m1001w02-qs

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The University of Sydney

MATH1001 Differential Calculus

(http://www.maths.usyd.edu.au/u/UG/JM/MATH1001/)

Semester1, 2012 Lecturer: H. Dullin, R. Marangell, D. Warren, Z. Zhang

Week 2 Tutorial

Objectives

By the end of Week 2 you should be able to

(a) interpret set notation and, in particular, the symbols ∪, ∩, \, ⊆, ⊇, ∈, ∈ and {. . . | . . . };

(b) recognise and use the following notation and terminology:

N = {0, 1, 2, . . . } (the set of all natural numbers),

Z = { . . . , −2, −1, 0, 1, 2, . . . } (the set of all integers),

Q = { a/b | a, b ∈ Z and b = 0 } (the set of all rational numbers),

R (the set of all real numbers),C = {a + ib | a, b ∈ R} (the set of all complex numbers);

(c) identify the set R with the real number line, recall the meanings of irrational number , the

absolute value of a real number, and the symbols <, ≤, >, ≥;

(d) recognise and use interval notation [a, b], [a, b), (a, b], (a, b), [a,∞), (a,∞), (−∞, a], (−∞, a)and (−∞,∞), where a, b ∈ R;

(e) apply the quadratic formula to solve equations of the form ax2 + bx + c = 0 to obtain either

real or complex solutions;

(f ) work with complex numbers written in Cartesian form a + bi, where a, b

∈ R and i2 =

−1,

and, in particular, be able to add, subtract, multiply and divide complex numbers in Cartesianform, and given a complex number z = a + bi, be able to write down the real part (Re z = a),

the imaginary part (Im z = b) and the complex conjugate ( z = a − bi) of z;

(g) represent the set C of complex numbers by points on the complex plane, including the real

and imaginary axes, and know the definition of the modulus of a complex number.

Preparatory questions to do before the tutorial

1. Let A = {1, 2, 3, 4} and B = {3, 4, 5, 6}. Find

(i) A ∪ B (ii) A ∩ B (iii) A \ B (iv) B \ A (v) ( A \ B) ∪ ( B \ A) .

Solution.

(i) The union of the sets A and B equals {1, 2, 3, 4, 5, 6} (consisting of all things that are

elements of A together with all things that are elements of B).

(ii) The intersection of the sets A and B equals {3, 4} (consisting of all elements that are

common to A and B).

(iii) The set consisting of the elements of A that are not in B is {1, 2}.

(iv) The set consisting of the elements of B that are not in A is {5, 6}.

(v) The union of the sets A \ B = {1, 2} and B \ A = {5, 6} is {1, 2, 5, 6}.

2. Let X = { n ∈ Z | n2 ≤ 5 }.

(i) Rewrite X as a list of numbers.

(ii) Decide which of the following statements are true and which are false:

(a) X ⊆ Z (b) X ⊇ Z (c) 5 ∈ X (d) −2 ∈ X .

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Solution.

(i) The set of integers between −√ 5 and

√ 5 (inclusive) is X = {−2, −1, 0, 1, 2}.

(ii) (a) True. All the elements of X are integers.

(b) False. It is not true that all integers are elements of X .

(c) False. It is not true that 52 ≤ 5; so 5 does not satisfy the condition for membership

of X .

(d) False. Since (−2)2 = 4 ≤ 5 the number −2 does satisfy the condition for

membership of X ; so −2 ∈ X .

3. If z = 2 − i and w = −4 + 3i, find

(i) z + w (ii) z − w (iii) | z| (iv) w (v) zw (vi) z

w.

Solution.

(i) z + w = (2 − i) + (−4 + 3i) = (2 − 4) + (−1 + 3)i = −2 + 2i.

(ii) z − w = (2 − i) − (−4 + 3i) = (2 + 4) + (−1 − 3)i = 6 − 4i.

(iii) |2 − i| = 22 + (−1)2 = √ 22 + 12 = √ 5.

(iv) w = −4 − 3i.

(v) zw = (2 − i)(−4 + 3i) = 2(−4 + 3i) − i(−4 + 3i) = (−8 + 6i) + (4i − 3i2) = −5 + 10i.

(vi) z

w=

2 − i

−4 + 3i=

(2 − i)(−4 − 3i)

(−4 + 3i)(−4 − 3i) =

−8 − 6i + 4i + 3i2

42 + 32 =

−11 − 2i

25= − 11

25 − 2

25i.

Questions to do in the tutorial class

4. Locate each of the following sets on the real number line and then express each as an interval

or as a union of intervals:

(i)

x ∈ R | 2 ≤ x ≤ 4

(ii)

x ∈ R | −1 < x ≤ 1 or x ≥ 5

(iii) [2, 5] ∩ (3, 6] (iv)

x ∈ R | | x − 1| > 2

.

Solution.

(i) This is the closed interval [2, 4].

0 1 2 3 4−1−2

(ii) This is the union of two disjoint intervals, (−1, 1] ∪ [5,∞).

0 1 2 3 4 5 6 7−1−2

(iii) This intersection of intervals can be written as a single interval, (3, 5].

0 1 2 3 4 5 6 7−1−2

(iv) Since | x − 1| can be interpreted as the distance on the number line between x and 1, the

set is made up of all numbers that are more than two units from 1 on the number line.

Hence

x ∈ R | | x − 1| > 2

= (−∞, −1) ∪ (3,∞).

0 1 2 3 4 5−1−2−3−4

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5. Express the following complex numbers in cartesian form:

(i) (2 + 3i) + (5 − 6i) (ii) (2 + 3i) − (5 − 6i)

(iii) (1 + i)(1 − i) (iv) 1 + i

1 − i

(v) (2 + 3i)(5 − 6i) (vi) 2 + 3i

5

−6i

(vii) 1i − 3i

1 − i (viii) i123 − 4i9 − 4i .

Solution.

(i) 7 − 3i.

(ii) −3 + 9i.

(iii) (1 + i)(1 − i) = 1 − i + i − i2 = 1 + 1 = 2.

(iv) 1 + i

1 − i=

(1 + i)(1 + i)

(1 − i)(1 + i) =

1 + 2i + i2

12 + 12 =

2i

2= i.

(v) (2 + 3i)(5−

6i) = 10−

12i + 15i−

18i2 = 28 + 3i.

(vi) 2 + 3i

5 − 6i=

(2 + 3i)(5 + 6i)

(5 − 6i)(5 + 6i) =

10 + 12i + 15i + 18i2

52 + 62 =

−8 + 27i

61= − 8

61 + 27

61i.

(vii) 1

i− 3i

1 − i= −i − 3i(1 + i)

12 + 12 = −i − −3 + 3i

2=

−2i + 3 − 3i

2= 3

2 − 5

2i.

(viii) Notice that i2 = −1 gives i3 = −i, and i4 = −i2 = 1. Now i5 is the same as i, and

i6 the same as i2, and i7 the same as i3, and so on. The sequence of powers of i goes

i, −1, −i, 1, i, −1, −i, 1, . . . , repeating with period 4. Indeed, since i4 = 1 it follows

that i4n = (i4)n = 1n = 1 for all n ∈ N. So i9 = i8i = i and i123 = i120i3 = i3 = −i,

since 8 and 120 are multiples of 4. Hence i123

−4i9

−4i =

−i

−4i

−4i =

−9i.

6. Solve the following equations over C:

(i) z2 + 3 z + 2 = 0 (ii) z2 + z + 1 = 0

(iii) 3 z2 − 4 z + 4 = 0 (iv) z4 = 16 .

Solution.

(i) z2 + 3 z + 2 = ( z + 1)( z + 2) = 0. Hence the solutions are z = −1 and z = −2.

(ii) Using the formula for the solutions of a quadratic equation, we get

z = −1

±√

1

−4

2 = −1

±√

−3

2 = −1

2 ±√

3

2 i .

(iii) By the quadratic formula again, we get

z = 4 ± √

16 − 48

6=

4 ± √ −32

6=

4 ± 4i√

2

6=

2

3± 2

√ 2

3i .

(iv) Since z4 = ( z2)2 it follows that z4 = 16 if and only if z2 = ±√ 16 = ±4. Now z2 = 4 if

and only if z = ±√ 4 = ±2, and z2 = −4 if and only if z = ±√ −4 = ±2i. So the set

{ z

| z4 = 16

} is

{2,

−2, 2i,

−2i

}.

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7. Sketch the following regions in the complex plane:

(i) { z ∈ C | | z| ≤ 3 } (ii) { z ∈ C | | z + i| > 2 }(iii) { z ∈ C | Re z < −1 } (iv) { z ∈ C | Im z ≥ −1 }(v) { z ∈ C | | z − i| ≤ | z − 1| } .

Solution.

(i) This is the set of all points in the complex plane whose distance from z = 0 is at most 3.It is the shaded set below:

i

2i

3i

−i

−2i

−3i

1 2 3 4 5

−1

−2

−3

−4

−5

(ii) This is the set of all points whose distance from −i exceeds 2. It is the shaded set

below (extending outwards infinitely in all directions) not including the broken line:

i

2i

−i

−2i

−3i

−4i

−5i

1 2 3 4 5−1−2−3−4−5

(iii) The shaded set, not including the broken line:

i

2i

3i

−i

−2i

−3i

1 2 3−1−2−3−4−5

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(iv) The shaded set:

i

2i

−i

−2i

1 2 3 4 5

−1

−2

−3

−4

−5

(v) The shaded set:

i

2i

3i

−i

−2i

−3i

1 2 3−1−2−3

In this example, the required set consists of all points in the complex plane that are closer

to z = i than to z = 1 or are at the same distance from z = i as from z = 1. Thus, if

z = x + iy, the solution of the inequality is the diagonal half-plane y

≥ x. Alternatively,

the inequality can be solved algebraically instead of graphically, as follows. Since z = x + iy gives

| z − i| = | x + ( y − 1)i| =

x2 + ( y − 1)2

and

| z − 1| = |( x − 1) + yi| =

( x − 1)2 + y2

the required inequality becomes

x2 + ( y − 1)2 ≤

( x − 1)2 + y2.

Since both sides are nonnegative, squaring both sides gives an equivalent condition:

x2 + ( y − 1)2 ≤ ( x − 1)2 + y2.

Expanding ( y − 1)2 and ( x − 1)2 shows that this is the same as

x2 + y2 − 2 y + 1 ≤ x2 − 2 x + 1 + y2,

which becomes −2 y ≤ −2 x after cancelling the terms that appear on both sides. Multi-

plying through by the negative number

−1/2 reverses the inequality, and so the solution

is y ≥ x (in agreement with the graphical method).

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8. (Suitable for group discussion.) In each case decide whether or not the statement is true.

Explain your answer.

(i) The square of an imaginary number is always real.

(ii) It does not make sense to write | z| > |w| when z and w are complex numbers because

the complex numbers are not ordered.

(iii) Real numbers cannot be graphed on the complex plane.

(iv) When a real number is divided by a complex number the answer can never be real.

Solution.

(i) True. The square of an imaginary number is always real and negative. For example,

(−7i)2 = −49, and (i/2)2 = −1/4. Every imaginary number has the form yi for some

y ∈R, and squaring we find that ( yi)2 = − y2 ∈ R.

(ii) False. Since the complex numbers are not ordered, a statement of the form z > w is

nonsense unless z and w are both real. For example, it makes no sense to ask which

z ∈ C satisfy z > 1 + i. However the modulus of a complex number is a real number.

So

| z

| and

|w

| are always real numbers, for all z, w

∈C, and so it is meaningful to write

| z| > |w|.(iii) False. All real numbers are complex numbers. The real numbers are the complex

numbers that lie on the real axis of the complex plane.

(iv) This is obviously false, since the complex numbers include the real numbers. A real

number divided by a nonzero complex number that happens to be real will give a real

number. Furthermore, if the real number 0 is divided by any nonzero complex number

then the answer is 0, which is real. (However, it is true that dividing a nonzero real

number by a complex number that is not real can never give a real number.)

Questions for further practice

9. On the complex plane, graph the solutions to the equations in Exercise 6. What pattern do

you notice in each pair of solutions? Explain why this pattern occurs.

Solution.

Below are the solutions to the equations in Question 6 plotted on the complex plane. As you

plot them, you should notice that when there are non-real solutions, these occur in pairs that

are the image of one another when reflected in the real axis. In the case of quadratics, it is

easy to see why this happens. Taking square roots in the quadratic formula gives terms of the

form a ± bi. Hence both solutions have the same real part but have imaginary parts of the

opposite sign but the same magnitude.

i

2i

3i

−i

−2i

−3i

1 2 3−1−2−3 x

iy

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10. Locate the following sets, which are given in interval notation, on the real number line.

Rewrite each set using {. . . | . . . } notation.

(i) [−7, −1) (ii) (−∞, −2] ∪ [2,∞) (iii) (2, 3) ∪ [5, 6] .

Solution.

(i) [−7, −1) = { x ∈R | −7 ≤ x < −1 }.

0 1 2−1−2−3−4−5−6−7−8

(ii) (−∞, −2] ∪ [2,∞) = { x ∈ R | x ≤ −2 or x ≥ 2 }.

0 1 2 3 4 5−1−2−3−4−5

(iii) (2, 3) ∪ [5, 6] = { x ∈R | 2 < x < 3 or 5 ≤ x ≤ 6 }.

0 1 2 3 4 5 6 7 8−1−2

11. Use a Venn diagram with three sets A, B and C to show the following:

(i) A ∪ B ∪ C (ii) A ∩ B ∩ C

(iii) ( A ∪ B) ∩ C (iv) A ∪ ( B ∩ C )

(v) ( A ∪ B ∪ C ) \ ( A ∩ B ∩ C )

(vi) A \ ( B ∪ C )

∪ B \ ( A ∪ C )

∪ C \ ( A ∪ B)

.

Solution.

(i) A B

C

(ii) A B

C

(iii) A B

C

(iv) A B

C

(v) A B

C

(vi) A B

C

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