Complex Analysis Homework 3 Steve Clanton David Holz March 3, 009 Problem 3 Solution. Let z = re iθ. Then, we see the mapping leaves the modulus unchanged while multiplying the argument by -3: Ω z = z re iθ z = re iθ = re 3iθ In particular, we note Ω r e iθ + e iθ = Ω r + e iθ = r + e 3iθ = r e 3iθ + e 3iθ = Ω r e iθ + Ω e iθ Now, we consider the images of z + ε and z + iε where Arg z = Arg ε. We note in Figure that the angles between the image points triple while the modulus stays the same. z i i z 3 z i z z Figure : Ω is not a conformal mapping The mapping is not conformal since Ω z + ε = Ω z + Ω ε but Ω z + iε Ω z + i Ω ε. The image Ω z Ω z + iε has a different modulus than and is not perpendicular to Ω ε. The mapping is neither conformal nor anticonformal since the lines in the image of a perpendicular intersection are not perpendicular.
Problem 4 Problem. The picture shows the shaded interior of a curve being mapped by an analytic function to the exterior of the image curve. If z travels round the curve counter-clockwise, then which way does its image w travel round the image curve? We have two particularly important vectors we may draw on each curve:. The yellow vector, which is tangent to the curve at zome point z and points in the direction of circulation.. The blue vector, which is perpindicular to the yellow vector and points towards the shaded region. In order to preserve the angle between these two vectors as well as their respective properties the the yellow vector must lay be degrees clockwise from the blue vector. By this we see that the direction of circulation on the mapped curve is clockwise. Figure : By preserving the angles between tangent and perpindicular vectors we may deduce the circulation of the new curve.
Problem 5 Problem. Consider fx + iy = x + y + iy/x. Find and sketch the curves that are mapped by f into a horizontal lines, and b vertical lines. Notice from your answers that f appears to be conformal. Show that it is not in two ways: i by explicitly finding some curves whose angle of intersection isn t preserved; and ii by using the Cauchy-Riemann equations. a Horizontal lines have a constant imaginary component: y I[fx + iy] = = const x b This describes all lines with constant slope that go through the origin, otherwise known as radial lines. Veritcal lines have a constant real component: This describes all circles centered about the origin. R[fx + iy] = x + y = const 3 4 0 0 3 0 4 4 6 8 0 Figure 3: Radial lines map to horizontal lines and origin centered circles map to vertical lines i Vertical line x = z = + t + it x = + t, y = t Horizontal line y = z = + t + i t x = + t, y = t Finding the intersection point: t = t t = ± x =, y = ± 3
We may check if they are conformal by looking at the derivatives of each line at the intersection. Intersection at: x = Horizontal line: y = x = 3/ Vertical line: y = x = The slopes of the tangent lines at the point of intersections are not perpindicular as they were before the mapping. Thus, this mapping is not conformal..5 0.5.5 0.5 0 0 0.5.5 0 0.5.5.5.5 3 3.5 4 4.5 5 Figure 4: An intersection of two lines which are not conformal. ii u = x + y, v = y x x u = x x = yv The mapping fx + iy does not satisfy the Cauchy-Riemann equations, thus it is not conformal. We see that the Cauchy-Riemann equations are satisfied when x = ± / since the equations reduce to the condition x u = y v x = x x v = y u y x = y x = 4
Problem 6 Problem. Continuing from the previous exercise, show that no choice of v can make fx + iy = x + y + iv analytic. u = x + y, x u = y v, v = v x v = y u x = x v, y = y v v =xy + fx 0 v = xy + gy 0 v =xy xy We must have made an incorrect assumption, namely that there is a value of v which can make fx+iy analytic. Alternatively, we may also demonstrate this by: Problem 4 f = x iy = x iy = z Geometrically, we know that all the sides of the rectangle will be rotated and amplified by some amplitwist f. We can apply the rotation first, and we note that the rotation transforms our original rectangle onto a congruent rectangle. Next we scale each side by f. This change our area to A new = f l f w = f lw = f A old ds dv du dr Figure 5: We can also derive this fact by looking at the determinant of the Jacobian: A old = dr x dr y ds x ds y A new = du x du y dv x dv y = drx dr J y = J ds x ds y dr x dr y ds x ds y = J A old J = xu x v y u y v = xu y v y u x v = x u + x v = x u + i x v x u i x v = f f = f 5
Problem 5 i b b a b r a b e b b R a b e Figure 6: The transformation of a square under z e z ii Call the square S and the image of the square S. The area of S is 4b, since the length of one side is b. We see the area of the image S is θ π of the disc area of S = θ πr π = b e a+b e a b = be a sinh b π and iii area of Λ = S area of S = ea sinh b b From exercise 4, we know that the local magnification for the factor for the area of the square is the square of the amplification. We know the amplification is e a since we found e z = e z in exercise 3. Accordingly, we expect the area of S to be e a times the area of S as b shrinks to nothing. iv lim Λ = lim e a sinh b b 0 b 0 b LH = e a lim b 0 cosh b = e a lim b 0 cosh b = e a 6
Problem 6 Figure 7: Consider some arbitrarily short chord of the origin-centered circle passing through z. Because intersections are preserved in inversion, we know that the image of the chord is a chord through the corresponding points in the inversion of its arc. Because the mapping is conformal, we know the image of the chord is a straight line and that the direction stays the same. In Figure 7, we see that the triangle formed by points 0, z, and z + ε is similar to the one formed by 0, I z, and I z + ε, since they are both isosceles and share a vertex. This tells us that the modulus of the amplitwist a = /z solves r = R = ar. Thus, we have /z / = /R = z 7
Problem 7 i Needham gives the formula x+iy = ii The Jacobian is x x +y i u = y x +y x u J = y u = x v y v on page 7 of the text. Thus, x x + y and v = y x + y y x x +y xy x +y xy y x x +y x +y We note that x u = y v and that x v = y u for all points in their domain. Thus, the Cauchy-Riemann equations are satisfied except where x + y = 0, which only occurs when x = y = 0. iii Recall the polar coordinate relationships x = r cos θ, y = r sin θ, and x + y =. Now, we put the Jacobian from part ii into polar form: J = = y x xy x +y x +y xy y x x +y x +y sin θ cos θ sin θ cos θ = cos θ sin θ sin θ cos θ sin θ cos θ sin θ cos θ = r sin θ r cos θ r cos θr sin θ r 4 r 4 r cos θr sin θ r sin θ r cos θ r 4 r 4 = cosθ sinθ sinθ cosθ This transformation is scaling by the factor / followed by one rotation by π by the negative sign and another rotation of θ by the matrix. iv f = x u + i x v = y x +ixy x +y = x iy x+iy x iy = x+iy = z We know that z = e iθ = e iθ is the same transformation as in part iii, scaling by the factor / followed by one rotation by π by the negative sign and another rotation of θ. 8