Problem Set # 02 MIT (18.04 spring 2020) Rodolfo R. Rosales (MIT, Math. Dept., room 2-337, Cambridge, MA 02139) February 22, 2020

Due Thursday March 5, 2020. Turn it in (by 3PM) at the Math. Problem Set Boxes, right outside ...... room 4-174. There is a box/slot there for 18.04. Be careful to use the right box (there are many slots). Contents

1 Other Problems 1 OT 99.2.1 Images of sets by the inverse of the cubic root ...... 1

2 S&S 2 problems 2 ss 2.3.04 The real and imaginary parts, and the absolute value, are not complex differentiable ...... 2 ss 2.3.16 The mean-value theorem does not apply to complex functions ...... 2

3 Roots, Branch Points, and Formulas #02 2 Formulas for arccoth and arcsech ...... 2

4 Polynomial roots; branches with geometry #01√ 2 4 Find the branch points (and design branch cuts) for√ z − 1 using geometrical arguments ...... 2 Example: Branch points and branch cuts for z2 − 1 using geometrical arguments ...... 2 Example: The danger of blindly using canned functions ...... 4

5 Contour Integration #01 4 Evaluate integral of arbitrary power around unit circle ...... 4

6 Contour Integration #04 5 Evaluate integral of z |z|2 over the “hyperbolic rectangle” contour ...... 5 List of Figures

4.1 Polars for product√ of two roots ...... 3 2 4.2 Branch cuts for z − 1...... √ 3 4.3 Contours to check the branch cuts of z2 − 1...... 3 6.1 The “hyperbolic rectangle” ...... 5

1 Other Problems

♣ [OT 99.2.1] Consider the multiple valued mapping in the complex plane given by: z −→ z−1/3. What are the images, under this map, of (1) The half plane: Re(z) > 0? (2) The quadrant: Re(z) < 0 and Im(z) < 0? π π (3) The wedge: − < arg(z) < ? 4 4 In each case: draw the initial set, the image set, and explain your answer.

1 2

2 S&S 2 problems

♣ [ss 2.3.04] Use the definition of complex derivative to show that each of the following functions is nowhere differentiable: (a) f(z) = Re(z). (b) f(z) = Im(z). (c) f(z) = |z|. This example shows how much more restrictive than the real derivative is the complex derivative. Here (a-b) are smooth (have partial derivatives of all orders) everywhere in the complex plane, while for (c) smoothness fails at z = 0 only. √ √ 3 ♣ [ss 2.3.16] Let f(z) = z + 1 and let z1 = (−1 + i 3)/2 and z2 = (−1 − i 3)/2. Show that there is no 0 point w in the straight line segment from z1 to z2 such that f(z2) − f(z1) = f (w)(z2 − z1). This shows that the mean-value theorem of calculus does not extend to complex functions!

3 Roots, Branch Points, and Formulas #02

Statement: Roots, Branch Points, and Formulas #02 Derive the following formulas: √ ! 1 z + 1 1 + 1 − z2 (a) coth−1(z) = log , and (b) sech−1(z) = log . (3.1) 2 z − 1 z

Recal that coth−1 and sech−1 denote the inverse functions to the hyperbolic cotangent and secant. Important note: Note that here the logarithm and the square root are meant in their fully multiple-valued glory — this is why we do not write ± in front of the square root. Obviously coth−1 and sech−1 are multiple valued as well. Choose above any allowed value for the logarithm (and the square root), and an allowed value for coth−1 and sech−1 follows.

4 Polynomial roots; branches with geometry #01

Statement: Polynomial roots; branches with geometry #01 √ A. Use the geometrical approach in example 4.1 to find the branch points for F (z) = z4 − 1, and devise appropriate branch cut geometries — propose at least two distinct ones (i.e.: they should not be merely distorted versions of the same geometry). B. Use the geometrical approach in example 4.1 to find the branch points for G(z) = (z4 − 1)1/4. C. Challenge: for case B, propose and analyze branch cut geometries (again, at least two distinct ones). Note: I am calling this a challenge because there is a potential trap you can be caught into. √ Branch points and branch cuts for z2 − 1 using geometrical arguments √ Example 4.1 Branch points and branch cuts for f(z) = z2 − 1 using geometrical arguments. 1 1 See remark 4.1. 3

Let z = r ei φn (n = 1, 2) be the vectors/complex numbers connecting n n √ 2 i (φ1+φ2)/2 the roots of z − 1 to an arbitrary z; as shown in figure 4.1. Hence f(z) = r1 r2 e . (4.1) This formula allows easy testing of branch points and branch cut geometries.

Figure 4.1: Example 4.1 The picture on the left shows the polar coordinates used √ √ 2 i (φ1+φ2)/2 to write f(z) = z − 1 as f(z) = r1 r2e . i φ1 Here z + 1 = r1 e is the polar representation of the i φ2 vector from −1 to z. Similarly z − 1 = r2 e is the polar representation of the vector from 1 to z.

Consider a tiny loop around some arbitrary complex number z, traverse it counterclockwise (once), and observe what happens with φ1 and φ2 — hence the value of f.

1. If z 6= ±1, the values of φn oscillate a little and return to their initial values. Hence so does f.

2. If z = 1 or z = −1, one of the φn oscillates a little and returns to its initial value. But the other φn increases by 2 π. Hence the value of f flips sign. 2 3. If z = ∞, both φn increase by 2 π. Hence f returns to its initial value.

It follows that: The branch points are z = ±1. (4.2) Possible branch cut geometries are displayed in figure 4.2.

Figure 4.2: Example 4.1

Examples√ of branch cuts for f(z) = z2 − 1.

The simplest is just the straight line segment joining the branch points (left panel). But one can use a curve segment as well (second from the left panel). For that matter, the “curve segment” can go through the point at infinity (third and fourth panels from the left) — but this is not needed. However, we must check that these proposed branch cut geometries actually prevent multiple values. They

Figure 4.3: Example 4.1 The pictures on the left show typical con- tours needed to check if the proposed branch cut√ geometry works. The value of f(z) = z2 − 1 should return to its initial value af- ter a full traversal of the contours. certainly prevent paths that go around a single branch point; but multiple values could arise when going around multiple branch points as well, and we need to show that such paths (if any) are forbidden by the cuts. In this case this is easy. Examples of all the possible paths are shown in figure 4.3. In each case φ1 + φ2 either does not change upon a full (counter-clockwise) traverse, or it increases by 4 π. Either way, f(z) returns to its original value. ♣

2 Note that tiny loop here means “tiny in terms of 1/z”. In fact, in terms of z, the loop is huge. 4

Remark 4.1 Note that the technique used in example 4.1 can be applied to any function that can be written in Qn aj the form f(z) = 1 (z − zj ) , where the {zj, aj} are complex constants. Examples of this are powers of rational functions: f(z) = (P (z)/Q(z))α, where P and Q are polynomials and α is a complex constant. ♣ Remark 4.2 What to do if an expression such as (4.1) is not available. Some general tools you can use to detect branch points follow. Consider a composite function f(z) = h(g(z)). Then:

a. g is analytic at z0, and h is analytic at g(z0). Then f is analytic at z0; hence z0 is not a branch point of f. 0 b. z0 is a branch point for g, h is analytic at g(z0), and h (g(z0)) 6= 0. Then z0 is a branch point of f. 0 c. g is analytic at z0, g(z0) is a branch point for h, and g (z0) 6= 0. Then z0 is a branch point of f. iθ Proofs. Let z = z0 +∆z, with ∆z small, ∆g = g(z)−g(z0), and ∆f = f(z)−f(z0). We will consider what happens as ∆z = r e traverses a small circle: θ = 0 → θ = 2π, with 0 < r arbitrarily small. 0 0 0 0 0 0 Case a. Then ∆f = h (g(z0)) ∆g = h (g(z0)) g (z0) ∆z, so f (z0) = h (g(z0)) g (z0). 0 Case b. Then ∆f = h (g(z0)) ∆g. Because z0 is a branch point for g, ∆g does not return to its initial value after one 0 traverse by ∆z. Hence neither does ∆f. Note that h (g(z0)) 6= 0 is crucial for this argument. 0 Case c. Then ∆g = g (z0) ∆z traverses a small circle as ∆z does. Hence, since f has a branch point at g(z0), ∆f does not 0 return to its initial value after one traverse by ∆z. Note that g (z0) 6= 0 is crucial for this argument.

In the following cases we need more detailed information about the behavior of g near z0, and h near g(z0) 0 d. z0 is a branch point for g, h is analytic at g(z0), and h (g(z0)) = 0. 0 e. g is analytic at z0, g(z0) is a branch point for h, and g (z0) = 0.

f. z0 is a branch point for g, and g(z0) is a branch point for h.

Finally: note that b, d, and f above make sense only if there is a value g(z0). But this need not be so; for example: log(z) has a branch point at z = 0, but log(0) is not defined. ♣

The danger of blindly using canned functions √ Example 4.2 Branch cut geometry for f(z) = z2 − 1 when using the principal value square root. Suppose √ that you compute f(z) = z2 − 1 using some program that implements the principal value of the square root. This gives you a single valued function, but what is the resulting branch cut geometry? Well, what happens † is that you get as branch cuts the straight line segment from −1 to 1 (as in the left panel of figure 4.2) plus the whole imaginary axis! This second cut is completely spurious and unnecessary. It splits the function f into: √ The function f(z) = r r ei (φ1+φ2)/2, defined for Re(z) > 0, with the cut {Im(z) = 0, 0 < Re(z) ≤ 1}; and √ 1 2 i (φ1+φ2)/2 The function f(z) = − r1 r2 e , defined for Re(z) < 0, with the cut {Im(z) = 0, −1 ≤ Re(z) < 0}. Note the spurious discontinuity introduced along the imaginary axis. ♣ † To see this, note that the “branch cuts” arise from the equation “z2 − 1 = real and negative”. Taking the imaginary part of this equation yields that either Re(z) = 0 or Im(z) = 0. The real part then supplies the rest.

5 Contour Integration #01

Statement: Contour Integration #01 Evaluate I = H zαdz once around the unit circle, counter-clockwise, where α is any complex number (consider all cases). When needed, choose a branch cut for zα that intersects the unit circle once. Does the integral depend of your choice of branch cut? 5

6 Contour Integration #04

Statement: Contour Integration #04 Z Evaluate I = z |z|2 dz, (6.1) Γ where Γ is the boundary (traversed once counter-clockwise) n o of the “hyperbolic rectangle” z = x + i y such that: |x2 − y2| ≤ 1, 1 ≤ x y ≤ 2, x > 0 . See figure 6.1.

Hyperbolic rectangle 2 Figure 6.1: Contour Integration #04

1.5 The picture on the left shows√ the “hyperbolic rectangle”, with 2 y boundary given√ by (i) y = x − 1 (solid red); (ii) y = 2/x (solid blue); (iii) y = x2 + 1 (dashed red); and (iv) y = 1/x (dashed 1 blue);

0.5 0.5 1 1.5 2 x

Hint: Let w = z2(= u + i v).

THE END