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Assignment 1

MATH3900 — Geometry and Topology

2022

Geometry

1. (a) [2 marks] Let A and B be points represented by complex numbers 1 + 2i and 3 + 3i respectively. Find the point C (corresponding with x + iy) such that triangle ABC has side-length AC = 4AB and ∠BAC is π clockwise.

Solution: It follows from the given conditions that zC − zA = 4e−iπ/6(zB − zA ), (1 mark)

(b) [2 marks] Let a = 1, b = 2 + i, c = −1 + 3i, d = 5 − 2i, e = x + iy , f = 2 + 7i be complex

= =⇒ =

Solving this equation for e, we obtain

e − (5 − 2i) (1 + i)(−2 − 3i) 1

−3 + 9i ( −2 + 3i)(−2 − 3i)

Hence a = and b = − . (1 mark)

2. (a) [2 marks] Find the equation (in the form R(z) = az + b) of the isometry R which is the rotation

around the point (1 + i) by an angle π anticlockwise.

Solution: The Rotation R is given by (1 mark)

R(z) = eiπ/3 (⃞z − (1 + i))⃞ + (1 + i); where eiπ/3 = cos π + i sin π = (⃞ 1 + √3i)⃞

Expanding and simplifying, we obtain:

R(z) = (⃞ 1 + √3i)⃞z + (1 + i) (⃞ 1 − √3i)⃞ = (⃞ 1 + √3i)⃞ z + (⃞ 1 + √3 + (1 − √3)i)⃞ . Hence a = (⃞ 1 + √3i)⃞ and b = (⃞ 1 + √3 + (1 − √3)i)⃞ . (1 mark)

(b) [2 marks] Let m be the line passing through the points 1 + i and 4 − 3i.

Solution: Since it is a reflection, write F(z) = a + b. Now the two points 1 + i and 4 − 3i on

the line m are fixed points; hence F(1 + i) = 1 + i and F(4 − 3i) = 4 − 3i.

This gives us the two equations: (1 mark) Now (2) − (1) gives

a(3+4i) = 3 − 4i so that a =3(3) −7 24i . Thus we have b = 1+i − a(1 −i) = (56+42i).

Hence F(z) =25(1) [⃞(−7 − 24i) + 14(4 + 3i)]⃞ . (1 mark)

(c) [2 marks] Find the equation of the isometry U which is the translation along the line m in part (b) by a length 2 (in the direction of increasing x), then reflecting across the line m.

Solution: A vector u along line m is u = (4 − 3i) − (1 + i) = 3 − 4i and ||u|| = √32 + 42 = 5.

F(z) + (3 − 4i). Hence

U(z) = (⃞(−7 − 24i) + 14(4 + 3i))⃞ + (3 − 4i) = (⃞(−7 − 24i) + 2(43 + i))⃞ . (1 mark)

(d) [2 marks] Let f(z) = −iz − 1 + i. Find all fixed points of f .

Solution: The fixed points of f are given by z = f(z) = −iz − 1 + i, hence (1 + i)z = −1 + i,

Because f is a direct isometry, it is a rotation with centre z = i. So it will have the form

f(z) = eiθ (z − i) + i for some rotation angle θ, and we infer that eiθ = −i, hence θ = − π . Thus,

3. [3 marks] Show that in an arbitrary quadrangle, the midpoints of the four sides are the vertices of a

parallelogram.

Solution: Let the vertices of the quadrangle A,B,C,D be represented by a, b, c, and d in C. The midpoints M1,M2 ,M3 and M4 of the sides, in order are given by

1 =

a + b 2 ,

2 =

b + c 2 ,

3 =

c + d

2 ,

m4 = 2 . (1 mark)

The displacements between midpoints are

b + c − a − b c − a

m3 − m2 = =

m4 − m3 = = m1 − m2 = −(m2 − m1 )

m1 − m4 =

a + b − d − a

= m2 − m3 = −(m3 − m1 ). (1 mark)

When two displacement vectors are non-zero real multiples of each other, then the corresponding segments are parallel. Hence M1M2M3M4 is a parallelogram. (1 mark)

Topology

4. [3 marks] Consider the following subsets of R2:

A = {⃞(a,0) : a ∈ R} ∪ {(0,b) : b ∈ R}⃞

B = {⃞(c,c) : c ∈ R} ∪ {(d, −d) : d ∈ R}⃞

C = {⃞(0,b) ∈ R2 : b ∈ R}⃞ .

(a) [2 marks] Show that A and B are homeomorphic by writing down an explicit homeomorphism

f : R2 → R2 such that f(A) = B .

[You should check that f is invertible, but you don’t need to check that f or f −1 are continuous]. Solution: The space A is given by the union of the x- and y-axes in the plane. The space B is

given by the union of the lines y = x and y = −x:

an angle of − π .

To obtain a precise formula for the function f, note that rotation about the origin is a linear

transformation which sends (1, 0) to (1, 1)/√2 and sends (0, 1) to ( −1, 1)/√2. It follows that f is

Alternatively f could rotate by any of ±4(1)π or ±4(3)π, and could scale by any positive factor.

(b) [1 mark] Give an argument which explains why A and C are not homeomorphic.

Solution: Remove the point (0, 0) from A; the resulting space has four connected components. But there is no point in C which leaves more than 3 components when removed.

5. [4 marks] Let S be an arbitrary set in the real line R.

• A point b ∈ R is called a boundary point of S if every ε-neighbourhood of b intersects both S and its complement R \ S .

• A point s ∈ S is called an interior point of S if there exists an ε-neighbourhood of s completely contained within S . The set of interior points of S is denoted int S .

• A point t ∈ S is called an isolated point of S if there exists a neighbourhood U of t such that U ∩ S = {t}.

• A point r ∈ R is called an accumulation point of S if every neighbourhood of r contains points of S distinct from r .

Find the boundary and interior of the following spaces: (i) (0 , 4) (ii) [ −1, 2] (iii) R (iv) ∅ .

Solution: (i) The interval (0, 4) has no isolated points and is its own interior; its boundary is {0} ∪ {4} and the set of accumulation points is the closed interval [0 , 4]. (1 mark)

(ii) The interval [ −1, 2] has no isolated points and is its own set of accumulation points; its interior is

the open interval ( −1, 2), with boundary being {−1, 2}. (1 mark)

(iii) The set R is its own interior and set of accumulation points; it has no boundary nor any isolated points. (1 mark)

(iv) The empty set ∅ has no points, so there is no interior, no boundary, nor any accumulation nor isolated points. (1 mark)

6. [8 marks] Join the maps in this atlas to get a PIE.

b ↓↓

d ↗

↓↓

↑↑ f

←

g ↗ ↖ e

↑↑

Taking into account identified edges, determine how many different vertices there are for the PIE, how many edges, the Euler characteristic, and classify the surface into the form mD + nT + rP, for some numbers m, n and r .

Solution: There are many ways to join the maps to get a PIE. Here’s one answer: move the polygons around a bit, flipping some over; make identifications (1 mark) and redraw (2 marks) :

e ↗