- Suppose you have U ⊆ ℂ an open set and f : U → ℂ is analytic but has only real values. Find all possible f with these properties.
- Suppose f is an entire function (analytic on ℂ) and suppose Ref is never 0. Show that f must
be constant. Hint: Consider U = ,V =. These are open and disjoint so one must be empty. If V is empty, consider 1∕e
^{f(z) }. Use Liouville’s theorem. - Suppose f : ℂ → ℂ is analytic. Suppose also there is an estimate
Show that f must be a polynomial. Hint: Consider the formula for the derivative in which γ

_{r}is positively oriented and a circle or radius r for r very large centered at 0,and pick large n. Then let r →∞.

- Define for z ∈ ℂ sinz ≡∑
_{k=0}^{∞}^{n}. That is, you just replace x with z. Give a similar definition for cosz, and e^{z}. Show that the series converges for sinz and that a corresponding series converges for cosz. Then show thatShow that it is not longer true that the functions sinz,cosz must be bounded in absolute value by 1. Hint: This is a very easy problem if you use the theorem about the zeros of an analytic function, Theorem 15.5.9.

- Verify the identities cos= cos z cosw + sinz sinw and similar identities. Hint: This is a very easy problem if you use the theorem about the zeros of an analytic function, Theorem 15.5.9.
- Consider the following contour in which the large semicircle has radius R and the small one has
radius r ≡ 1∕R.
The function z →

is analytic on the curve and on its inside. Therefore, the contour integral with respect to the given orientation is 0. Use this contour and the Cauchy integral theorem to verify that ∫_{0}^{∞}dz = π∕2 where this improper integral is defined asThe function is actually not absolutely integrable and so the precise description of its meaning just given is important. You can use dominated convergence theorem to simplify some of the limits if you like but it is possible to establish the needed estimates through elementary means. I recommend using the dominated convergence theorem. To do this, show that the integral over the large circle of ∫

_{CR}dz → 0 as R →∞ and verify that you get something else like −π for the integral over the small integral as r → 0. - A set U is star shaped if there exists a single point z
_{0}∈ U such that every segment from z_{0}to z is contained in U. Now suppose that U is star shaped and f : U → X is analytic. Show that f has a primitive on U. - Let U be what remains of ℂ after (−∞,0] is deleted. Explain why U is star shaped. Letting γbe the straight line segment from 1 to z, let f= ∫
_{γ(1,z) }dw. Explain why f^{′}=, f= 0. Now explain why f is analytic and why e^{f(z) }= z for all z ∈ U. Also formulate an assertion which says f= z for suitable z. This f is the principal logarithm, denoted log. - Explain why one could delete any ray starting at 0 and obtain a function fwhich is a primitive of 1 ∕z.
- For z ∈ ℂ ∖ (−∞,0], let arg ≡ θ ∈such that z =e
^{iθ}. Show that - Suppose f= u+ ivis analytic. Show that both u,v satisfy Laplace’s equation, u
_{xx}+ u_{yy}= 0. - Suppose you have two complex numbers z = a + ib and w = x + iy. Show that the dot product of the
two vectors ⋅is Re= Re.
- ↑Suppose you have two curves t → zand s → wwhich intersect at some point z
_{0}corresponding to t = t_{0}and s = s_{0}. Show that the cosine of the angle θ between these two curves at this point isNow suppose z → f

is analytic. Thus there are two curves t → fand s → fwhich intersect when t = t_{0}and s = s_{0}. Show that the angle between these two new curves at their point of intersection is also θ. This shows that analytic mappings preserve the angles between curves. - Suppose z = x + iy and f= u+ ivwhere f is analytic. Explain how level curves of u and v intersect in right angles.
- Suppose Γ is a simple closed rectifiable curve and that γ is an oriented parametrization for Γ which is
oriented positively, n= 1 for all z on the inside of Γ. Now letbe a simple closed rectifiable curve on the inside of Γ and letbe an orientation ofalso oriented positively. Explain why, if z is on the inside ofand f is analytic on the inside U
_{i}of Γ, continuous on U_{i}∪ Γ, then ∫_{ˆγ }dw = ∫_{γ}dw and if z is on the inside of Γ but outside of, Then ∫_{ˆγ }dw = 0 while ∫_{γ}dw = fand if z is outside of Γ then both integrals are 0. - Let Γ be a simple closed rectifiable curve in ℂ. Let γ be a parametrization of Γ which has positive
orientation. Thus n= 1 for all z inside Γ. Also suppose f is an analytic function on a connected open set containing Γ and its inside U
_{i}. Suppose f is not identically zero and has no zeros on Γ. Explain why f has finitely many zeros on the inside of Γ. A zero a has multiplicity m if f=^{m}gwhere g≠0 on U_{i}. Let the zeros of f in U_{i}bewhere there might be repeated numbers in this list, zeros of multiplicity higher than 1. Show that(15.34) Thus you can count the zeros of an analytic function inside a simple closed curve by doing an integral! Hint: First of all, m is finite since if not, Theorem 15.5.9 implies that f

= 0 for all z since there would be a limit point or else a zero of infinite order. Now argue that f= ∏_{k=1}^{m}gwhere gis analytic and nonzero on U_{i}. Use the product rule to simplify. Then use the fact that n= 1. - Give another very short proof of the fundamental theorem of algebra using the result of Problem 16
above. In fact, show directly that if pis a polynomial of degree n then it has n roots counted according to multiplicity. Hint: Let pbe a polynomial. Then by the Euclidean algorithm, Lemma 1.11.2, you can see that there can be no more than n roots of the polynomial phaving complex coefficients. Otherwise the polynomial could not have degree n. You should show this. Now there must exist Γ
_{R}, a circle centered at 0 of radius R which encloses all roots of p. Letting m be the number of roots,Now write down in terms of an integral on

and let R →∞ to get n in the limit on the right. Hence n = m. - Suppose now you have a rectifiable simple closed curve Γ and on Γ
^{∗},>where f,g are analytic on an open set containing Γ^{∗}. Suppose also that f has no zeros on Γ^{∗}. In particular, f is not identically 0. Let λ ∈.- Verify that for λ ∈, f + λg has no zeros on Γ
^{∗}. - Verify that on Γ
^{∗},≤ C. - Use Theorem 15.3.8 to show that for γ a positively oriented parametrization of
Γ,
is continuous.

- Now explain why this shows that the number of zeros of f + λg on the inside of Γ is the same as the number of zeros of f on the inside of Γ. This is a version of Rouche’s theorem.

- Verify that for λ ∈
- Give an extremely easy proof of the fundamental theorem of algebra as follows. Let γ
_{R}be a parametrization of the circle centered at 0 having radius R which has positive orientation so n= 1 . Let pbe a polynomial a_{n}z^{n}+ a_{n−1}z^{n−1}++ a_{1}z + a_{0}. Now explain why you can choose R so large that>for all≥ R. Using Problem 18 above explain why all zeros of pare inside γ_{R}^{∗}and why there are exactly n of them counted according to multiplicity. - The polynomial z
^{5}+ z^{4}− z^{3}− 3z^{2}− 5z + 1 = phas no rational roots. You can check this by applying the rational root theorem from algebra. However, it has five complex roots. AlsoBy graphing, observe that x

^{5}−> 0 for all x ≥ 2.4. Explain why the roots of pare inside the circle= 2 .4. - This problem will feature the situation where the radius of the simple closed curve is sufficiently
small. The zero counting integral can be used to prove an open mapping theorem for analytic
functions. Suppose you have f= f+ ϕ
^{m}for z ∈ V an open set containing z_{0}and ϕ= 0 ,= 2 r≠0, and m ∈ ℕ. Let Cdenote the positively oriented circle centered at a which has radius ρ.- Explain why there exists δ > 0 such that if = δ, then B⊆ V and
Therefore, if

< rδ, then if= δ,≠0. - Use continuity of w →∫
_{C(z,δ) 0 }dz for< δand Problem 16 to conclude that there exists ε < rδ such that if< ε there is one zero for ϕ− w in B. In other words, ϕ⊇ B. Then also ϕ^{m}⊇ B. Hint: If you have w ∈ B, then there are m m^{th}roots of w equally spaced around B. Thus these roots are on a circle of radius less than ε. Pick one. Call it ŵ. Then there exists z ∈ Bsuch that ϕ= ŵ. Then ϕ^{m}= w. Fill in details. - Explain why f⊇ f+ Band why for w ∈ f+ Bthere are m different points in B,z
_{1},,z_{m}such that f= w.

- Explain why there exists δ > 0 such that if
- ↑Let Ω be an open connected set. Let f : Ω → ℂ be analytic. Suppose fis not a single point. Then pick z
_{0}∈ Ω. Explain why f= f+^{m}gfor all z ∈ V an open ball contained in Ω which contains z_{0}and g≠0 in V,ganalytic. If this were not so, then z_{0}would be a zero of infinite order and by the theorem on zeros, Theorem 15.5.9, f= ffor all z ∈ Ω which is assumed not to happen. Thus, every z_{0}in Ω has this property that near z_{0},f= f+^{m}gfor nonzero g. Now explain why f= f+ ϕ^{m}where ϕ= 0 but ϕ^{′}≠0 and ϕis some analytic function. Thus from Problem 21 above, there is δ such that f⊇ f+ B. Hence fis open since each fis an interior point of f. You only need to show that there is Gsuch that G^{m}= gand then ϕ≡Gwill work fine. When you have done this, Problem 21 will yield a proof of the open mapping theorem which says that if f is analytic on Ω a connected open set, then fis either an open set or a single point. So here are some steps for doing this.- Consider z →. It is analytic on the open ball V and so it has a primitive on V . In fact, you could take h≡∫
_{γ(z0,z) }dw. - Let the primitive be h. Then consider
^{′}. Show this equals 0. Then explain why this requires it to be constant. Explain why there is a + ib such that g= e^{h(z) +a+ib}. Then use the primitive h+ a + ib instead of the original one. Call it h. ThenYou can then complete the argument by letting g

^{1∕m}≡ e^{h(z) ∕m}and

- Consider z →
- If you have an open set U in ℂ show that for all z ∈ U,< sup. In other words, z →never achieves its maximum on any open set U ∈ ℂ.
- Let f be analytic on U and let B⊆ U. Let γ
_{r}be the positively oriented boundary of B. Explain, using the Cauchy integral formula whyShow that if equality is achieved, then

must be constantly equal to m_{r}on γ_{r}^{∗}. - The maximum modulus theorem says that if Ω is a bounded connected open set and f : Ω → ℂ is
analytic and f : Ω → ℂ is continuous, then if achieves its maximum at any point of Ω then f is equal to a constant on Ω. Thusachieves its maximum on the boundary of Ω in every case. Hint: Suppose the maximum is achieved at a point of Ω, z
_{0}. Then let B⊆ Ω. Show that if f is constant on B, then it equals this constant on all of Ω using Theorem 15.5.9. However, if it is not constant, then from the open mapping theorem of Problem 22, fis an open set. Then use Problem 23 above to obtain a contradiction. - Let f : ℂ → ℂ be analytic with f
^{′}≠0 for all z. Say f