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WA 4: Solutions WA 4: Solutions Problem 1. Show that for any z such that |Im z| ≥ δ, with some δ > 0, 1 1 1 2 1 2 |tan z| ≤ 1 + , |cot z| ≤ 1 + . sinh2 δ sinh2 δ Solution. We have | sin z|2 sin2 x + sinh2 y 1 + sinh2 y 1 | tan z|2 = = ≤ = 1 + . | cos z|2 cos2 x + sinh2 y sinh2 y sinh2 y Since sinh2 y is an even function, it suffices to show that sinh2 y ≥ sinh2 δ when y ≥ δ. This follows from the fact that (sinh2 y)0 = 2 sinh y cosh y > 0 when y > 0, and thus sinh2 y is increasing for positive y. Therefore, 1 1 | tan z|2 ≤ 1 + ≤ 1 + . sinh2 y sinh2 δ The second inequality is proved analogously. −1 −1 π Problem 2. (a) Show that the identity sin z +cos z = 2 holds for all z ∈ C and the principal values of sin−1 z and cos−1 z. (b) Is this identity true for sin−1 z and cos−1 z considered as multivalued func- tions? Solution. (a) For z ∈ (−1, 1), the principal values of sin−1 z and cos−1 z coincide with the values of these functions as defined in trigonometry. Indeed, for x ∈ (−1, 1) we have p sin−1 x = −iLog[ix + (1 − x2)1/2] = −iLog[ix + 1 − x2] p p = −i(ln 1 + iArg(ix + 1 − x2)) = Arg(ix + 1 − x2), √ 2 π π −1 and since Arg(ix+ 1 − x ) ∈ (− 2 , 2 ), the statement for sin x follows. Similarly, p cos−1 x = −iLog[x + i(1 − x2)1/2] = −iLog[x + i 1 − x2] p p = −i(ln 1 + iArg(x + i 1 − x2)) = Arg(x + i 1 − x2), √ and since Arg(x + i 1 − x2) ∈ (0, π), the statement for cos−1 x follows. Since the interval (−1, 1) belongs to the domain of analyticity for the principal branches of functions sin−1 z and cos−1 z, and all points of this interval are accumulation −1 −1 π points, by the uniqueness theorem the identity sin z + cos z = 2 holds for all z ∈ except when 1 − z2 ∈ (−∞, 0] or when the Log is applied to a negative C √ number. Extending the principal value of the square root, P.V.(reiΘ)1/2 = reiΘ/2, −π < Θ < π, to Θ = π by (one-sided) continuity using the same formula, i.e., assigning the argument π to negative numbers, and doing the same with Log, we observe that the identity holds for all z ∈ C. (Notice that Log is never applied to 0 in the formulas for sin−1 z and cos−1 z.) −1 −1 π (b) The identity sin z + cos z = 2 + 2πn, n ∈ Z, must replace the identity −1 π −1 in part (a). However, the identity sin z = 2 − cos z still holds for multivalued functions if it is understood as the equality of the two sets of values. 1 2 Problem 3. Find the linear fractional transformation that maps the points z1 = −i, z2 = 0, z3 = i to the points w1 = −1, w2 = i, w3 = 1, respectively. What is the image of the imaginary axis Re z = 0 under this transformation? Solution. The LFT is uniquely determined by the equation (w − w )(w − w ) (z − z )(z − z ) 1 2 3 = 1 2 3 , (w − w3)(w2 − w1) (z − z3)(z2 − z1) i.e., by (w + 1)(i − 1) (z + i)(−i) = . (w − 1)(i + 1) (z − i)i Solving for w, we obtain z − 1 w = −i . z + 1 Since for every y ∈ R, 2 2 2 iy − 1 y + 1 |w(iy)| = −i = = 1, iy + 1 y2 + 1 the imaginary axis Re z = 0 is mapped to the unit circle |w| = 1, with the point w(∞) = −i excluded. (The extended imaginary axis is transformed to the whole unit circle.) Problem 4. Find the linear transformation w = Az + B, A 6= 0, mapping the strip {z = x + iy : kx + b1 ≤ y ≤ kx + b2}, where k, b1, b2 are real constants and b1 < b2, onto the strip {w = u + iv : 0 ≤ u ≤ 1}, so that z = ib2 is mapped to w = 0. Solution. The boundary lines are mapped to the boundary lines under the linear transformation w = Az + B. Since z = ib2 belongs to the line y = kx + b2 (for x = 0), and 0 belongs to the line u = 0, we conclude that the line y = kx + b2 is mapped to u = 0, and the line y = kx + b1 is mapped to the line u = 1. The first condition can be written as Re (A(x + i(kx + b2)) + B) = 0, and the second condition can be written as Re (A(x + i(kx + b1)) + B) = 1. Since z = ib2 is mapped to w = 0, we obtain 0 = Aib2 + B, i.e., B = −iAb2. Then the first condition becomes Re (A(1 + ik)) = Re A − kIm A = 0, i.e., Re A = kIm A. Then the second condition becomes Re (A(x + i(kx + b1 − b2)) = Re (iA(b1 − b2)) = Im A(b2 − b1) = 1, i.e., Im A = 1 . Therefore, b2−b1 k + i k + i A = ,B = −ib2 , b2 − b1 b2 − b1 3 and k + i w = Az + B = (z − ib2). b2 − b1 Problem 5 (a) Suppose that a linear fractional transformation maps the real line into itself. Show that the images of any two points symmetric about the real line are also symmetric about the real line. (b) Extend the result of (a) to an arbitrary pair of straight lines, i.e., show that if a linear fractional transformation maps a line L1 into a line L2 then the images of any two points symmetric about L1 are symmetric about L2. Solution. (a) We have that for every x ∈ R, w(x) − w(x) = 0, i.e., ax + b ax¯ + ¯b (ac¯ − ca¯)x2 + (ad¯− ad¯ )x + (bd¯− ¯bd) 0 = − = . cx + d cx¯ + d¯ |cx + d|2 Since a nontrivial quadratic polynomial can have at most two real zeros, the equality above can hold for all x ∈ R only if ac¯ − ca¯ = ad¯− ad¯ = bd¯− ¯bd, which holds only if a, b, c, and d all have the same argument, say θ. Multiply- ing them all by e−iθ if necessary, we obtain that the transformation w(z) can be described as az + b w(z) = cz + d with a, b, c, d ∈ R. Then az¯ + b az + b− w(¯z) = = = w(z), cz¯ + d cz + d i.e., w maps symmetric points about the real line into symmetric points about the real line. (b) Let w = T (z) be a given linear transformation which maps a line L1 to a line L2. Then one can represent both lines as the images of the real line under linear transformations: z = αZ + β, w = γW + δ, α 6= 0, γ 6= 0, i.e., z ∈ L1 if and only if Im Z = 0, and w ∈ L2 if and only if Im W = 0. Then the linear fractional transformation defined implicitly by γW + δ = T (αZ + β) maps the real line to the real line. By part (a), it has the property that the images of any two points symmetric about the real line are also symmetric about the real line. Since linear transformations are combinations of shifts, scalings, and rotations, and each of these operations maps two symmetric points about a line to two points symmetric about the image line, the original transformation w = T (z) also has this property, i.e., any two points z1 and z2 symmetric about L1 are mapped by the transformation T to w1 and w2 symmetric about L2. Problem 6. Find the image domains of the unit disk and its upper half under 5−4z the linear fractional transformation 4z−2 . 5−4z Solution. The transformation w = 4z−2 maps the real line to itself. The unit circle intersects the real line at two points z1 = 1 and z2 = −1, whose images are w1 = 1/2 and w2 = −3/2. Therefore, the image of the unit circle is a circle through w1 and w2 (it cannot be a line, since this line would be the real line whose 4 pre-image is the real line itself). Since the unit circle is symmetric about the real line, so is its image (see Problem 5(a)). Therefore, the line segment [−3/2, 1/2] is the diameter of the image circle, and its center is at w0 = −1/2. We conclude that the image of the unit circle is the circle |w + 1/2| = 1. Since w(0) = −5/2 does not belong to the circle |w + 1/2| = 1, the unit disk is mapped to the exterior of 5−4i the circle |w + 1/2| = 1. Since w(i) = 4i−2 = −1.3 − 1.2i has a negative imaginary part, the LFT maps the upper half plane onto the lower half plane. Therefore, the image of the upper half of the unit disk is the portion of the exterior part of the circle |w + 1/2| = 1 which lies in the lower half plane..
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