Complex Variables 05-3, Exam 1 Solutions, 7/14/5 Question 1

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1 Complex Variables 05-3, Exam 1 Solutions, 7/14/5 Question 1 Let a circle S in the plane be given, centered at (2, 3) and of radius 5 units. Find the complex equation of the circle. The equation is: z 2 3i = 5, (z 2 3i)(z 2 + 3i) = 25, zz (2 3i)z (2 + 3i)z + 13 = 25, 0 = zz (2 3i)z (2 + 3i)z 12. Find the image of the circle under the transformation z z i. If w = z i, then z = w 1 i. Substituting in the equation of the circle, we get: 0 = (w 1 i)(w 1 + i) (2 3i)(w 1 i) (2 + 3i)(w 1 + i) 12 = ww (3 4i)w (3 + 4i)w = (w 3 4i)(w 3 + 4i) 25 = w 3 4i So the image circle is centered at (3, 4) and has radius 5. Find the image of the circle under the transformation z (1 + i)z. If w = (1 + i)z, then z = (1 i) 2 w. Substituting in the equation of the circle, we get: 0 = (1 i) 2 (1 + i) ww (2 3i)(1 i)w 1 (2 + 3i)(1 + i)w 12 2 = 1 2 ww 1 2 (2 3i)(1 i)w 1 (2 + 3i)(1 + i)w 12, 2 0 = ww (2 3i)(1 i)w (2 + 3i)(1 + i)w 24 = ww + (1 + 5i)w + (1 5i)w 24 = (w + 1 5i)(w i) 50 = w + 1 5i 2 (5 2) 2. So the image circle is centered at ( 1, 5) and has radius 5 2.

2 Find the image of the circle under the transformation z 1 z. If w = z 1, then z = w 1. Substituting in the equation of the circle, we get: 0 = 1 w 1 w (2 3i) 1 w (2 + 3i) 1 w 12, 0 = 12ww + (2 + 3i)w + (2 3i)w 1, 0 = ww (2 + 3i)w (2 3i)w 1 12 = (w (2 3i))(w (2 + 3i)) = w i 4 2 ( 5 12 )2. So the image circle is centered at ( 1, 1 5 ) and has radius Also sketch the circle and its three images on the same complex plane.

3 Question 2 Compute the following limits, or explain why the limit in question does not exist. lim z i z z 2 + 3iz 2 We have: lim z i z z 2 + 3iz 2 = lim z i z 2 2i lim z 1+i z 2 2z + 2 We have: lim z 0 I(z) z. lim z 1+i (z i)(z + i) (z + 2i)(z + i) = lim z i z 2 2i z 2 2z + 2 = lim z 2 (1 + i) 2 z 1+i (z 1) (z 1 i)(z i) = lim z 1+i (z 1 i)(z 1 + i) = lim (z i) z 1+i (z 1 + i) = 1 + i i 1 + i 1 + i = 2 + 2i = 1 i. 2i Put z = x + iy with x and y real. Then we need: y lim x+iy 0 x + iy. Take x = 0 and approach the origin along the y-axis. Then we get: y lim y 0 iy = i. Take y = 0 and approach the origin along the x-axis. Then we get: lim x 0 0 = 0. (z i) (z + 2i) = i i i + 2i = 2. Since by two different approaches to the origin, we obtain two different limits, the required limit does not exist.

4 Question 3 Show that z = 2i is a root of the polynomial f(z) = z 4 + 2z 3 + 6z 2 + 8z + 8. Hence factor the polynomial f(z) and plot the roots of the polynomial on the complex plane. Also show that g(z) = f(z + 1) and h(z) = z 4 f( 1 ) are polynomials and z plot the roots of all three polynomials f, g and h on the same complex plane. We have: f(2i) = (2i) 4 + 2(2i) 3 + 6(2i) 2 + 8(2i) + 8 = 16 8i i + 8 = 0. So z = 2i is a root of f(z). Since the polynomial is real for each root, its complex conjugate is also a root. So z = 2i is a root of f(z). So f(z) has a factor of (z 2i)(z + 2i) = z We see that: f(z) = z 4 + 2z 3 + 6z 2 + 8z + 8 = z 4 + 6z z 3 + 8z = (z 2 + 4)(z 2 + 2) + 2z(z 2 + 4) = (z 2 + 4)(z 2 + 2z + 2) = (z 2 + 4)((z + 1) 2 + 1) = (z + 2i)(z 2i)(z i)(z + 1 i). So the roots of f(z) are 2i, 2i, 1 i and 1 + i. Then we have: g(z) = f(z + 1) = (z i)(z + 1 2i)(z i)(z + 2 i). This is a product of polynomials so is a polynomial. Explicitly, we have: g(z) = ((z + 1) 2 + 4)((z + 2) 2 + 1) = (z 2 + 2z + 5)(z 2 + 4z + 5)

5 = z 4 + 6z z z The roots of g(z) are 1 2i, 1 + 2i, 2 i and 2 + i. Next we have: h(z) = z 4 f( 1 z ) = z4 ( 1 z + 2i ) ( 1 z 2i ) ( 1 z i ) ( 1 z + 1 i ) = (1 + 2iz)(1 2iz)(1 + (1 + i)z)(1 + (1 i)z). This is a product of polynomials, so is itself a polynomial. Explicitly, we have: h(z) = z 4 f ( ) ( (1 ) 4 1 = z z z ( ) z = 1 + 2z + 6z 2 + 8z + 8z 4. ( ) z ( ) ) z The roots of h(z) are the reciprocals of the roots of f(z), so the set of roots is: { 1 2i, 1 2i, 1 1 i, i } = { i 2, i 2, 1 2 (1 i), 1 (1 + i)}. 2

6 Question 4 Give parametrizations for the boundaries of the following regions in the complex plane and identify each region as open, closed, or neither and as connected or disconnected. Also sketch each region in the complex plane. R = {z C : z 3i < 3}. This is a standard open connected bounded disc centered at 3i and of radius 3, so has the parameterization: z = 3i + 3e it, 0 t 2π. S = {z C : (2 + i)z + (2 i)z < 4}. Clearly S is an open connected unbounded half-plane bounded by the straight line (2 + i)z + (2 i)z = 4. A point of this line is z = 1. Another point is z = 2i. So the line is z = 1 + t( 2i 1), where t varies over all of R. This line passes below the origin, and the origin lies in S (since 0 < 4 is true), so S is the half-plane lying above the line. T = {z C : z > 3 and z 4 5}. This region is the region outside the closed disc of radius 3, center the origin and inside the open disc centered at 4 of radius 5. The region is bounded and connected, but is neither closed nor open. Sketching, we see that the boundary consists of two arcs, one on the circle z = 3 and the other on the circle z 4 = 5, meeting at ±3i.

7 Alternatively we may solve for the meeting points of the boundary circles: z = 3, z 4 = 5, zz = 9, (z 4)(z 4) = 25, zz = 9, zz 4z 4z + 16 = 25, x 2 + y 2 = 9, z + z = 0, x = 0, y 2 = 9,, (x, y) = (0, ±3), z = ±3i. A parameterization for the boundary arcs is then: z = 3e it, π 2 t π 2. z = 4 + 5e is, a s a. Here the endpoint parameter value a is given by: 4 + 5e ia = 3i, 5e ia = 3i 4, e ia = 4 + 3i 5 = cos(a) + i sin(a), cos(a) = 4 5, sin(a) = 3 5, tan(a) = 3 4, ( ) 3 a = π arctan = radians = degrees. 4

8 Question 5 Let u(x, y) = x 3 + 4x 2 + x(6 3y 2 ) 4y 2, defined for any real (x, y). Prove that u is harmonic. Differentiating, we have: u = x 3 + 4x 2 + x(6 3y 2 ) 4y 2 u x = 3x 2 + 8x + 6 3y 2, u y = 6xy 8y, u xx = 6x + 8, u yy = 6x 8. So u xx + u yy = 6x + 8 6x 8 = 0 and u is harmonic, as required. Find the harmonic conjugate v of u. We first need: v y = u x = 3x 2 + 8x + 6 3y 2. Integrating with respect to y, we get: Finally we need: v = 3x 2 y + 8xy + 6y y 3 + g(x). 0 = u y + v x = 6xy 8y + 6xy + 8y + g (x) = g (x). So g(x) = C, a constant and the harmonic conjugate of u is v = 3x 2 y + 8xy + 6y y 3 + C, with C an arbitrary real constant. If f(z) = u(x, y) + iv(x, y), where z = x + iy, explain why f(z) is analytic and write a formula for f (z) in terms of the derivatives of u and v. Since the derivatives u x, u y, v x and v y are everywhere defined and continuous and since the Cauchy-Riemann equations hold everywhere, by our basic theorem on differentiability, we know that f(z) is differentiable everywhere, so is analytic (and entire). Then we have, by the same theorem: f (z) = f x = u x + iv x = 3x 2 + 8x + 6 3y 2 + i(6xy + 8y).

9 Write f (z) directly as a polynomial in z. By the above, we have: f (z) = 3x 2 + 8x + 6 3y 2 + i(6xy + 8y) = 3(x 2 y 2 + 2ixy) + 8(x + iy) + 6 = 3(x + iy) 2 + 8(x + iy) + 6 = 3z 2 + 8z + 6. By integrating your expression for f (z), or otherwise, express f(z) as a polynomial in the variable z. We have 3z 2 + 8z + 6 = d dz (z3 + 4z 2 + 6z), so we get: f(z) = z 3 + 4z 2 + 6z + A, where A is constant. Note that f(0) = A = u(0, 0) + iv(0, 0) = ic. Check: z 3 + 4z 2 + 6z + C = (x + iy) 3 + 4(x + iy) 2 + 6(x + iy) + ic = x 3 3xy 2 + 4x 2 4y 2 + 6x + i(3x 2 y + 8xy + 6y + C) = u(x, y) + iv(x, y).

10 Question 6 Consider the function, defined for all points z = x + iy in the complex plane, where x and y are real: f(x, y) = 2xy + i(x 2 3y 2 ). Find, with proof, all points where the function f is differentiable, giving its derivative at each such point and find all such points at which the function is analytic. We note that the real and imaginary parts of f are polynomials, so are everywhere continuous and differentiable, with continous derivatives. Therefore by the main theorem on differentiability, f is differentiable at a point iff it obeys the Cauchy-Riemann equations at that point and then the value of the derivative is f x evaluated at that point, where f x is the partial derivative of f with respect to x. Computing the partial derivatives, we have: f x = 2y + 2ix, f y = 2x 6iy. The Cauchy-Riemann equations are then: 0 = f x + if y = 2y + 2ix + i(2x 6iy) = 8y + 4ix. So the Cauchy-Riemann equations hold only when y = 0 and x = 0. So by our main theorem, f is differentiable only at the origin, where it has derivative f x = [2y + 2ix] x=0,y=0 = 0. Finally, since there is no open region on which f is differentiable, f is analytic no-where.