Indian Institute of Technology Indore MA 205: Complex Analysis PDF
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Indian Institute of Technology Indore
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M. A. Khan
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This document is an assignment for Mathematics-Complex Analysis course at Indian Institute of Technology Indore. The document contains various questions covering topics of complex analysis and their solutions.
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INDIAN INSTITUTE OF TECHNOLOGY INDORE Autumn Semester Assignment CA# 2 MA 205: Complex Analysis 1. Explain the geometrical meaning of the relations given in probl...
INDIAN INSTITUTE OF TECHNOLOGY INDORE Autumn Semester Assignment CA# 2 MA 205: Complex Analysis 1. Explain the geometrical meaning of the relations given in problems. |z − z0 | < R, |z − z0 | > R, |z − z0 | = R. |z − 2| + |z + 2| = 5. Solution: An ellipse with foci at the points z = ±2 and major semi-axis 5/2. Rez ≥ m Solution: The straight line x = m and the half-plane on the right of it. Rez + Imz < 1 Solution: The half-plane bounded by the straight line x+y = 1 and containing the coordinate origin. Arg z = θ0. −π < Arg z < π. π 6 < Arg z < π 4. 1 < |z| < 3. 2. Find the real numbers p and q such that the complex numbers z = p + iq, w = p + i 1q be equal. 3. Given a non-zero complex number z0 and a natural number n ∈ N, find all distinct complex numbers w such that z0 = wn. 4. Find the principal argument of the following numbers: 5i Solution: The half-plane bounded by the straight line x+y = 1 and containing the coordinate origin. π Ans: 0 1 2 + 2 i Ans: π 4 −1 2 − 2 i 1 + √i3 Ans: π 6 1√ 1+i 3 Ans: −π 3 √ − 41 + i 4 3 Ans: −π 3 +π = 2π 3 3 1+i1√3 Ans: −π 3 × 3 + 2π = π Page 1 of 3 Assignment CA#3 M. A. Khan − 1−i 1+i Ans: −π 2 5. Find all the values of the following roots √ 1. 3 1 Ans: 1 √ 2. 3 + 4i Ans: ±(2 + i) p √ 2 3. 4 (−1) Ans: ± 2 (1 ± i) 3 4. 1+i1√3 Ans: − 213 + i0 √ 5. 3 i 6. Let z0 be a limit point of S. Then prove that every neighbourhood of z0 contains infinitely many points of S. 7. Find a parametric representation γ : [a, b] → C for the following curves: 1. The straight-line segment from 0 to 4 − 7i Ans: γ(t) = (4 − 7i)t, 0 ≤ t ≤ 1 2. The upper half of |z − 4 + 2i| = 3 Ans: γ(t) = 4 − 2i + 3(cos t + i sin t), 0 ≤ t ≤ π 1 1 3. y = x from (1, 1) to (4, 4) Ans: γ(t) = t + ti , 1 ≤ t ≤ 4 4. |z + 3 − i| = 5, counter clockwise Ans: γ(t) = −3 + i + 5(cos t + i sin t), 0 ≤ t ≤ 2π 5. |z + 3 − i| = 5, clockwise Ans: (2π) − (−3 + i + 5(cos t + i sin t)), 0 ≤ t ≤ 2π 8. Check if the following functions can be prescribed a value at z = 0, so that they become continuous: |z|2 1. f (z) = z Solution: Yes, as limz→0 f (z) exists and is equal to 0. Therefore, we need to take f (0) = 0 z 2. f (z) = z Solution: Ans: No as limz→0 f (z) does not exists. Re z 3. f (z) = z Solution: Ans: No z 4. f (z) = |z| Solution: Ans: No zRe z 5. f (z) = |z| Solution: Yes, f (z) = 0. Prove limx,y→(0,0) √ xy p 2 = 0 using | x2 + y 2 | ≥ |x| x +y 2 1 9. Examine the continuity of the function (a) f (z) = 1−z on the domain D = {z : |z| < 1}, and (b) f (z) = |z|2 on C. 10. Which of the following subsets of C are domain. Page 2 of 3 Assignment CA#3 M. A. Khan {z : |z − 1| < 1} ∪ {z : |z − 3| < 1}. {z : |z − 1| < 1} ∪ {z : |z − 3| < 1} ∪ {1}. {z : Re z × Im z > 0} ∪ {0}. {z : Re z × Im z > 0}. 11. Let zk = rk (cos θk + i sin θk ), k = 1, 2,... , n. Show that 1. z1 z2 = r1 r2 [cos(θ1 + θ2 ) + i sin(θ1 + θ2 )]. 2. z1 z2 · · · zn = r1 r2 · · · rn [cos(θ1 + θ2 + · · · + θn ) + i sin(θ1 + θ2 + · · · + θn )]. 3. De Moiver’s formula: z n = [r(cos θ + i sin θ)]n = rn (cos nθ + i sin nθ), where n is an integer. Page 3 of 3