# EE2007 Tutorial 7 Complex Numbers, Complex Functions, Limits and Continuity

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1 EE27 Tutorial 7 omplex Numbers, omplex Functions, Limits and ontinuity Exercise 1. These are elementary exercises designed as a self-test for you to determine if you if have the necessary pre-requisite for this complex variable module. These exercises will not be discussed in the tutorial class. If you find that there is a knowledge gap, then you should do something to help yourself, e.g. read relevant textbooks, ask your friends or consult your tutors. (a) Let z x + iy. Find Im (1/z), Im (z 4 ), (1 + i) 16 [ y/(x 2 + y 2 ); 4x 3 y 4xy 3 ; 256 ] (b) Represent each of the following in polar form and plot each of them in the complex plane. 2b + 2ai ( a bi )2, (c) Find and plot all the roots: i 3 + 3i, 2 + i 5 3i, 7 5i 4i [ 4; (1/ 18)e πi/4 ; 5/34(cos i sin 1.47); 37/8e i ] i, 3 216, 4 4, i (d) Solve the equations [ 6 2(cos kπ kπ 12 + i sin 12 ), k 1, 9, 17; 6 and 3 ± 3 3i; ±(1 ± i); ±(3 + 4i) ] (a) z 2 + z + 1 i [ i, 1 i ] (b) z 4 3(1 + 2i)z i [ ±(2 + i) and ±(1 + i) ] (e) Find the principal values of (a) i 3i and (b) [ e 2 ( 1 + i 3)] 3πi [ e 3π 2 ; e 3πi[1+i 2π 3 ] ]

2 Exercise 2. These questions were discussed in the lectures. Ask your tutor if you still have doubts. (a) Let f(z) xy x 2 +y 2 + ixy. Does lim x,y f(z) exist? (b) Does lim z ( z z z z ) exist? (c) Is f(z) Re (z 2 )/ z 2 continuous at the origin? Im (z) (d) Is f(z) 1+z z continuous at z? (e) Is f(z) zre (z) z continuous at z? [ No; No; No; Yes; Yes ] Exercise 3. (a) [October/November 23 G264/Q3] Find all the values of ln(i 1/2 ). [ iπ( n), n, ±1, ±2,... ] (b) [October/November 22 G264/Q5, (Rephrased)] Find all the values of i i and show that they are all real. Hence, or otherwise, find all the points z such that z i is real. [ e (π/2±2nπ) ;e ±nπ e iθ, n, 1,... ] (c) [April/May 25 EE27/Q3] Find all the values of e (ii). [ e e ( π 2 +2nπ) ] Exercise 4. Find out whether f(z) is continuous at the origin if f() and for z, the function f(z) is equal to (a) Re z z (b) Im z 1+ z [ No; Yes ] Exercise 5. The function f(z) is equal to Im z z for z and f(z) for z. Determine if f(z) is continuous at z, z 5 and z 5 + i. [ No;Yes;Yes ]

3 Solutions to Exercises Exercise 3(a) ln(i 1/2 ) ln([e i(π/2+2nπ) ] 1/2 ) ln(e iπ(1/4+n) ) iπ( 1 + n), 4 n, ±1, ±2,...

4 Exercise 3(b) (a) Let y i i. Then ln y i ln i i ln e i(π/2±2nπ) (π/2 ± 2nπ) where n, 1, 2,.... Hence y i i e (π/2±2nπ) is real. (b) Now, let y z i. Following the same technique as above, we have ln y i ln z i ln re i(θ+2nπ) i ln r (θ ± 2nπ) In other words, z i e i ln r e (θ±2nπ) For z i to be real, one particular solution is to choose r 1, θ and n, i.e. z 1 and we have 1 i 1 which is real. In general, we need Im (e i ln r ) or sin(ln r). This gives r e ±nπ, n, 1, 2,.... Thus (e ±nπ e iθ ) i, n, 1, 2,... is real.

5 Exercise 3(c) Let y i i. Then ln y i ln i i[ln 1 + i( π 2 + 2nπ)] ( π 2 + 2nπ) Hence i i e ( π 2 +2nπ) and e (ii) e π e ( 2 +2nπ)

6 Exercise 4. A function f(z) is said to be continuous at z z o if (a) f(z o ) is defined, and (b) lim z zo f(z) f(z o ). In this question, condition (a) is always satisfied at the origin. condition (b). So we need only to check Let z re iθ. (a) f(z) Re z z reiθ lim f(z) lim Re z r re iθ lim r cos θ { cos θ 1 if θ, 2π,... if θ π/2, 3π/2,... r r Hence f(z) is not continuous at the origin. (b) f(z) Im z 1+ z lim f(z) lim z r Im lim r reiθ 1 + re iθ r (i sin θ) 1 + r along all directions. Hence f(z) is continuous at the origin. Exercise 4

7 Exercise 5. We have { Im z f(z) z, z, z If lim z zo f(z) f(z o ), then f(z) is continuous at z zo. Note that f(), f(5) Im 5 5+i 5, and f(5 + i) Im 5+i 1 26 Now, to check if the limit exists at z z o, let z z o + re iθ. Then lim f(z) lim Im z o + re iθ z z o r z o + re iθ (1) At z o, we have reiθ lim Im r re iθ lim r sin θ r r sin θ Hence, the limit does not exist and f(z) is not continuous at z z o. At z o 5, we have 5 + reiθ lim Im r 5 + re iθ lim r sin θ r 5 Hence f(z) is continuous at z 5. At z o 5 + i, we have f(5) 5 + i + reiθ lim Im r 5 + i + re iθ lim 1 + r sin θ r 5 + i + re iθ 1 f(5 + i) 26 Hence f(z) is continuous at z 5 + i. In fact, from eq.( 1), it can be seen that in general the function is continuous for all z o. Exercise 5

8 EE27 Tutorial 8 Differentiabiliy and Analyticity of omplex Functions, auchy-riemann Equations and omplex Line Integration Exercise 1. These questions were discussed in the lectures. Ask your tutor if you still have doubts. (a) Discuss the differentiability of f(z) z. (b) Show that f(z) (x 3 3xy 2 ) + i(3x 2 y y 3 ) is differentiable for all z and find its derivative. (c) Discuss the analyticity of the function f(z) x 2 + iy 2. (d) Evaluate zdz where is given by x 3t, y t2, 1 t 4. (e) Evaluate (z z o) m dz where is a W circle of ρ with centre at z o. Exercise 2. Find the constants a and b such that f(z) (2x y) + i(ax + by) is differentiable for all z. Hence, find f (z). [ a 1 and b 2, f (z) 2 + i ] Exercise 3. Are the following functions analytic? (a) f(z) Re(z 2 ) (b) f(z) i z 4 (c) f(z) z z (d) f(z) e x (sin y i cos y) [ No; Yes, for z ; No; Yes, for all z ] Exercise 4. (a) [EE27, April/May 25, Q1] Find f (z), the derivative of f(z) 2xy ix 2. State clearly the point (or points) where f (z) exists. [ on x-axis, f (z) i2x ] (b) [G264, April/May 24, Q3] Using the auchy-riemann equations, determine the analyticity of the function f(z) z 2 2z + 3 and find its derivative. [ z, f (z)2z-2 ]

9 Exercise 5. Evaluate f(z) dz where (a) f(z) Re z, the parabola y x 2 from to 1 + i. (b) f(z) 4z 3, the straight line segment from i to 1 + i. (c) f(z) e z, the boundary of the square with vertices, 1, 1 + i, and i (clockwise) (d) f(z) Im(z 2 ), as in problem (c). [ i 2 3, 1 + i4,, 1 i ]

10 Solutions to Exercises Exercise 2. For f(z) u(x, y) + iv(x, y) to be differentiable, u, v and their partial derivatives must be continuous and satisfies -R equations: Then u x v y and u y v x f (z) u x + iv x For this problem, we have u 2x y, v ax + by u x 2, v y b b 2 u y 1, v x a a 1 Thus f(z) (2x y) + i(x + 2y) is differentiable for all z and f (z) 2 + i. In fact, with a 1 and b 2, f(z) (2x y) + i(x + 2y) 2z + iz and it is no surprise that f(z) is differentiable z and f (z) 2 + i. Exercise 2

11 Exercise 3. We shall make use of auchy-riemann equations u x v y, u y v x or u r 1 r v θ, v r 1 r u θ to check for analyticity. (a) f(z) Re(z 2 ) Re((x + iy) 2 ) x 2 y 2 Therefore and u(x, y) x 2 y 2, v(x, y) u x 2x, u y 2y, v x, v y We see that the -R equations are satisfied ONLY at the origin, hence f is nowhere analytic. (b) Since u(r, θ) 1 sin 4θ, r4 v(r, θ) 1 cos 4θ r4 and u r 4 r 5 sin 4θ, u θ 4 r 4 cos 4θ, v r 4 r 5 cos 4θ, v θ 4 sin 4θ. r4 Except for z, the functions u and v and their partial derivatives are continuous and the -R equations are satisfied. Hence f(z) is analytic except at z. (c) Since u(x, y), v(x, y) 2y Same reasoning as (a) and the function is not analytic. (d) Since u(x, y) e x sin y, v(x, y) e x cos y and u x e x sin y v y u y e x cos y v x The function is analytic everywhere in the complex plane. Exercise 3

12 Exercise 4(a) For f (z) to exist, we need the auchy-riemann equations u x v y and u y v x to be satisfied. In this case, u 2xy and v x 2, and their partial derivatives are u x 2y, v y and u y 2x, v x 2x Hence, -R equations are only satisfied when y (i.e on the x-axis), and on the x-axis, f (z) u x + iv x i2x

13 Exercise 4(b) Substituting z x + iy, we have f(z) z 2 2z + 3 (x + iy) 2 2(x + iy) + 3 (x 2 y 2 2x + 3) + i(2xy 2y) u(x, y) + iv(x, y) The partial derivatives are and u x 2x 2 v y u y 2y v x Since -R equations are satisfied for all x, y, and u, v and their partial derivatives are continuous for all x, y, f(z) is analytic in the entire complex plane. The derivative of f(z) is f (z) u x + iv x 2x 2 + i2y 2z 2 The result confirms that polynomials in z are analytic in the entire complex plane (see p.53 of lecture notes) and the usual differentiation formulae applies.

14 Exercise 5. (a) The path is represented by z(t) t + it 2, t 1 Then dz (1 + i2t) dt, f[z(t)] t Hence, Re z dz 1 t(1 + i2t) dt i2 3 (b) The path is represented by z(t) t + i, t 1 Hence, (4z 3) dz 1 (4t + 4i 3) dt 1 + 4i Alternatively, since f(z) 4z 3 is analytic, the integral is path independent. Hence (4z 3) dz 2z 2 3z 1+i zi 1 + 4i (c) The four sides on the square are represented by 1 : z(t) it t 1 2 : z(t) t + i t 1 3 : z(t) 1 + it t 1 4 : z(t) t t 1 Hence, e z dz e it idt + e t+i dt 4 1 e 1+it idt + 1 e t dt Alternatively, since e z is entire, the integral is zero from auchy s theorem. (d) Im(z 2 ) dz i dt + 2t dt ti dt + 1 dt 1 i Exercise 5

15 EE27 Tutorial 9 auchy Integral Theorem, auchy Integral Formula and Real Integrals Exercise 1. These questions were discussed in the lectures. Ask your tutor if you still have doubts. (a) Evaluate (b) Evaluate 2π (c) Show that dz z 2 +1 where is the circle z 3. dθ 2 cos θ. dx π 1+x Exercise 2. Integrate 1 z 4 1 (a) z 1 1, (b) z 3 1 counterclockwise around the circle [ πi 2, ] Exercise 3. Evaluate the following integrals where is any simple closed path such that all the singularities lie inside (W). (a) (b) 5z z 2 +4 dz z+e z z 3 z dz (c) 3z 2 23z+5 (2z 1) 2 (3z 1) dz [ 1πi, 2πi( 1 + e+e 1 2 ), 5πi ] Exercise 4. Evaluate the following real integrals using the complex integration method. (a) 2π (b) 2π dθ 5 3 sin θ cos θ cos 2θ dθ [ π/2, ] Exercise 5. Evaluate the following improper integrals using the complex integration method. (a) x (x 2 2x+2) 2 dx (b) 1 dx (4+x 2 ) 2 [ π/2, π/16 ]

16 Solutions to Exercises Exercise 2. The function is not analytic at z 4 1, or z ±1, ±i. (a) The path is a circle of radius 1 and centered at z 1. enclosed. Factor z 4 1 as The singular point z o 1 is z 4 1 (z 1)(z 3 + z 2 + z + 1) and make use of the auchy s Integral formula f(z) dz 2πif(z o ) z z o In this case, f(z) (z 3 + z 2 + z + 1) 1 and f(z o ) 1/4. Hence, the answer is πi/2. (b) The answer is because ±1 and ±i are outside the circle. Exercise 2

17 Exercise 3. (a) z gives simple poles at z ±2i. Hence the integral is 5z z dz 5z 5z z+2i z 2i dz + z 2i z + 2i dz 5z 2πi( z + 2i z2i + 5z z 2i z 2i) 1πi (b) z 3 z gives simple poles at z and z ±1. Hence the integral is z + e z z 3 z dz z+e z z+e z (z+1)(z 1) z(z+1) dz + z z 1 dz + z + e z 2πi( (z + 1)(z 1) z + z + ez z(z + 1) z1 + 2πi( e e 1 2 z+e z z(z 1) z + 1 dz z + ez z(z 1) z 1) ) 2πi( 1 + e + e 1 ) 2 (c) The integrand has a second-order pole at z 1/2 and simple pole at z 1/3. Hence the integral is 3z 2 3z 23z z+5 3z 2 23z+5 (2z 1) 2 (3z 1) dz 3(2z 1) 2 z 1/3 dz + 4(3z 1) (z 1/2) 2 dz 2πi( 3z2 23z + 5 3(2z 1) 2 z1/3 + d 23z + 5 dz (3z2 ) 4(3z 1) z1/2 ) 2πi( ) 5πi Exercise 3

18 Exercise 4. (a) Let z e iθ and substitute sin θ 1 2i (z z 1 ) and dz iz dθ we have 2π dθ 5 3 sin θ 1 dz 5 3 2i (z z 1 ) iz 2 (3z i)(z 3i) dz 2 2πi 3(z 3i) zi/3 π 2 2 (3z 2 1iz 3) dz 2 3(z 3i) z i/3 dz since only the simple poles z i/3 is inside, the unit circle. (b) Let z e iθ and substitute we have cos 1 2 (z + z 1 ), cos 2θ 1 2 (z2 + z 2 ) and dz iz dθ 2π cos θ cos 2θ dθ 1 2 (z + z 1 ) dz 13 6(z 2 + z 2 ) iz z i(6z 4 13z 2 + 6) dz z i(3z 2 2)(2z 2 3) dz z i(2z 2 3) z 2 2/3 dz z i(2z 2 3)(z+ 2/3) z 2/3 + since only the simple poles z ± 2/3 are inside the unit circle. z i(2z 2 3)(z 2/3) z + 2/3 Exercise 4

19 Exercise 5. (a) onsider the function f(z) z (z 2 2z + 2) 2 z ((z 1 + i)(z 1 i)) 2 which has second-order poles at z 1 ± i. Only z 1 + i is inside the upper half plane (UHP). Hence the integral is x (x 2 2x + 2) 2 dx (b) onsider the function f(z) UHP 1 (4 + z 2 ) 2 1 (z + 2i) 2 (z 2i) 2 z ((z (1 i))(z (1 + i))) 2 dz 2πi d dz ( z (z (1 i)) 2 ) z1+i 2πi i 4 π 2 which has a second order pole z 2i inside the UHP. Hence the integral is dx (4 + x 2 ) 2 UHP 1 (z + 2i) 2 (z 2i) 2 dz 2πi d dz ( 1 (z + 2i) 2 ) z2i 2πi i π 16 Exercise 5

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