Download Solutions to assignment 9 File

MTH4110/MTH4210
Mathematical Structures
Solutions to assignment 9
Autumn 2016
The solutions to homework problems don’t just give you the answer; they do two
other things. It is your job to disentangle these!
(a) They give some pointers on how to attack an unfamiliar question.
(b) They give specimens of how to write out solutions to maths problems.
1 (∗) Let f : C → C be given by
f (z) = z2 + i .
(a) Is f a function?
(b) Is it injective?
(c) Is it surjective?
(a) Yes, it is a function. Given a complex number z as input, f unambiguously
outputs a complex number (namely z2 + i).
(b) We need to check whether f (z) = f (w) implies z = w or not. Suppose that
f (z) = f (w). Then z2 = w2 , so z2 − w2 = 0, which implies (z − w)(z + w) = 0. Thus
z = w or z = −w. In other words, f is not injective. For a concrete counterexample,
you can take any complex number z with z 6= 0, since then
z 6= −z,
but
f (−z) = (−z)2 + i = z2 + i = f (z) .
Even more succinctly, we could have written: f is not injective, since f (−1) = f (1).
(c) We need to determine whether the range of f is all of C. Put differently, we
need to check whether, given any w ∈ C we can find z ∈ C such that f (z) = w. This is
the case, since, given any w ∈ C, we can find z such that
z2 + i = w ,
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either by appealing to the Fundamental Theorem of Algebra (which tells you that the
polynomial p(x) = x2 + i − w has a complex zero, that is, there is a z ∈ C satisfying
p(z) = 0) or by using the quadratic formula (which holds in the complex case as well,
see Proposition 8.2.1 in the Supplementary material to Notes 8). Thus, f is surjective.
2 (†) Find the real and imaginary part of
1+
1+ 1+i
1−i
1+i
1− 1−i
1−
1+ 1+i
1−i
1+i
1− 1−i
.
Perhaps the simplest way of untangling this is to observe that
1 + i (1 + i)(1 + i) (1 + i)2 1 + 2i + i2 1 + 2i − 1 2i
=
=
= = i.
= 2
=
1 − i (1 − i)(1 + i)
|1 − i|2
1 + 12
2
2
Thus
1+
1−
1+ 1+i
1−i
1− 1+i
1−i
1+i
1+ 1−i
1+i
1− 1−i
=
1 + 1+i
1−i
1−
1+i
1−i
=
1+i
= i,
1−i
so the real part is 0 and the imaginary part is 1.
3 (†) Use De Moivre’s Theorem to express cos 4θ as a polynomial in cos θ , and to
express sin 4θ as cos θ multiplied by a polynomial in sin θ .
De Moivre’s theorem tells us that
cos 4θ + i sin 4θ = (cos θ + i sin θ )4
= cos4 θ + 4i cos3 θ sin θ − 6 cos2 θ sin2 θ − 4i cos θ sin3 θ + sin4 θ
= (cos4 θ − 6 cos2 θ sin2 θ + sin4 θ ) + 4i(cos3 θ sin θ − cos θ sin3 θ ),
using the Binomial Theorem for n = 4 and the fact that i2 = −1 in the second line; so
cos 4θ = cos4 θ − 6 cos2 θ sin2 θ + sin4 θ ,
sin 4θ = 4 cos θ sin θ (cos2 θ − sin2 θ ).
These are not quite what was asked for; we have to do a bit further work.
To get cos 4θ as a polynomial in cos θ , we use the fact that sin2 θ = 1 − cos2 θ ; so
cos 4θ = cos4 θ − 6 cos2 θ (1 − cos2 θ ) + (1 − cos2 θ )2
= 1 − 8 cos2 θ + 8 cos4 θ ,
2
and similarly
sin 4θ = 4 cos θ sin θ (1 − 2 sin2 θ ).
4 (†) One of the following is not a function; which one, and why? In the other cases,
is the function one-to-one? is it onto?
(a) The function F : C → R given by F(z) = arg(z).
(b) The function F : R → R given by F(x) = x3 − x.
(c) The function F : R → R given by F(x) = ex .
Before getting into the nitty-gritty of the question itself, a quick remark on terminology. The terms ‘one-to-one’ and ‘onto’ are synonyms for ‘injective’ and ‘surjective’, respectively. Thus, a function is injective if and only if it is one-to-one, and it is
surjective if and only if it is onto.
(a) This is not a function: the argument of the complex number 0 is undefined.
(b) This is a function, since for every input x there is an output x3 − x.
It is not one-to-one, because F(−1) = F(0) = F(1) = 0.
It is onto: this is most easily seen from the graph of the function:
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The function grows unboundedly as x → ∞, and so every positive real number however
large is a value of the function. Similarly for negative values.
A more formal argument is the following. The function is continuous. Moreover,
limx→−∞ F(x) = −∞ and limx→∞ F(x) = ∞, so by the Intermediate Value Theorem, F
takes all values in (−∞, ∞) = R.
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(c) This is a function, since for every input x there is an output ex .
It is one-to-one: if F(x1 ) = F(x2 ), that is, ex1 = ex2 , then taking logarithms we see
that x1 = x2 .
It is not onto, since ex is always positive, so zero and negative values do not occur.
5 (†) Is this proof valid?
Theorem cos2 x + sin2 x = 1.
Proof Let f (x) = cos2 x + sin2 x.
Differentiating, f 0 (x) = 2 cos x(− sin x) + 2 sin x cos x = 0. Hence f (x) is constant.
But f (0) = 1; so f (x) = 1 for all x.
There is nothing wrong with the logic of the proof. It is a good idea to check what
facts from calculus we are using:
(a) the formulae for differentiating sums, products, sines and cosines;
(b) the fact that an everywhere differentiable function whose derivative is zero is
constant;
(c) the values of cos(0) and sin(0).
6 (From the summer exam 2016) Let F : C → C be a function. A complex number
w is said to be a fixed point of F, if F(w) = w.
(a) Show that 2 − i is a fixed point of the function f given by f (z) = z2 − 3z + 5.
(b) Does every polynomial of degree greater 2 have a fixed point? Give reasons
for your answer.
(a) We have
f (2 − i) = (2 − i)2 − 3(2 − i) + 5 = 4 − 4i − 1 − 6 + 3i + 5 = 2 − i ,
so 2 − i is a fixed point of f .
(b) Yes. If p is a polynomial of degree greater 2, then h(z) = p(z) − z is a nonconstant polynomial, so must must have a zero w by the Fundamental Theorem of
Algebra. Thus p(w) − w = 0, so w is a fixed point of the polynomial p.
OFB 30/11/16
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