Hence the velocity u points along the tangent of the level curves of ψ. Thus the fluid flows along the level curves of ψ, which are called streamlines. An example of such a flow is provided by such a function f(z) =(z + a2 z )V, where a>0and V are constants. The corresponding flow is flow past a (cross section of) circular cylinder at z = a, see graph 6.5 of text on page