SUGGESTIONS FOR ANALYSIS I - SERIES 3
Exercise 9
3+2i
• It suffices to eliminate the denominators by multiplying for 1+31
1+3i and 3+2i
and then do the computations.
• Write 8 − 6i in the form ρeα where ρ = k8 − 6ik and α is the phase of the
complex number 8 − 6i. Now, to extract the root it suffices to compute
the roots of ρ, which is a positive real number, while for the rooth of the
exponential consider just a fraction of α.
1+i
and
• It is similar to point (b). Here at first eliminate the denominator by 1+i
α
then use the decomposition of a complex number in the form ρe . Finally,
compute the power ρk and ekα .
• Use the formula for the solution of equations of degree 2 and then play with
the techniques seen in the previous points.
Exercise 10
• If A = ∅ there is nothing to say. Now, suppose that A 6= ∅. To show that
sup(A) ≤ sup(B) it suffices to show that sup(B) ≥ x for each x ∈ A.
To do this, suppose, by contraddiction, that there exists x0 ∈ A such
that x0 > sup(B). Since A ⊆ B it follows that x0 ∈ B. However, from
x0 > sup(B) we have x0 ∈
/ B (explain better this step). So that, x0 ∈ B
and x0 ∈
/ B which is a contraddiction.
The second part with inf is similar. Please explain better and not just say
that follows from the previous part.
• If A or B are empty it is quite easy. You have just to remember that
inf(∅) = +∞ by definition. Moreover, if inf(A) = −∞ or inf(B) = −∞
also inf(A ∪ B) = −∞. To see this just use step (a). Now, let us assume
inf(A) ∈ (−∞, +∞) and inf(B) ∈ (−∞, +∞) and let us suppose A 6= ∅
and B 6= ∅. Since A ⊆ A ∪ B and B ⊆ A ∪ B we can deduce from the
previous part that inf(A ∪ B) ≤ inf(A) and inf(A ∪ B) ≤ inf(B). Hence,
inf(A ∪ B) ≤ min {inf(A), inf(B)}
so that we have just to prove the inequality
inf(A ∪ B) ≥ min {inf(A), inf(B)} .
(1)
Finally, to show (1) use the definition of inf.
• Use the definition of sup and inf and the one of the set −A.
• From part (a)
!
[
sup
Ai ≥ sup(Ai )
i∈I
for all i ∈ I and then
!
sup
[
Ai
≥ sup {sup(Ai ) |∈ I} .
i∈I
Hence, we just have the inequality
!
[
sup
Ai ≤ sup {sup(Ai ) | i ∈ I}
i∈I
to prove.
1
2
SUGGESTIONS FOR ANALYSIS I - SERIES 3
Exercise 11
• To prove the first part it is sufficient to show that the following system of
n + 1 equations in the n + 1 variables c0 , . . . , cn


 p(z0 ) = w0
..
.


p(zn ) = wn
admits one and only one solution. Pay attention that here the variables
of your system are exaclty c0 , . . . , cn . If you develop the computation you
have the following system

c0





 c0 + (z1 − z0 )c1
c0 + (z2 − z0 )c1 + (z2 − z1 )(z2 − z0 )c2

.


 ..


c0 + (z2 − z0 )c1 + (z2 − z1 )(z2 − z0 )c2 + · · · + (zn − z0 )(zn − z1 ) · · · (zn − zn−1 )cn
Last, you have to explain why this system of n + 1 equations in the n + 1
variables c0 , . . . , cn admits one and only one solution. As a suggestion,
notice that the homogeneous (n + 1) × (n + 1) associated matrix has nonzero determinant.
• To show that q(z) is a good interpolation it is now sufficient to show that
1 if l = k
Lk (zl ) =
0 if l 6= k
and then plug in this result to conclude that q(zl ) = wl for l = 0, . . . , n + 1.
• Finally to show that p = q use the polynomial identity theorem over the
complex field.
Exercise 12
• The part (a) is known as Cantor’s diagonal principle. Look in the literature
for more details.
• If A is the set of polynomials with integer coefficients, write A in the form
[
A=
An
n∈N
where An is the set of polynomials with integer coefficients of degree less
then or equal than n. Now, use point (a) and the fact that each An is
countable. (You have to explain better this step).
• If B is the set of algebraic numbers, this set will be the union of the set
of solutions of polynomials with integer coefficients. For each of those
polynomial we have at maximum a finite number of solution (why?). So
that, we can again express B as countable union of countable sets.
• We proved that the set B of algebraic numbers is countable. Moreover, B is
contained in R. Since the cardinality of R is greater than countable, there
exists an element σ ∈ R \ B. In particular, since σ ∈
/ B we conclude that σ
is a trascendent number.
=
=
=
..
.
w0
w1
w2
..
.
= wn