14.13. EXERCISES 371

16. The polynomial z5 + z4− z3− 3z2− 5z+ 1 = p(z) has no rational roots. You cancheck this by applying the rational root theorem from algebra. However, it has fivecomplex roots. Also

∣∣z4− z3−3z2−5z+1∣∣≤ |z|4+ |z|3+3 |z|2+5 |z|+1. By graph-

ing, observe that x5−(x4 + x3 +3x2 +5x+1

)> 0 for all x ≥ 2.4. Explain why the

roots of p(z) are inside the circle |z|= 2.4.

17. Let f be analytic on U and let B(z,r)⊆U . Let γr be the positively oriented boundaryof B(z,r). Explain, using the Cauchy integral formula why

| f (z)| ≤max{| f (w)| : w ∈ γ∗r} ≡ mr.

Show that if equality is achieved, then | f (w)| must be constantly equal to mr on γ∗r .

18. Let f :C→C be analytic with f ′ (z) ̸= 0 for all z. Say f (x+ iy) = u(x,y)+ iv(x,y).

Thus the mapping (x,y)→(

u(x,y)v(x,y)

)is a C1 mapping of R2 to R2. Show that

at any point∣∣∣∣ ux uy

vx vy

∣∣∣∣ ̸= 0. Therefore, by the inverse function theorem, Theorem

4.8.7, this mapping is locally one to one. However, the function does not need to beglobally one to one. Give an easy example which shows this to be the case. Hint:You might want to consider something involving the exponential function.

19. Let Γ be a simple closed rectifiable curve and let { fn} be a sequence of functionswhich are analytic on Ui, the inside of Γ and continuous on Γ∗. Then if γ is aparametrization of Γ with n(γ,z) = 1 for z ∈ Ui, then fn (z) = 1

2πi∫

γ

fn(w)w−z dw. This

is by the Cauchy integral formula presented above. Suppose fn converges uniformlyon Γ∗ to a continuous function f . Show that then, for z ∈Ui, and f (z) defined asf (z) ≡ 1

2πi∫

γ

f (w)w−z dw. It follows that fn (z)→ f (z) for each z ∈ Ui and also f is

analytic on Ui. Hint: You might use Theorem 14.4.7. This is very different thanwhat happens with functions of a real variable in which uniform convergence ofpolynomials pn to f does not necessarily confer differentiability on f . For example,to approximate f , a continuous function having no derivatives or even a very easyfunction like f (x) = |x− (1/2)| for x ∈ [0,1].

20. Sketch an example of two differentiable functions defined on [0,1] such that theirproduct is 0 but neither function is 0. Explain why this never happens for the setof analytic functions defined on an open connected set. In other words, if you havef g = 0 where f ,g are analytic on D an open connected set, then either f = 0 org = 0. For those who like to classify algebraically, this says that the set of analyticfunctions defined on an open connected set is an integral domain. It is clear thatthis set of functions is a ring with the usual operations. The extra ingredient is thisobservation that there are no nonzero zero divisors. Hint: To show this, considerD\ f−1 (0) an open set. If f−1 (0) = D, then you are done. Otherwise, you have g is0 on an open set. Now use Theorem 14.9.2.

21. For D ≡ {z ∈ C : |z|< 1} , consider the function sin( 1

1−z

). Show that this function

has infinitely many zeros in D. Thus there is a limit point to the set of zeros, but itslimit point is not in D. It is good to keep this example in mind when consideringTheorem 14.9.2.

14.13. EXERCISES 37116.17.18.19.20.21.The polynomial 2 + z+ — 23 — 322 —5z+ 1 = p(z) has no rational roots. You cancheck this by applying the rational root theorem from algebra. However, it has fivecomplex roots. Also |z4 — z3 — 3z? — 5z+1] < |z|*+|z|> +3 |z|? +5 |z|+1. By graph-ing, observe that x° — (x4 +23 43x? + 5x4 1) > 0 for all x > 2.4. Explain why theroots of p(z) are inside the circle |z| = 2.4Let f be analytic on U and let B(z,r) CU. Let y, be the positively oriented boundaryof B(z,r). Explain, using the Cauchy integral formula whyIP (2)| < max {|f (v)] sw € 78} =m.Show that if equality is achieved, then | f (w)| must be constantly equal to m, on Y;.Let f : C > C be analytic with f’ (z) £0 for all z. Say f (x +iy) =u(x,y) +iv(x,y).Thus the mapping (x,y) > mee) is aC! mapping of R? to R?. Show thatat any point x y #0. Therefore, by the inverse function theorem, Theoremx Vy4.8.7, this mapping is locally one to one. However, the function does not need to beglobally one to one. Give an easy example which shows this to be the case. Hint:You might want to consider something involving the exponential function.Let I be a simple closed rectifiable curve and let {f,} be a sequence of functionswhich are analytic on Uj, the inside of I’ and continuous on _ men if yisaparametrization of [ with n(y,z) = 1 for z € Uj, then f, (z) = x5 ae ; dw. Thisis by the Cauchy integral formula presented above. Suppose f,, converges uniformlyon I* toa continuous function f. Show that then, for z € U;, and f(z) defined asf@= oo Io — LW) dy, It follows that f,(z) + f(z) for each z € U; and also f isanalytic on U;. Hint: You might use Theorem 14.4.7. This is very different thanwhat happens with functions of a real variable in which uniform convergence ofpolynomials p, to f does not necessarily confer differentiability on f. For example,to approximate f, a continuous function having no derivatives or even a very easyfunction like f (x) = |x—(1/2)| for x € [0,1].Sketch an example of two differentiable functions defined on [0,1] such that theirproduct is 0 but neither function is 0. Explain why this never happens for the setof analytic functions defined on an open connected set. In other words, if you havefg =0 where f,g are analytic on D an open connected set, then either f = 0 org = 0. For those who like to classify algebraically, this says that the set of analyticfunctions defined on an open connected set is an integral domain. It is clear thatthis set of functions is a ring with the usual operations. The extra ingredient is thisobservation that there are no nonzero zero divisors. Hint: To show this, considerD\ f—! (0) an open set. If f~! (0) = D, then you are done. Otherwise, you have g is0 on an open set. Now use Theorem 14.9.2.For D = {z € C: |z| < 1}, consider the function sin (;4,) . Show that this functionhas infinitely many zeros in D. Thus there is a limit point to the set of zeros, but itslimit point is not in D. It is good to keep this example in mind when consideringTheorem 14.9.2.