Problems and Solutions in - Mathematics - University of Hawaii

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2.1 1989 April 2 COMPLEX ANALYSIS It remains to check that 2I = ∞ −∞ Indeed, γR f(z) dz → 0, as R → ∞, which will allow us to conclude that f(x) dx = lim R→∞ ΓR f(z) dz − f(z) dz γR = cos z γR (1 + z2 dz ) 2 ≤ This inequality holds for all R > 1, so, letting R → ∞, we have I = ∞ 0 γR f(z) dz = lim f(z) dz = R→∞ ΓR π . (32) 2e 1 (R2 − 1) 2 ℓ(γR) πR = (R2 . − 1) 2 γR cos x (1 + x2 π dx = ) 2 4e . 42 f(z) dz → 0. Therefore, by (32), ⊓⊔
2.2 1991 November 21 2 COMPLEX ANALYSIS 2.2 1991 November 21 INSTRUCTIONS: In each of sections A, B, and C, do all but one problem. TIME LIMIT: 2 hours 1. Where does the function SECTION A (Do 3 of the 4 problems.) f(z) = zRez + ¯zImz + ¯z have a complex derivative? Compute the derivative wherever it exists. Solution: Writing f in terms of the real and imaginary parts of z = x + iy, we have f(x + iy) = (x + iy)x + (x − iy)y + x − iy = x 2 + xy + x + i(xy − y 2 − y) = u(x, y) + iv(x, y), where u(x, y) = x 2 + xy + x and v(x, y) = xy − y 2 − y are the real and imaginary parts of f. Therefore, ux = 2x + y + 1 vy = x − 2y − 1 (33) uy = x vx = y. (34) If f is holomorphic in some region, the Cauchy-Riemann equations (ux = vy, uy = −vx) must hold there. By (33) and (34), this requires 2x + y + 1 = x − 2y − 1 and x = −y. Substituting the second equation into the first yields −y + 1 = −3y − 1, or y = −1. Then, since x = −y, we must have x = 1. Therefore, f has a complex derivative at (x, y) = (1, −1), or z = 1 − i. For any region Ω ⊆ C, we define the linear functional ∂ : H(Ω) → C by ∂ = 1 2 f ∈ H(Ω), then the derivative of f is given by f ′ (z) = (∂f)(z), z ∈ Ω. In the present case, ∂f = 2x + y + 1 + iy, ∂x ∂f ∂y = x + i(x − 2y − 1). ∂ ∂ ∂x − i ∂y , and recall that, if Therefore, ∂f(x + iy) = 1 1 2 [(2x + y + 1 + iy) − i(x + i(x − 2y − 1))] = 2 [(3x − y) + i(y − x)], and finally, f ′ (1 − i) = 1 (4 − 2i) = 2 − i. 2 2. (a) Prove that any nonconstant polynomial with complex coefficients has at least one root. (b) From (a) it follows that every nonconstant polynomial P has the factorization P (z) = a N (z − λn), n=1 43 ⊓⊔
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Page 79 and 80: B List of Symbols F an arbitrary fi
Page 81: INDEX INDEX Riemann mapping theorem

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