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The math 534 homework #1 from the autumn 2008 semester. The assignments cover various topics such as power series expansions, complex analysis, and calculus. Students are required to prove theorems, find the convergence of series, and define complex exponentials. Some problems involve geometric interpretations and others require the use of limits.
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Math 534 Homework # Autumn 2008
Some are routine, but take time. A creative idea on others can lead to simple solutions. Learning how to write accurate, succinct and subtantiated arguments is part of this course, so please write your solutions carefully. You may discuss the problems with others, but you must write the solutions in your own words.
n=0 nz
n (^) convergent? Same question for
∑^ ∞
n=
z 1 + z
)n .
Draw pictures for the regions.
n=
zn n!.^ Show that this series converges for all^ z, that^ e
z (^) ew (^) = ez+w, that
eiθ^ = cos θ + i sin θ where cos θ and sin θ are defined by their series expansion as you learned them in calculus. Finally show |ez^ | = eRez^.
j=0 |aj^ |
(^2) < ∞. Show f (z) = ∑∞ j=0 aj^ z
j (^) is analytic in {z : |z| < 1 }. Compute
lim r↗ 1
∫ (^2) π
0
|f (reiθ)|^2
dθ 2 π
(Prove your answer).
|z + w|^2 + |z − w|^2 = 2(|z|^2 + |w|^2 ).
In geometric terms, the equality says that the sum of the squares of the lengths of the diagonals of a parallelogram equals the sum of the squares of the lengths of the sides. It is perhaps a bit easier than a proof using high school geometry.
A =
b − a
∫ (^) b
a
f (x)dx,
be the average of f over the interval [a, b]. (a) Show that if |f (x)| ≤ |A| for all x ∈ [a, b], then f is constant. (A picture might help) (b) Show that if |A| = (1/(b − a))
∫ (^) b a |f^ (x)|dx,^ then arg^ f^ is constant.
Challenge problem: 7. Suppose f is analytic in a convex open set U. Define f ′(ζ) to be the
coefficient of z − ζ in the power series expansion of f based at ζ (hence f ′(ζ) = limz→ζ f^ (z z)−−fζ^ (ζ)). Suppose that for each z, w ∈ U there exists a point ζ on the line segment between z and w with
f (z) − f (w) z − w
= f ′(ζ).
Prove f is a polynomial of degree at most 2. (The point is that you have to be careful: not all calculus theorems extend to similar complex versions.)