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·+a1z+a0 be a complex polynomial

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1. Homework 8

(1) In this exercise, you are going to prove the fundamental Theorem of algebra: any nonconstant complex polynomial must have a root in C. Let us prove it by contradic- tion. Let p(z) = anzn+· · ·+a1z+a0 be a complex polynomial. Assume that p(z) 6= 0 for all z ∈ C. Define a function f : [0, ∞) × [0, 2π] → R by f (r, θ) = 1/p(re). Define I : [0, ∞) → C by

I(r) = Z

0

f (r, θ)dθ.

(a) Prove that fθ = irfr and that I0(r) = 0.

(b) Prove that lim

r→∞I(r) = 0. (Hint: show that when r is sufficiently large, |p(re)| >

|an|rn/2.) (c) Prove that lim

r→0+I(r) = I(0). Find I(0). (Hint: use the continuity of f (r, θ) to estimate |I(r) − I(0)| when r > 0.)

(d) Use the above results to show that such p does not exist.

(2) Let U be an open subset of R2 and [a, b] × [c, d] be a compact subinterval of U.

Suppose f : U → R is C1. Prove that d

dt Z t

a

f (x, t)dx = Z t

a

ft(x, t)dx + f (t, t).

(3) For t > 0, we define

F (t) = Z t

0

log(1 + tx) 1 + x2 dx.

(a) Compute F0(t).

(b) Prove that F (t) = 1

2log(1 + t2) tan−1t for t > 0.

(c) Evaluate Z 1

0

log(1 + x) 1 + x2 dx.

(4) Prove that

Z 0

(e−x− e−tx)dx

x = log t

for t > 0 by justifying differentiation under the integral sign. Evaluate Z

0

(e−ax− e−bx)dx x . (5) Prove that

Z 0

sin(tx)

x(x2+ 1)dx = π

2(1 − e−t) for t > 0 by justifying differentiation under the integral sign.

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