Solution Key to Selected Exercises of MATH 423/823

Sec.1.2.

14. Prove that |z_{1}z_{2}| = |z_{1}||z_{2}|. Proof: Since z^{2} = z\overline{z} and \overline{z_{1}z_{2}} = \overline{z_{1}} \overline{z_{2}}, then by the commutativity and associativity of multiplication we have |z_{1}z_{2}|^{2} = (z_{1}z_{2})(\overline{z_{1}z_{2}}) = (z_{1}\overline{z_{1}})(z_{2}\overline{z_{2}}) = |z_{1}|^{2}|z_{2}|^{2}, which implies the identity.

Sec.1.3.

16. Prove that ||z_{1}| − |z_{2}|| ≤ |z_{1} − z_{2}|. Proof: Let z = z_{1} − z_{2} and w = z_{2}. Since z + w = z_{1}, we have by the triangle inequality that |z_{1}| = |z+w| ≤ |z|+|w| = |z_{1}−z_{2}|+|z_{2}|. Hence, |z_{1}|−|z_{2}| ≤ |z_{1}−z_{2}|. Exchanging the roles of z_{1} and z_{2}, we have −(|z_{1}| − |z_{2}|) = |z_{2}| − |z_{1}| ≤ |z_{2} − z_{1}| = | − (z_{1} − z_{2})| = |z_{1} − z_{2}|, equivalently, −|z_{1} − z_{2}| ≤ |z_{1}| − |z_{2}|. Together, the inequalities −|z_{1} − z_{2}| ≤ |z_{1}| − |z_{2}| and |z_{1}| − |z_{2}| ≤ |z_{1} − z_{2}| is equivalent to ||z_{1}| − |z_{2}|| ≤ |z_{1} − z_{2}|. The end of proof.

Sec.1.6.

10. Prove that the closed disk |z − z_{0}| ≤ ρ is bounded. Proof: For every z satisfying |z − z_{0}| ≤ ρ we have ||z| − |z_{0}|| ≤ |z − z_{0}| ≤ ρ which is −ρ ≤ |z| − |z_{0}| ≤ ρ. The second inequality becomes |z| ≤ |z_{0}| + ρ := N, showing that z is bounded by N.

13. Let S be a subset of ℂ. Prove that S is closed if and only if its complement S^{c} := ℂ \ S is open. Proof: Denote by ∂D the boundary of a set D. It is straightforward to show by definition that ∂S = ∂S^{c} because ∀z ∈ ∂S and ∀ϵ > 0, D_{ϵ}(z) ∩ S ≠ ∅ and D_{ϵ}(z) ∩ S^{c} ≠ ∅ which by definition is equivalent to z ∈ S^{c}. To show S^{c} is closed if S is open we only need to show ∀z ∈ ∂S^{c} ⇒ z ∈ S^{c}. This is true because if z ∈ S then ∃D_{ρ}(z) ⊂ S for some ρ > 0 since S is open, an contradiction. To show S is open if S^{c} is closed we only need to show for each z ∈ S there is a ρ so that D_{ρ}(z) ⊂ S. This is true because z ∉ S^{c} ⇒ z ∉ ∂S^{c} = ∂S ⇒ z ∉ ∂S ⇒ there must be one ρ > 0 so that D_{ρ}(z) ⊂ S. Otherwise, we would have for all small ϵ > 0 there must be at least one point in D_{ϵ}(z) ∩ S^{c}, a contradiction.

Sec.1.7.

8. Second part: give a geometric explanation of χ(z, w) = χ(1/z, 1/w). Outline of Explanation: Let R_{2} be the reflection operation with respect to the x_{1}x_{3}- plane in ℝ^{3}, R_{2}(Z) = R_{2}(x_{1}, x_{2}, x_{3}) = (x_{1}, −x_{2}, x_{3}) and similarly R_{3} be the reflection operation with respect to the x_{1}x_{2}-plane with R_{3}(Z) = R_{3}(x_{1}, x_{2}, x_{3}) = (x_{1}, x_{2}, −x_{3}). Let Z = P (z) denote the stereographic projection. We note first that both R_{2} and R_{3} preserve the Euclidean distance: d(Z, W) = d(R_{i}(Z), R_{i}(W)). We note second that the x_{1}x_{3}-reflection reserves the stereographic projection, i.e. R_{2}(P (z)) = P (R_{2}(z)), because R_{2} transforms z, Z = P (z) and its defining line through the north pole to the corresponding reflections. We should also note that the two reflection operations map points from the unit sphere to points on the unit sphere. Let Z = P (z), Z′ = R_{2}(Z), Z′′ = R_{3}(Z′) and similar notations for w. We know d(Z, W) = d(Z′, W′) = d(Z′′, W′′). It only remains to show this claim that P (\frac{1}{z}) = Z′′ = R_{3}(R_{2}(P (z))), which implies χ(z, w) = d(Z, W) = d(Z′′, W′′) = χ(1/z, 1/w). Use the polar coordinate form z = re^{iθ} to get 1/z = \frac{1}{r}e^{−θ}, a point on the radial plane with the phase angle −θ in ℝ^{3}. The reflected point z′ = R_{2}(z) and Z′ = R_{2}(Z) lie on the same radial plane as well. The claim is proved by using Euclidean geometry as depicted in the diagram below. It is a cross-section view on the radial plan Argz = −θ for the case of r ≥ 1. From which we see the following: ΔNOD ∼ ΔNSZ′′ because they are right triangles and share a common angle. Also ΔNSZ′′ ∼ ΔNOr because they are right triangles and because the angle ∠NSZ′′ = ∠SNr as they are from the isosceles triangle ΔNSr. Therefore ΔNOD ∼ ΔNOr which implies D/1 = 1/r by similarity. Thus P (1/z) = Z′′. For the case of r < 1, the same diagram works after exchanging r with 1/r and Z′ with Z′′. This completes the outline.

ℂ x_{3} N 1 O S Z′ Z′′ r = |R_{2}(z)| D \frac{1}{r}

Sec.2.2.

Theorem 1(ii): If lim_{z→z_{0}} f(z) = A, lim_{z→z_{0}} g(z) = B, then lim_{z→z_{0}} f(z)g(z) = AB. Proof: We first show that If lim_{z→z_{0}} f(z) = A then f is bounded by a constant M in a disk D_{ρ}(z_{0}) for some ρ > 0. In fact, by definition for ϵ = 1 there is a ρ > 0 so that |z −z_{0}| < ρ ⇒ |f(z)−A| < ϵ = 1 which implies ||f(z)|−|A|| ≤ |f(z)−A| < 1 and |f(z)| < |A|+1 := M follows. Thus, without loss of generality (wlog), we can assume |f(z)|, |g(z)| ≤ N for z ∈ D_{r}(z_{0}) for some number N > max |A| + 1, |B| + 1 and r > 0.

We first use the assumption to have ∀ϵ > 0 ∃ δ_{1} > 0 and δ_{2} > 0 with δ_{1,2} < r s.t.(such that) |z − z_{0}| < δ_{1} ⇒ |f(z) − A| < ϵ/(2N) and |z − z_{0}| < δ_{2} ⇒ |g(z) − B| < ϵ/(2N). Now let δ = min{δ_{1}, δ_{2}} then ∀|z − z_{0}| < δ ≤ δ_{1}, δ_{2} both |f(z) − A| < ϵ/(2N) and |g(z) − B| < ϵ/(2N).

Now for the product rule, we have ∀z with |z − z_{0}| < δ

|f(z)g(z)−AB| = |f(z)g(z)−Ag(z)+Ag(z)−AB| ≤ |f(z)−A||g(z)|+|A||g(z)−B|

≤ N(|f(z) − A| + |g(z) − B|) < N[ϵ/(2N) + ϵ/(2N)] = ϵ.

This proves the result.

22. Show that if lim_{n→∞} f(z_{n}) = w_{0} for every sequence {z_{n}}_{1}^{∞} converging to z_{0} (z_{n} ≠ z_{0}), then lim_{z→z_{0}} f(x) = w_{0}. Proof: Suppose on the contrary that lim_{z→z_{0}} f(x) ≠ w_{0}. Then by definition there must be a number ϵ_{0} > so that for every δ > 0 the inequality |f(z) − w_{0}| < ϵ_{0} does not hold for every point |z − z_{0}| < δ. That is there is some point z_{δ} with |z_{δ} − z_{0}| < δ but |f(z_{δ})−w_{0}| ≥ ϵ_{0}. In particular, for each integer n we can find a point z_{n} satisfying |z_{n} − z_{0}| < 1/n but |f(z_{n}) − w_{0}| ≥ ϵ_{0}. This contradicts the assumption because by construction this sequence z_{n} → z_{0} but lim_{n→∞} f(z_{n}) ≠ w_{0}.

Sec.2.4.

Prove the product rule of derivative: [f(z)g(z)]′ = f′(z)g(z) + f(z)g′(z) Proof: We will use the fact that a function is differentiable at a point is also continuous at the point. By definition, and a standard trick of adding and subtracting a same term, f(z)g(z + Δz), in the numerator below, we have

[f(z)g(z)]′ = lim_{Δz→0} \frac{f(z + Δz)g(z + Δz) − f(z)g(z)}{Δz}

= lim_{Δz→0} \frac{f(z + Δz)g(z + Δz) − f(z)g(z + Δz) + f(z)g(z + Δz) − f(z)g(z)}{Δz}

= lim_{Δz→0} [\frac{f(z + Δz) − f(z)}{Δz}g(z + Δz) + f(z)\frac{g(z + Δz) − g(z)}{Δz}]

= lim_{Δz→0} \frac{f(z + Δz) − f(z)}{Δz} lim_{Δz→0} g(z + Δz) + f(z) lim_{Δz→0} \frac{g(z + Δz) − g(z)}{Δz}

= f′(z)g(z) + f(z)g′(z).

The last equality follows from the assumption and the fact that all the limits involved exist, as well as the production and summation rules of limit. This proves the result.

11 Suppose that f(z) and \overline{f(z)} are analytic in a domain. Show that f(z) is constant in D. Proof: Let f(z) = u(x, y) + iv(x, y) and g(z) = f(z) + \overline{f(z)} = 2u(x, y). Then by the summation rule, g is also analytic in D. Since Img(z) = 0, by the C-R equations we have u_{x} = u_{y} = 0 and hence u(x, y) is constant, which in turn implies v(x, y) is constant, and hence the proof.

Sec.3.3.

Lecture example: Find the domain of the function w = z^{2} whose range is the plane with the branch cut L = {Rew = 1, Imw ∈ [0, +∞)}. Solution: The branch cut L has two sides. On the left side w ∈ L_{−}, we have argw = Argw ∈ [0, π/2). On the right side w ∈ L_{+}, we have argw = Argw + 2π ∈ [2π, 2π + π/2). To find the preimage of L_{±} is to find the image of the inverse map z = w^{1/2} = |w|^{1/2}e^{i(argw)/2} = |w|^{1/2}e^{i(Argw/2+kπ)}. Let w = u + iv and Argw =

θ. Then for w ∈ L_{−} with k = 0, |w| = \sqrt{1 + v^{2}}, and cos(θ/2) = \sqrt{\frac{1+cos(θ)}{2}},

sin(θ/2) = \sqrt{\frac{1−cos(θ)}{2}} by the half-angle formula. As tan θ = v/u = v, we have cos(θ) = 1/\sqrt{1 + v^{2}} (and sin(θ) = v/\sqrt{1 + v^{2}}). Therefore x + iy = z = w^{1/2} = (1 + v^{2})^{1/4}[cos(θ/2) + i sin(θ/2)] = (1 + v^{2})^{1/4}[\sqrt{\frac{\sqrt{1+v^{2}}+1}{2\sqrt{1+v^{2}}}} + i\sqrt{\frac{\sqrt{1+v^{2}}−1}{2\sqrt{1+v^{2}}}}], which

is simplified to x = s(v) = \sqrt{\frac{\sqrt{1+v^{2}}+1}{2}} and y = t(v) = \sqrt{\frac{\sqrt{1+v^{2}}−1}{2}} parameterized by parameter v ∈ [0, ∞). This obtains L_{−} corresponding side of a cut in the domain D. For the L_{+} side of the cut in the range, since argw = Argw + 2π, the image under w^{1/2} differs from that of L_{−} exactly by a negative sign, that is x = −s(v), y = −t(v). See figure, in which the additional cut {0 ≤ u ≤ 1, v = 0} is needed for the definition of the inverse function z = w^{1/2}. Also the corresponding subregions in domain and range are color coded.

x yu v1 1 −1 w = z^{2} z = w^{1/2} 0 0 L_{−} L_{+}

Sec.3.3.

10. Show that the function Log(−z) + iπ is a branch of log z in the domain D_{0} consisting of all points in the plane except those on the nonnegative real axis. Solution: By definition, Log(−z) + iπ = log |z| + i(Arg(−z) + π). There are two cases as Arg(z) ∈ (−π, π]. For z in the quadrant I or II, −z is in quadrant III or IV, Arg(−z) = Arg(z) − π. For z in the quadrant III or IV, −z is in quadrant I or II, Arg(−z) = Arg(z) + π. Therefore, Arg(−z) + π = Arg(z) = arg_{0}(z) ∈ (0, π) if z is in quadrant I or II, and Arg(−z) + π = Arg(z) + 2π = arg_{0}(z) ∈ (π, 2π) if z in the quadrant III or IV. Together, this shows Log(−z) + iπ = log_{0}(z).