COMPLEX ANALYSIS, Extra Problems.    Spring 1999	(Prof. King)

This list is current as of
	12Apr1999

Use _ for subscript, ^ for superscript, and braces ({}) for bracketing
a compound object.  Thus b_{23} and b_2 are "b sub 23" and "b sub 2",
respectively.  And 2^{11} equals 2048.
  Use z% for the complex-conjugate of z, and use Re() and Im() for
the real-part imaginary-part functions.  Use |z| for the absolute
value of z.
  Use "zip" to mean the zero-polynomial; all its coefs are zero.

:::::::::::::::: Drill Problems ::::::::::::::::

D1: For a non-zero complex number w, express Im(1/w) in terms of
Im(w) and |w|.

::::::::::::::::

 /------------------------------------------------------------\
/   E1: Suppose f and g are two complex polynomials.  Prove    \
    that if f and g describe the same function (i.e. for all z
    in C, f(z)=g(z)) then, f and g have the same coefficients.

  SOLUTION: Letting h := f-g, the hypothesis is that h is
the zero-fnc, and we wish to conclude that -as a
polynomial- all of h's coefs are zero.
  Let N denote the degree of h.  Consider any N+1 points in
the complex plane; h() must be zero at each of them.  Thus

(*)	h() has more roots than its degree.

The only poly with this property is zip.

  Now, To prove this latter --that each non-zip poly of degree N has
at most N roots-- one can use "synthetic division".
  First, if your poly is not zip, then you can multiply by a non-zero
constant so that your poly is monic (that is, the coef of its
high-order term is 1).
  Now for each root r, divide (z - r) into your poly, thus reducing
its degree by 1.  You can divide *at most N* times, for after N such
divisions, the resulting poly is monic, and of degree 0; so it
definitely has no more roots.
----------------------------------------------------------------

E2: Describe all the multiplicative subgroups in the complex plane.
----------------------------------------------------------------

E3: If (a,b,c) are the coordinates on the Riemann Sphere of sigma(z)
for some z in C, write a, b, and c in terms of rational-functions of z
and z-bar.  Here, "sigma" is stereographic projection.

  ANSWER: Let z% denote z-bar, the complex conjugate of z, and let M
denote the reciprocal of [z * z% ]+1.   Then
	(a,b,c) =   M * ( z + z% ,  [z - z%]/i , [z * z% ] - 1 ).
In terms of the Re&Im parts of z, we can write
	(a,b,c) =   [1/[x^2 + y^2 + 1]] * ( 2x , 2y , [x^2 + y^2] - 1 ).
Notice, as z goes to infinity in C*, that (a,b,c) does indeed head to
(0,0,1), the NorthPole.
  [We worked out a solution to this problem in the extra class.]
----------------------------------------------------------------

E4: Let g(x) := exp(-1/(x^2)); so Dom(g) is the
punctured line, R \ {0}.  Now

		/
	       | g(x),	for x neq 0
  let f(x) :=  |
	       | 0,	for x=0.
		\

  PROVE: f is infinitely differentiable at 0.  Furthermore, for all n
the nth derivative of f evaluated at 0 equals 0.
  HINT: By induction on n, prove that for each n there is a
rational-function r (depending on n) such that

		/
	       | r(x)*g(x),	for x neq 0;
 f^{(n)}(x) =  |
	       | 0,		for x=0.
		\

(Note the distinct uses of = and of := above.)
----------------------------------------------------------------

E5: Suppose g is a complex-polynomial, which is not zip.  Prove that g
is relatively prime to its derivative, g', IFF g has no multiple roots
(over C).
----------------------------------------------------------------

E6: (a) Suppose f is analytic.  Prove that z |-> f-bar(z-bar) is
analytic.

  (b) More generally, suppose g is the N-fold composition
	f_{N} o f_{N-1} o ... o f_2 o f_1 ,
and that each f is either analytic or anti-analytic.  Let A denote
the number of anti-analytic fncs in the composition.  Prove that g is
analytic/anti-analytic as A is even/odd.  (Hint: Induction on N.)
----------------------------------------------------------------

E7: Use (N; m_1, m_2, ..., m_L) for the multinomial coefficient
[defined in class] of "N choose m_1 comma m_2 comma ... m_L".

(a) What is the coef of
	x^5 * y^2 * z^6
in (x + y + z)^{13} ?  Express your answer as a product of powers of
prime numbers.

(b) Show that the multinomial coef (N; a,b,c) equals this product of
binomial coefs:
	(N; a, b+c) * (b+c; b,c).
Use this, together with induction on L to get a general formula for
	(N; m_1, m_2, ..., m_L)
as a product of [L-1] many binomial coefficients.  Now express this
formula in terms of factorials.

(c) Use the Elem. Binomial Thm (as stated in class; ditto the
multinomial version of it) to get a nice formula for the two
alternating sums below, where C(N; j) means "N choose j", that is,
the multinomial coef (N; j, N-j).

  Alternating Sum:
C(N; 0) - C(N; 1) + C(N; 2) - C(N; 3) + ... + [(-1)^N * C(N; N)] .

  Another Alternating Sum:
C(N; 0) - 2C(N; 1) + 4C(N; 2) - 8C(N; 3) + ... + [(-2)^N * C(N; N)] .

(d) Invent an interesting problem using multinomial coefficients.
----------------------------------------------------------------

E8: [Alteration of P128#1(d)]
  Let M be a fixed real number in (-1,1) and let f(z) be
tan(z)/[z-M]^2.  With Gamma being the rectangle with vertical edges
at -1 and 2, and horizontal edges at i and -i, can you compute the
anti-clockwise integral of f around Gamma?  Let FRED denote the
result
  First compute the integral of f around a small circle centered at
M; call this number ABBY.  Let BERT be integral of f around a small
circle centered at Pi/2.  Argue that FRED equals ABBY plus BERT.
  Argue that to compute BERT, the main issue is computing the
integral of 1/cos(z) around a small circle centered at Pi/2; call
this last integral DAVE.  Argue that BERT equals DAVE times
sin(z)/[z-M]^2 evaluated at z=Pi/2.
  Can you compute DAVE?

----------------------------------------------------------------
Lemma L9:  Suppose h is entire, and K is a non-negative integer.  If
the derivative h^(K+1) is identically zero, then h is a
polynomial of degree at most K.
  If h is not zip, then deg(h)=M, where M is the *smallest* integer
such that h^(M+1) is identically zero.
----------------------------------------------------------------

E9.  (For this problem, you may use lemma L9, without proof.)
  Suppose f is entire, K in naturals, and:

 e9.1:  For all z in complexes,
		|f(z)| <= 19 + 17*|z|^K .

Using Cauchy's Estimate, prove that f is a polynomial, and has
degree at most K.
  [Hint: Fix w and show that f^(K+1)(w) is zero.  Notice that a
disk of radius R about w is enclosed by disk centered at 0, but
with a slightly larger radius.]

----------------------------------------------------------------
E10:  Prove lemma L9 by induction on M.

----------------------------------------------------------------
E11: Let
  [A]	h(z):= Sum_{k=0}^{+oo} z^{k!}.
This power series has RoC=1.

ell.1:	For each rational number r, let omega := cis(r*2Pi). Call
such a point a "rational point", and call all other points on the
unit-circle "irrational points".
  Prove that as t increases to 1, this limit

  [B]       LIMIT h(t*omega)	=  oo (in the Riemann Sphere).
          as t increases to 1

[Hint: Write r as P/Q with Q a positive integer.  Now break the sum
in (A) into the first Q terms, and the remaining terms starting with
k := Q.  What do you know about omega^Q ?]


e11.2:	Now let omega be an arbitrary point on the unit-circle.
Consider the limit of h(z) as z goes to omega, with z staying in
the unit ball.  Prove that this limit does not exist.
  [Hint: This follows from (e11.1).  Consider rational points
which are near omega.  Note: For *some* points omega, the
radial-limit (B) will exist.]

  The upshot is that h() cannot even be extended *continuously*
(let alone ANALYTICALLY) to the unit-circle.  It can't even be
extended at one point!

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