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MATH 316 (1)
John Toth (1)
Final

# Practice exam solution

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School
McGill University
Department
Mathematics & Statistics (Sci)
Course
MATH 316
Professor
John Toth
Semester
Fall

Description
SAMPLE PROBLEMS WITH SOLUTIONS FALL 2012 1. Let f(z) = y▯2xy+i(▯x+x ▯y )+z where z = x+iy is a complex variable de▯ned 0 in the whole complex plane. For what values of z does f (z) exist? Solution: Our plan is to identify the real and imaginary parts of f, and then check if the Cauchy-Riemann equations hold for them. We have f(z) = y ▯ 2xy + i(▯x + x ▯ y ) + x ▯ y + 2ixy 2 = x ▯ 2xy + y ▯ y + i(▯x + 2xy + x ▯ y ); 2 2 and so 2 2 2 2 u(x;y) = x ▯ 2xy + y ▯ y ; v(x;y) = ▯x + 2xy + x ▯ y : We compute the partial derivatives of u and v as ux(x;y) = 2x ▯ 2y; vx(x;y) = ▯1 + 2y + 2x; uy(x;y) = ▯2x + 1 ▯ 2y; vy(x;y) = 2x ▯ 2y: We see that the Cauchy-Riemann equations ux= v ;y vx= ▯u ; y hold all x and y, which means that f (z) exists for all values of z, i.e., the function f is an entire function. For completeness, we can compute the derivative f (z) = u + iv = 2x ▯ 2y + i(2x + 2y ▯ 1) = 2z + 2iz ▯ i: x x Alternative solution: Another way to solve this would be to notice that f(z) = z + iz ▯ iz; which reveals that f is entire since f is a polynomial (in z). 2. In the preceding question, take f(z) = cosx ▯ isinhy. Solution: This time the Cauchy-Riemann equations are quicker: 0 0 (cosx) x ▯sinx; (▯sinhy) =x0; 0 0 (cosx) y 0; (▯sinhy) = ycoshy: The equation ▯sinx = ▯coshy is never satis▯ed because sinx ▯ 1 and coshy > 1 0 except at y = 0. So f (z) does not exist anywhere. 1 2 FALL 2012 3 3. Show that f(z) = (z ▯+ 1) ▯ 3z ▯ is nowhere analytic. Solution: Expanding the cubic, we get f(z) = (x + 1 ▯ yi) ▯ 3(x ▯ yi) 3 2 2 3 = (x + 1) ▯ 3(x + 1)y ▯ 3(x + 1) yi + y i ▯ 3x + 3yi 3 2 3 2 = (x + 1) ▯ 3(x + 1)y ▯ 3x+i(y + 3y ▯ 3(x + 1) y): | {z } | {z } u(x;y) v(x;y) Let us compute the partial derivatives of u and v. u = 3(x + 1) ▯ 3y ▯ 3; v = ▯6(x + 1)y; x x 2 2 u y ▯6(x + 1)y; vy= 3y + 3 ▯ 3(x + 1) : We see that they satisfy exactly the opposite of what we want. For instance, we have v = u , rather than v = ▯u . So v = ▯u holds, only when v = u = 0. This x y x y x y x y means that 6(x + 1)y = 0, i.e, x = ▯1 or y = 0. Therefore f(z) has a chance of being di▯erentiable only at the lines x = ▯1 and y = 0. But then f(z) cannot be analytic, as any neighbourhood of each point on those lines will contain a point not on any of the lines, at which f is not di▯erentiable. d (e2tsin(2t)) 4. Find dt . Solution: Since sin(2t) is the imaginary part of e 2i, e sin(2t) is the imaginary part 2t(1+i) of e . Hence we can di▯erentiate the latter 5 times and then take the imaginary part of the result to ▯nd what is asked. We compute 5 2t(1+i) d e 5 5 2t(1+i) 5 = 2 (1 + i) e : dt p i▯=4 In order to take the 5-th power of 1 + i, we write it in polar form as 1 + i = 2e , and compute 5 5=2 5▯i=4 5=2 ▯1=2 (1 + i) = 2 e = ▯2 2 (1 + i); resulting in 5 5 2t(1+i) 5 2 2t 2 (1 + i) e = ▯2 2 (1 + i)e (cos2t + isin2t): 7 2t The imaginary part of this is ▯2 e (cos2t + sin2t), i.e., d (e sin(2t)) = ▯2 e (cos2t + sin2t): dt5 n n 5. What is the value of the integer n if x ▯ y is harmonic? Solution: Recall that a function u(x;y) is called harmonic if u +u = 0. So x ▯y n n▯2 n▯2 xx yy is harmonic if n(n ▯ 1)(x ▯ y ) = 0, which means that n = 0, n = 1, or n = 2. 6. Find the harmonic conjugate of e cosy + e cosx + xy. Solution: With u(x;y) = e cosy + e cosx + xy, we need to ▯nd a function v(x;y) such that f = u + iv is analytic, that is, u x v any v = ▯uy. We haye
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