Complex Prelim Practice

8/10/2019 Complex Prelim Practice

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SOLUTIONS TO COMPLEX PRELIM PROBLEMS

CIHAN BAHRAN

In this document, I will collect my solutions to some (not all, probably not even half)of the Complex Prelim problems that were asked in the past in UMN.

Some theorems get used a lot. I will copy some of them from David C. Ullrichswonderful book Complex Made Simple.

1. Open mapping theorem

This is very useful in general. Its easy to forget the connectedness assumption, so Iwill state it precisely. H(V) denotes the set of analytic maps from an open set V to C.

Open Mapping Theorem. LetVbe open andf H(V). Also letWbe an open andconnected set contained inV. Thenf(W) is either a singleton (that is, f is constantonW) or open inC.

Fall 2011, 7 and Fall 2010, 3. Prove that there is no one-to-one conformal mapof the punctured disk{z C : 0


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SOLUTIONS TO COMPLEX PRELIM PROBLEMS 2

contained in D {0}, we have

g(V)g(W) =g(V)

g({0})gW {0}=f(V)

{} f

W {0}

=

f(V) {} f(V)fW {0}=f(V) {}={} .

This is a contradiction.

Spring 2011, 2. Let h be a holomorphic function on a connected open set V.Prove that ifh(z)2 =h(z) for allzV then h is constant on V. Find all possiblevalues for such h.

First of all, V is nonempty

1

. Let f=h

3

H

(V). So for all zV,f(z) =h(z)2

h(z) =h(z)h(z) =|h(z)|2 R. That is,f(V) R. Supposef is not constant. Then the openmapping theorem yields f(V)Int(R) = (we are considering R as a subspace ofChere), a contradiction. Thus for allzV,h(z)3 =f(z) =a for somea C. Thereforeif we let =e

2i3 , we have

h(V) {|a| 13 , |a| 13 , 2|a| 13} ,so h(V) is finite. Nonempty open sets in C are infinite, thus h(V) is not open. Againby the open mapping theorem we deduce that h is constant. The rest is the kind ofstuff Calc 1 students fail to do correctly quite often: basic algebra. For some x, y R,h(z) =x+iy for allzV. So

(x+iy)2 =x+iy

x2 y2 +i(2xy) =xiy .

Thusx2y2 =x and 2xy=y. The second equation givesy = 0 orx =12

. Going tothe first equation, the first case yields x2 =x and the second case yields 14y2 =12 .So the solutions are 0,1,12 + i

32 ,12i

32 .

2. Linear fractional transformations

I think the most useful linear fractional transformations for the prelims are the onesthat map a half plane to the unit disk. And, thanks to Ullrichs book, I know thatthere is a way to do this which is really cool and impossible to forget. Let U be theupper half plane and D be the open unit disk. The neat geometric observation is that

1Why? Because the correct definition of connectedness excludes the empty space. Otherwise we

wouldnt get a unique decomposition of every topological space into its connected components. The

same reason as why we dont count 1 as a prime number: to have unique factorization.


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z Uif and only ifz is closer toi than it is toi. This yieldsz U |zi|


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z1, z2, z3 determine a unique circle C in C; thus S1(R) = C. In a similar fashion,

we get (S T)1(R) =Csince 1, 2, 3 uniquely determine C. Thus we haveT(C) =T((S T1)(R)) = (T(S T)1)(R) =S1(R) =C .

Fall 2012, 7. Suppose f(z) is analytic on the punctured unit disk D {0}, andthe real part off(z) is positive. Prove that fhas a removable singularity at 0.

Let H ={z C : Re(z) > 0}. We may writef : D {0} H. Also by the abovediscussion, we have an invertible analytic map

: H Dz z1

z+ 1

with the inverse 1(z) = z+ 1

z+ 1

=z+ 1

1

z

. Note that g := f is analytic and mapsD {0}to D. In particular,g is bounded. Thus

limz0 zg(z) = 0

and hence 0 is a removable singularity ofg. So there exists an analytic function

h: D Csuch that h|D{0} = g. Since h(D {0}) D, by continuity we get h(D) D. Thereare two cases by the open mapping theorem:

(1) h is constant. Then picka D {0}, soh(0) =h(a) =g(a) D.(2) h(D)

Int(D) = D.

In any case, we may write h : D D. So forming the composition1 his legal, and(1 h)|D{0} = 1 g= f. We effectively removed the singularity offat 0.

Fall 2009, 5. Letf(z) be an analytic function on C which takes value in the upperhalf plane U. Show that f is constant.

Same deal. (z) =ziz+i

maps U conformally to D, thus f is an entire functionwhich takes values in D. Thus by Liouvilles theorem, f is constant. is invertible,so fis also constant.

Spring 2009, 3. Find an explicit conformal equivalence which maps the open setbounded by

z 12i = 12 and|zi|= 1 onto the upper half plane U.

I am too lazy to learn drawing circles in LaTeX now, so the reader should draw them.Let D1 ={z C :

z 12

i < 1

2} and D2 ={z C :|zi| < 1}. We want to map

the open set V := D2D1 conformally onto U. Consider the circlesC1 = D1 andC2 = D2. Note that C1C2 ={0}. Lets pick two other points on C2: 2iand 1 +i.We can sendC2 to the real line in various ways, but its better to pick an LFT so that


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0 . This will ensure C1 maps to a line as well. The assignments1 +i0

2i10

can be realized by the LFT

(z) =(z(1 +i))(2i 0)(2i (1 +i))(z0)

=(z1i)2i

(i 1)z=

2iz+ 22i(i 1)z .

So mapsC2 to the real line R. And since

(i) =2 + 22i

(i 1)i = 2i(i 1)i

= 2i 1 = 21i =2(1 +i)2 = 1 +i U ,mapsD2 onto the upper half plane U. Lets see what does toC1. We already know(0) = and (i) = 1 +i. Lets choose another point on C1: i+12 . We have

1 +i

2

=

2i1+i2

+ 22i

(i 1) 1+i2

=

i 1 + 22i1 =1 +i

So we deduce that (C1) ={z C : Im(z) = 1}. And since

i2

= 2i i2

+ 22i(i 1) i2

=12i1i2

= 24i1i =

2 + 4i1 +i

=(2 + 4i)(1i)

2

=2 + 6i

2 = 1 + 3i ,

we deduce that (D1) is the half plane{z C : Im(z)> 1}. Thus(V) =(D2D1) ={z C : Re(z)>0 and Im(z)< 1} .

Consider the translation(z) =zi. We have

( )(V) ={zC

: Re(z)> 0 and Im(z)0} .

In other words, if we let = , (V) is the fourth quadrant. Hence (i)(V) isthe first quadrant. Since zz2 sends the first quadrant onto the upper half plane U,(i )2(V) =2(V) = U. Being a composition of conformal maps,2 is conformal.


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Spring 2008, 6. Find an explicit conformal equivalence from the regionR1 to theregionR2, where

R1 ={z C : Re(z)>0, Im(z)>0, |z|< 1}and

R2 ={z C : Re(z)>0, Im(z)>0} .

Note thatR1is contained in the unit disk D andR2is contained in the upper half planeU. We noted above that

(z) =ziz+i

maps U(conformally) onto D. Note that for any a R,

(ia) = ia iia +i

=a 1a + 1

lies on the real axis R. Thusmaps the imaginary axisiR to R. Therefore(UiR) =DR and sinceR2is a connected component ofUiR,(R2) is a connected componentofD

R. Lets pick a point, say 1 + i

R

2, to see which component (R

2) is. We have

(1 +i) = 1

1 + 2i=

12i5

.

Therefore(R2) ={z D : Re(z) > 0} =: V. So we will be done if we can map R1onto V. And this is easy. Note that applying (z) =z2 yields

(R1) ={z D : Im(z)> 0} .And finally, rotation by /4 (a.k.a multiplying by i) maps this to V. Thus z 1(i(z)) maps R1 toR2.

3. Rouches theorem

Rouches Theorem. Suppose thatis a smooth closed curve in the open setV suchthatInd(, z) is either 0 or 1 for allz C and equals0 for allz C V, and let ={zV : Ind(, z) = 1}. Iff, gH(V) and

|f(z)g(z)|


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The more classical Rouches theorem states the result for the stronger condition

|f(z)g(z)|


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whereis are the roots off. This is possible since C is algebraically closed. Comparingthe constant terms, we get

1 =a(1) (n) ,thus

|1| |n|= 1

|a

|

.

Therefore if 1|a| 2n, at least one ofis should satisfy|i| 2. So this handles thecase|a| 12n . For|a| < 12n , we compare f(z) and z+ 1 in Rouches theorem on thedisk|z|1 .For|z|= R we have

|f(z)zn|=|an1zn1 +an2zn2 . . . a1z+a0|

|an1zn

1

|+|an2zn

2

| . . . |a1z|+|a0|=|an1|Rn1 +|an2|Rn2 . . . |a1|R+|a0| |an1|Rn1 +|an2|Rn1 . . . |a1|Rn1 +|a0|Rn1=Rn1 (|an1|+|an2|+ +|a1|+|a0|)< Rn =|zn|

Alsof(z) andzn are entire functions. Therefore by Rouches theorem, in|z|< R,f(z)and zn have the same number of roots, which is n.

Fall 2009, 1. How many roots (counted with multiplicity) does the function

g(z) = 6z3 +ez + 1

have in the unit disk|z|< 1?

This time we dont have a polynomial. g is of course entire though and for|z|= 1,|g(z)6z3|=|ez + 1| |ez|+ 1 =eRe(z) + 1e|z|+ 1 =e+ 1 < 6 =|6z3| .

Therefore by Rouches theorem,f(z) and 6z3 have the same number of zeros in|z|< 1,which is three.


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Spring 2009, 4. How many zeros does the function f(z) =z6 + 4z2ez+1 3 havein the unit diskD(0, 1)?

Lets try to apply Rouche for the entire functions f(z) and 4z2ez+1. For|z|= 1,|f(z)4z2ez+1|=|z6 3| |z6|+ 3 = 4

On the other hand, again for|z|= 1,|4z2ez+1|= 4|ez+1|= 4e|ez|= 4eeRe(z) 4ee1 = 4

because on the unit circle, Re(z) 1. Thus, we get that|f(z)4z2ez+1| |4z2ez+1|

on|z| = 1. The problem is we didnt get the strict inequality that Rouches theoremrequires. However, looking at how we arrived at this inequality, the two terms beingequal can only occur if|z6 3| = 4 and Re(z) =1 for some zwhere|z| = 1. Thelatter yields z=1, but|(1)6 3|= 2= 4. So equality can never occur. Thus, f(z)and 4z2ez+1 have the same number of zeros in|z| 1, show that the equation z+ ez = has exactly onesolution with positive real part.

Let, by our experience from IVT questions in calculus, f(z) =z+ ez. Consider thecircle centered at the origin with radius R, cut it into two with the y-axis and let theright semicircle beR. It suffices to show that, for every R > + 1,fhas a single zeroinside R (because the interiors of such semicircles cover the entire right half plane).Heref(z) andz are entire functions, and when z is on R, we have

|f(z)(z)|=|ez|= eRe(z) =eRe(z) 1 and >0), therefore falso has


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a single root inside R. It seems that >0 is enough for this question. Maybe I didsomething wrong.

Fall 2007, 7. Determine the number of zeros ofez2 4z2 in the open unit disk.

Let f(z) = ez2

4z2 and g(z) =

4z2. f(z) and g(z) are entire functions, and on

|z|= 1, we have|f(z)g(z)|=|ez2|= eRe(z2) e|z2| = e 0} .Suppose that|f(z)|


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Fall 2012, 8c. Let f be a holomorphic function in the unit disk D and supposethat f(0) = 0 and|f(z)| 1. Prove that for any integer m1, f(z)2mzm hasexactlymzeros (counting multiplicity) in the disk|z|


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It should be stated in this question that f maps Dto itself, otherwise the statement isfalse (consider f(z) = 2z2).Note that, ifDris the disk centered at 0 with radius r


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lemma. And has to be 1 since f(q) =q.For the general case, similar to the previous question, let

(z) = pz1pz.

Then the analytic functiong = f : D Dsatisfies

g(0) =(f((0))) =(f(p)) =(p) = 0and

g((q)) = (f(q)) =(q)

since = 1. As p=q, we have (q)= 0 and hence by the first case g = idD. Thusf= idD.

5. Cauchys Estimates

These are the only named estimates I know. This probably speaks more about myignorance in analysis than the lack of such estimates. Anyway, here they are:

Cauchys Estimates. Suppose thatfis holomorphic in a neighborhood of the closeddiskD(z0, r) and|f| M inD(z0, r). Then

|f(n)(z0)| Mn!rn

for everyn N.Liouvilles theorem, that is, the fact that entire and bounded functions are constant isan easy corollary. Examples:

Fall 2012, 5 and Fall 2010, 6. Suppose that g : C

C is a holomorphic

function,k andnare integers, and (2 +|zk|)1g(n)(z) is bounded on C. Prove that g is a polynomial. Estimate the degree ofg in terms of the integers k and n.

Let h= g(n), so his entire and|h(z)| M(2 +|zk|) for some M >0. Fix r >0. Nowfor|z| r, we have|h(z)| M(2 +rk), therefore we can apply Cauchys estimates toD(0, r) to get

|h(m)(0)| M(2 +rk)m!

rm

for every m N. Sending r above yields g(m+n)(0) = h(m)(0) = 0 for m > k.As an entire function, g has a globalpower series representation around 0; thus g is apolynomial of degree at most k +n.

Spring 2012, 3. Let f(z) be an entire function. Suppose that there are positivereal numbers a, b and k such that|f(z)| a +b|z|k for all z C. Prove that f(z)is a polynomial.

Same thing. Fix r >0. For|z| r, we have|f(z)| a +brk .


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That is, D(z0, |z0|) D(0, 1) and hence|f| M in this closed disk. Thus byCauchys estimates, we get

|f(z0)| M |z0|.

This is true for every 0 < 0}. Show that f isconstant.

The only nontrivial part is to recall that Hcan be mapped conformally onto the unitdisk D. Indeed, the linear fractional transformation

(z) =z

i

z+i

does precisely that because z D if and only if z is closer to i than it is toi,if and only if|zi| 0 such that

|z| L

f(z)

1 +|z|1/2 M .

Well, f(z)

1 +|z|1/2 is also bounded on|z| L by compactness, so we may assume

f(z)

1 +|z|1/2 M .

for all z

C. Now we are in business. Fixr >0. If

|z

| r, we have

|f(z)| M(1 +|z|1/2)M(1 +r1/2) .Thus applying Cauchys estimates to the closed disk D(0, r), we get

|f(n)(0)| M(1 +r1/2)n!

rn

for everyn N. Sendingr above, we getf(n)(0) = 0 for everyn1. Being entire,fhas aglobalpower series representation around 0, and therefore all the coefficients inthis power series except the constant term is 0. Thus f is constant.


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6. Residue Calculus

I wont state the Residue theorem, because its statement is much more confusing thanthe usual applications and I am lazy.Let me note the following useful lemma for calculating residues:

Lemma 5. Letg be analytic aroundz0 andf(z) =

g(z)

(zz0)n . Then

Res(f, z0) =g(n1)(z0)

(n1)! .

Proof. Res(f, z0) is the coefficient of the 1

zz0 term in the Laurent series expansion off aroundz0, so it is equal to the coefficient of the (z z0)n1 term in the power seriesexpansion ofg aroundz0.

Fall 2012, 3.Evaluate the following integral for any >0

0

ln(x)

x2 +2

Carefully justify your steps.

Pickr, R such that 0< r < < R. Consider the following curves:

R : [0, ] CtReit

r,R: [R, r]C

tt

r : [0, ] Ctrei(t)

r,R: [r, R] Ctt

These look like a random selection of functions because I am trying to avoid drawing

things. Drawing shows that these curves can be juxtaposed consecutively to yield aclosed curve R,r which has a semicircular shape with a small bump inside. Now letV = C {z C : Re(z) = 0 and Im(z) 0}. Note that every point which havenonzero winding number (which must then be 1) with respect to R,r lies in V. Also,there is a continuous angle (or argument) function : V (/2, 3/2) which givesrise to the analytic branch of logarithm

L: V Czln |z|+i(z)


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Let f(z) = L(z)

z2 +2. The only singularity offwhich lies inside R,r is iand if we let

g(z) = L(z)

z+i, Lemma 5 yields

Res(f,i) =g(i) =L(i)

2i

=ln(|i|) +i(i)

2i

=ln() +i2

2i .

Therefore by the Residue theorem,

R,r

f(z)dz= 2i Res(f,i)

R

f(z)dz+R,r

f(z)dz+r

f(z)dz+R,r

f(z)dz= ln()

+i 2

2. ()

Now we look at how the four integrals in () behave. When z is onR, we have

|L(z)|=

[ln |z|]2 + [(z)]2=

[ln R]2 + [(z)]2

[ln R]2 + 3

22

ln R+32

and

|z2 +2| |z2| |2|

=R2 2 .

Thus by the ML-inequality, we have

R

f(z)dz

ln R+ 3

2

R2 2 (length ofR) =ln R+ 3

2

R2 2 R0 as R .

Similarly, we have

r

f(z)dz

ln r+ 32

2 r2 r0 asr0


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since limr0 r ln r= 0. Taking these limits in (), we get

ln()

+i

2

2= lim

r0 limR

R,r

f(z)dz+R,r

f(z)dz

= limr0 limR

rR

f(x)dx+ Rr

f(x)dx

= limr0 limR

rR L(x)x2 +2 dx+

Rr

L(x)x2 +2 dx

= limr0 limR

rR

ln |x|+ix2 +2

dx+ Rr

ln |x|x2 +2

dx

= limr0 limR

rR

ln |x|x2 +2

dx+rR

i

x2 +2dx+

Rr

ln |x|x2 +2

dx

= limr0 limR

irR

x2 +2dx+ 2

Rr

ln |x|x2 +2

dx

.

Taking the real parts of both parts (note that Re(z) is a continuous function so itcommutes with limits), we get

ln()

= lim

r0 limR

2 Rr

ln x

x2 +2dx

.

Thus 0

ln x

x2 +2 =

ln()

2 .

I dont think Ill put any other solutions of integration problems. They are LONG, andtoday is the prelim day.

7. Miscellaneous

These are problems which dont fit in one of the above categories.

Fall 2012, 2. Let fbe a complex valued function in the unit disk D such thatg= f2 andh = f3 are both analytic. Prove that f is analytic.

Note that the zeros off , g, hin Dare the same. Lets denote this common zero set byZ. We may assumefis not identically zero, so the analytic functiong is not identicallyzero; hence its zero set Z is isolated. Therefore, as g and h are analytic, every aZhas a neighborhood such that

g(z) = (za)m

g(z)h(z) = (za)nh(z)

where m, n are positive integers and g, h are analytic functions which dont vanisharound a. Observe that a is a zero of the analytic map g3 = h2 of order 3m = 2n.Since 2 and 3 are relatively prime, there exists a positive integer k such that m = 2kand n = 3k. The function

f(z) = (za)k h(z)g(z)


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is well-defined and analytic around a. When z=a, we have f(z) = h(z)g(z)

=f(z). And

f(a) = 0 = f(a). Thus f= f is analytic at a. Since aZwas arbitrary, we concludethatfis analytic onZ. Finally as 1/gis analytic on DZ,f=h/g is analytic on DZ.

Spring 2012, 4. Prove that there is no function fthat is analytic on the punctureddisk D

{0}

, andf has a simple pole at 0.

Suppose, to the contrary, that there is such a function. So there is an open ball Bcontaining 0 such that inB {0}, we have

f(z) = c

z+g(z)

wherec= 0 and gH(B). Since B is simply connected, there exists GH(B) suchthat G = g. Thus the function

1

c(Gf) H(B {0}) is an antiderivative of 1/z.

This is impossible because there are closed curves in B {0} on which the integral of1/z is nonzero.

Fall 2011, 2. Let h be a nowhere zero, entire holomorphic function. Prove thatthere exists an entire holomorphic function g such that eg =h.

Observe that h

h is entire, therefore has an antiderivative f (C is simply connected).

Consider the entire function c(z) =h(z)ef(z). We have

c(z) =h(z)ef(z) h(z)ef(z)f(z) =ef(z) (h(z)h(z)f(z)) = 0 ,therefore c(z) is constant, say c(z) c. Note that c= 0 since h is nowhere zero.Thereforec = ea for some a

C. So if we let g(z) =a +f(z), we have

h(z) =cef(z) =eaef(z) =ea+f(z) =eg(z) .

Fall 2011, 5. Let f(z) be an entire holomorphic function. Suppose thatf(z) =f(z+ 1) and|f(z)| e|z| for all z C. Prove that f(z) must be constant.

NOTE: My friend Theo solved this problem. He was very concerned that I wont givecredit to him when I put the solution here :(The idea is to writef(z) =g(e2iz) and transfer the growth condition on fto a growthcondition on g. Let E(z) = e2iz . E is locally invertible, that is, every z C {0}has a neighborhood Vz such that there is an analytic map z :Vz

C which satisfies

Ez = id. Note that if: V Cand : W C are two such local inverses ofE,for zVW we have

e2i(z) =E((z)) = z= E((z)) =e2i(z)

therefore(z) =(z)+nfor somen Z. Sincefis invariant under integer translations,we have f = f . Thus, the map

g: C {0} Cz(f z) (z)


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is analytic. We havegE = f, as desired. Now for every z C {0}, there exists Cwith 0< Re()< 1 such that z= e2i . We have

|g(z)|=|f()| e||= e

[Re()]2+[Im()]2 < e

1+[Im()]2 e1+| Im()| . ()Note that

|z|=|e2i |= eRe(2i) =e2 Im()hence

e Im(w) =|z| 12 and eIm() =|z|12 .Thus|g(z)| e max{|z| 12 , |z|12 }by (). So for R >1, since g is entire, we can applyCauchys estimates to the closed disk D(0, R) and get

|g(n)(0)| R 1

2 n!

Rn =

n!

Rn 1

2

0 as R

for everyn; thus g(n)(0) = 0 for every n1. This implies that g is constant (becausebeing entire,g has aglobalpower series representation around 0). Thus fis constant.

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Complex Prelim Practice