# MAT 334 Complex Variables Assignment 3

Question # 00079864 Posted By: echo7 Updated on: 07/05/2015 01:52 PM Due on: 08/04/2015
Subject Mathematics Topic General Mathematics Tutorials:
Question
 @u@x + i@x@v

MAT 334 Complex Variables

Assignment 3

Due: Thursday July 2 in Class

Please write your arguments clearly, as marks may be taken off if a solution is not well written. Numbers in [ ] indicate how many marks a (part of a) question is worth. The assignment is out of 50.

1.[6] (a) [2] Verify the Cauchy-Riemann equations forf(z) = z3explicitly. (In other words, letz=x+iyand write z3 as u(x; y) + iv(x; y), where u and v are real-valued functions of two real variables x; y, and

then verify that

 @u = @v and @u = - @v : @x @y @y @x

Note that the equations must hold since f is analytic.)

(b) [2] Being a polynomial, f is easy to differentiate: f0(z) = 3z2. Verify that the formula f0(z) =

(c) [2] By checking whether Cauchy-Riemann equations hold, determine if the function f(z) = (z)2 is ana-lytic.

2.[14] The goal of this problem is to give evidence that the condition needed to be satisfied by a functionin order to be analytic is a strong one, and that consequently analytic functions are rather special.

(a) [0] Convince yourself that the statement below is true. A picture might be helpful. You do not need to write anything for this part in your solutions.

”If U is an open connected subset of C, and z; w 2 U, then there is a polygonal curve from z to w all of whose line segments are either horizontal or vertical.”

(b) [14] Suppose f is analytic on an open connected set U C. Write f = u + iv where u and v are real-valued functions. Show that in each of the following scenarios f is constant.

(i) u is constant on U.

(ii) v is constant on U.

(iii) jf(z)j is constant on U.

(iv) f(z) has a fixed argument on U.

(v) f0(z) is zero on U.

(Note that for instance, (ii) implies that a real-valued analytic function with connected domain is constant.)

Remark.You may want to look at Theorem 2 of Section 2.1 of the textbook.

 Suggestions:For (i), the hypothesis implies@uand@uare both zero onU. Use Cauchy-Riemann to @x @y conclude @v and @v are also both zero on U. Conclude that u and v, and hence f = u + iv, are constant @x @y

along horizontal and vertical line segments in U. Now given two points z; w 2 U, part (a) tells us there is a polygonal curve from z to w with horizontal or vertical line segments. Does it follow f(z) = f(w)?

(ii) is similar to (i). For (v), use f0(z) =@u@x + i@x@v and conclude@u@x =@x@v = 0 on U. Now use Cauchy-Riemann

1

2 ASSIGNMENT 3

to conclude that the other two partial derivatives are also zero, and then proceed as in (i). (iv) follows imme-diately from (ii) (or (i)) if you perform a clever trick. For (iii), first note that the assertion is clear if jf(z)j = 0.

Suppose jf(z)j 6= 0. The hypothesis implies u2 + v2 is constant, so that@x@ (u2 + v2) =@y@ (u2 + v2) = 0. From this try to show that the partials of u and v are all zero, and then proceed as in (i). An alternative approach

is given in the proof of Theorem 2 of Section 2.1 of the textbook. (You may give that argument, but you have to write it in your own words, not just copy from the textbook.)

3.[8] (a) [1] Supposefandgare analytic on an open connected setU. Letf0(z) =g0(z)for allz2U. Showthat f - g is a constant. Suggestion: Consider the function f - g. What is its derivative on U? Use Problem 2(v).

(b) [3] Find the domain and derivative of -Log(1 - z). Express the derivative as a power series for jzj < 1. (Recall that Log is the principal branch of logarithm.)

(c)[2] Show that the radius of convergence of the power series

X

 1 zn z2 z3 = z + + + : : : n 2 3

n=1

is 1. (Hence by Abel’s theorem, which is included at the end of this document for your reference, it is analytic on the open unit disk.) Express its derivative as a power series.

X1zn

-Log(1 - z) =

n

n=1

for jzj < 1.

4.[9] (a) [2] State Clairaut’s theorem on symmetry of second order mixed partial derivatives. Be careful tostate the hypotheses precisely. (You can look it up in your multi-variable calculus textbook.)

(b) [3] Suppose f(z) = u(x; y) + iv(x; y) (with u = Re(f), v = Im(f), x = Re(z), y = Im(z) as usual) is

 analytic. Suppose moreover that all first and second order partial derivatives (i.e. @u, @u , @v , @v , @2u , etc.) @x@y exist and are continuous. Use Cauchy-Riemann equations to show that @x @y @x @y (1) @2u @2u + = 0 @x2 @y2 and @2v @2v + = 0: @x2 @y2

(Suggestion: Think about this for a few minutes. Then you may want to look at the top of page 81 of the textbook.)

(c)[1] Explicitly verify that the real part u of z2 satisfies Eq. (1).

(d) [3] Is there an analytic function f(z) such that its real part is

u(x; y) = x2 + y2 ?

(Again as usual, we are writing z = x + iy where x and y are real.)

Remark.A real valued functiongof two real variablesx; yis calledharmonicif it satisfiesLaplace’sequation, i.e.

@2g @2g

@x2+@y2=0:

Harmonic functions appear naturally in different places in math and physics. After having solved this question you have shown that real and imaginary parts of an analytic function, assuming they have con-tinuous 1st and 2nd order partials, are harmonic. (We’ll see later in the course that the assumption on continuity of partials is redundant, as it will automatically hold.)

 ASSIGNMENT 3 3

 5.[6] (a) [3] Letr > 0andc2C. Suppose an(z - c)n converges on the open disk D = fz : jz - cj < rg. n=0 1 Let f(z) =an(z - c) n P on D. We know by Abel’s theorem that f is analytic. Briefly explain why for every n=0 1 (n) (n) th P n 0, the n derivative of f (denoted by f ) exists and is analytic. Express f (z) on D as a power series. (By convention, f(0) = f.) f(n)(c) (b) [2] Show that an = . n! (c) [1] Conclude that if an(z - c)n = bn(z - c)n on an open disk centered at c, then an = bn for all n. n=0 n=0 1 1 (When we write the equality it is understood that we are assuming both power series converge.) P P

6. [7] LetU C be open, andfbe a function onU. Given a pointc 2U, we sayfhas power series expansion

P1

near c or at c if there is r > 0 such that f(z) is equal to a power series an(z - c)n on the open disk

n=0

fz : jz - cj < rg.

(a) [1] Suppose f has power series expansion near every point in U. Does it follow from Problem 5(a) that for every n, f(n) exists and is analytic on U? Briefly explain.

(b) [6] Suppose moreover that U is connected, and that there is z0 2 U such that f(n)(z0) = 0 for all n 0. (I.e. f and all its derivatives vanish at a point z0 2 U.) Prove that f is the zero function.

Suggestions for (b):Take for granted the following two topological facts. You can use them freelythroughout the course.

(i) An open set U C is connected if and only if there are no two disjoint nonempty open sets V; W such that U = V [W. (Disjoint means empty intersection. Draw a picture to convince yourself that the statement intuitively makes sense.)

(ii) Given a function f : U ! C and a subset V C, the preimage of V under f is defined to be the set

fz 2 U : f(z) 2 Vg;

i.e. the subset of U consisting of all points that get mapped to V by f. The standard notation for this set is

f-1(V). Don’t confuse this notation with that of an inverse function. The function f here does not have to be one-to-one. For instance, if f : C ! C is the constant function f(z) = 1, f-1(f1g) = C and f-1(f0g) = ;. The promised fact, which you can freely use, is that if U is open, a function f on U is continuous if and only if for every open V C, f-1(V) is open. (In short, f is continuous iff the preimage under f of every open set is open.)

Equipped with these, try to argue as follows: Let

V = fz 2 U : f(n)(z) 6= 0 for some n 0g

and

W = fz 2 U : f(n)(z) = 0 for all n 0g:

Note that clearly V W = ; and U = V [ W. Also the statement of the question tells us z0 2 W, so that W is nonempty. Now your strategy should be to show that both V and W are open. Once you establish that, you can use Fact (i) above together with connectedness of U to conclude that V has to be empty. This would give f(z) = 0 for all z 2 U, as desired.

To show V is open, write

V = V0 [ V1 [ V2 [ : : : ;

where

Vn = fz 2 U : f(n)(z) 6= 0g:

Realize Vn as the preimage under f(n) of a certain open set. Then use continuity of f(n) and Fact (ii) above to conclude that Vn is open. Now use the fact that a union of open sets is open to conclude V is open.

As for W, try to show that it is open by showing that every point in it is an interior point. Here you’ll need to use the hypothesis that f has power series expansion at every point in U. Problem 5(b) is also useful.

4 ASSIGNMENT 3

Reading assignment.Review what you learned in multi-variable calculus about integration along paths.Review Green’s theorem.

Practice Problems.The following problems are for your own practice. Please do not hand them in.

1.Letf(z) be a branch ofza. Show thatf0(z) =af(zz). Does this agree with what you expect from calculus?Suggestion: Write f(z) = ealog(z) where log(z) is some branch of logarithm. Now use the chain rule and the fact that the derivative of log(z) is1z .

From 2.1: 1(a),(b); 3, 4, 6, 8, 9, 16

From 2.2: 1-6, 7, 9, 11, 14, 15, 16, 18

P1

THEOREM. (Abel) For every power series an(z - c)n there is R 2 R0 [ f1g (called the radius of con-

n=0

vergence) such that the series converges absolutely if jz - cj < R, and diverges if jz - cj > R. Moreover, if R > 0,

P1

the function f(z) = an(z - c)n is analytic on fz : jz - cj < Rg, and the derivative is obtained by term by term

n=0

differentiation, i.e.

X1

f0(z) = nan(z - c)n-1;

n=1

and the radius of convergence of the series of f0(z) is also R.

Comments on notation and convention:(1) HereR0denotes the set of non-negative real numbers.

(2) In the statement of the theorem, we are adopting the convention that 1 is bigger than any real number. So if the radius of convergence is 1, it means the series convergence for all z 2 C and it defines an entire function.

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1. ## Solution: MAT 334 Complex Variables Assignment 3

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