complex numbers

Year 12

Complex numbers

Complex Numbers

1 The arithmetic form of a complex number1.1 The number i1. Denition: i is the number satisfying the condition i2 = 1 i can be considered as the rst complex number. 2. Powers of i: Evaluate i2 = i3 = i4 = i5 = Evaluate in : To nd the value of in where n is an integer, n = 4k + r, then 1 i n 4k+r r i =i =i 1 i rst use long division to rewrite if if if if r r r r =0 =1 =2 =3

3. The complex number i and quadratic equations with real coecients: Consider the equation ax2 + bx + c = 0 where a, b and c are real numbers. Let = b2 4ac, then: b If > 0 the equation has two real roots x = , 2a b If = 0 the equation has a double real root x = , 2a b i . If < 0 the equation has two complex roots x = 2a Petrus Ky College 1 Y12 Mathematics Extension 2

Year 12

Complex numbers

Solution:

Example: Solve the following equation x2 x + 1 = 0 = i 3.

1i 3 1+i 3 The equation has two roots x1 = and x2 = . 2 2

= 3 = 3i2 , therefore

1.2 Arithmetic form of a complex number zz = a + ib where a and b are real numbers. a is the real part and b is the imaginary part of z, denoted as a = Re(z) and b = Im (z) Notes: The arithmetic form of complex numbers is also called the a, b form. A real number is also a complex number where the imaginary part is 0. It is called purely real complex number. If the real part of a complex number is 0, it is called a purely imaginary complex number. Example: 3, 0 and -4 are purely real, and 4i, i are purely imaginary. The set of all complex numbers is denoted by C Equality of complex numbers: Given two complex numbers z1 = a + ib and z2 = x + iy, z1 = z2 if and only if a = x and b = y.

1.3 Addition, subtraction and multiplicationGiven two complex numbers z1 = a + ib and z2 = x + iy, then z1 + z2 = (a + x) + i(b + y) z1 z2 = (a x) + i(b y) z1 z2 = (ax by) + i(ay + bx) Note that to evaluate z1 z2 , rst expand and simplify the product (a + ib)(x + iy) = ax + i(ay + bx) + i2 by then use the denition i2 = 1 to rewrite the expression to clarify the real part and imaginary part as required. Example: Given that z1 = 2 3i, z2 = 3 + 4i and z3 = 1 + i. Find Petrus Ky College 2 Y12 Mathematics Extension 2

Year 12

Complex numbers

(a) z1 + z2 , Solution:

(b) z1 z2 ,

(c) z2 + z3 z1 ,

(d) z1 z2 ,

(e) z1 z2 z3 .

1.4 The conjugate of a complex number1. Denition: The conjugate z of a complex number z = a + ib is z = a ib. Examples: If z = 2 2i then z = 3 + 2i. w is purely real if and only if w = w 2. Properties: For any complex number z = a + ib: (a) z + z = 2Re(z) = 2a. (c) z z =2iIm (z) = 2bi (b) z = a2 + b2 . z (d) z = z = a + ib If = 3i then = 3i.

For any 2 complex numbers z1 = a + ib and z2 = x + iy: (e) z1 z2 = z1 z2 . (f) z1 z2 = z1 + z2 .

Examples Prove the properties (a), (b), (e) and (f). Solution:

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Y12 Mathematics Extension 2

Year 12

Complex numbers

1.5 DivisionSuppose that z1 = a + ib and z2 = x + iy, the quotient z = dened by z= Examples 1. Suppose that z = 2 3i nd (a) z 2 2 , and 1 . z (b) z 2 2 z1 is a complex number z2

z1 z2 z1 z2 1 z1 = = 2 = z1 z2 2 2 + y2 z2 z2 z2 x +y x

2. If z = 2 3i and = 3 + i, nd

1.6 Square root of a complex number1. Denition: suppose that z = a + ib, a square root that 2 = z. That means = z is a complex number such

z 2 = z

Note that each complex number z = 0 has two square roots. 2. Problem 1: Finding the square roots of a complex number z = a + ib. Solution: Step 1: Let = z = x + iy, then (1)

2 = z (x + iy)2 = a + ib (x2 y 2 ) + i2xy = a + ib Step 2: Equate the real parts and imaginary parts of (1): x2 y 2 = a 2xy = b

(2)

Step 3: Solve (2) simultaneously for real numbers x and y to get two solutions(x1 , y1 ) and (x2 , y2 ), then write down the roots 1 = x1 + iy1 and 2 = x2 + iy2 .

Solution:

Example: Find the square roots of 8 + 6i

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Year 12

Complex numbers

3. Problem 2: Solving quadratic equations with complex coecients Solve the equation az 2 + bz + c = 0 where a, b and c are complex numbers. Solution: Step 1: Evaluate the discriminant = b2 4ac. Step 2: Find the square roots 1 and 2 of . Step 3: The solutions to the equation are z1 = b + 1 2a and z2 = b + 2 2a

Solution:

Example: Solve the equation 2z 2 + (1 i)z + (1 i) = 0

Exercises 1. If z1 = 2 3i and z2 = 1 + 4i, evaluate 1 z1 3 3 2 2 (a) , (b) , (c) z1 + z2 , (d) z1 z2 z1 z2 2. (a) Prove that az = a for all complex number z and all real number a. z (b) Hence show that if a2 +b+c = 0, where a, b, c are real, then a 2 +b +c = 0. (This result shows that if is a complex root of a quadratic equation with real coecients, then is also a root of that equation.) 3. (a) z C such that Im(z) = 2 and z 2 is real. Find z. 4. z C such that (a) -16,

(b) z C such that 2Im(z) = Re(z) and z 2 4i is real. Find z. z is real. Show that z is purely imaginary. zi (c) i, (d) 4 + 3i,

5. Find the square roots of the following complex numbers: (b) 3i,

(e) 5 12i.

6. Solve the following quadratic equations: (a) x2 x + 1 = 0 (b) 2x2 4x + 3 = 0 (c) 4x2 4(1 + 2i)x (3 4i) = 0 (d) ix2 2(i + 1)x + 10 = 0

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Complex numbers

2 The modulus argument form of a complex number2.1 Geometrical representation of a complex number as a pointConsider z = a + ib.In a number plane (also called a Argand diagram) the point P (a, b) represents the complex number z. Example: Plot the points K, M, N, P, Q, R, S and T representing the complex numbers 2 3i, 2 + 3i, 1 + i, 1 2i, i -3, 2 and 2i respectively. y M 3

-2

O

x

2.1.1

The modulus of a complex number

1. Denition: If the point P represents z = a + ib on an Argand diagram, then the distance OP is the modulus of z, denoted by |z| or r. 2. Formula If z = a + ib then |z| = a2 + b2 or r = a2 + b2

Note that the complex number z = a + ib lies on the circle centred at the origin, and the radius is r = |z| = a2 + b2 . 2.1.2 The arguments of a complex number

1. Denition Suppose that P is the point representing the complex numberz = a + ib. Angles formed by the positive x-axis and the ray OP as shown on the diagram are called arguments of z, denoted by arg(z). The principle argument is the unit argument lying between and , denoted by Arg (z) arg (z)= angle formed by the positive x-axis and OP . = Arg (z) if is the argument satisfying < . arg (z)= Arg (z) + k2 for some integers k. Petrus Ky College 6 Y12 Mathematics Extension 2

Year 12

Complex numbers

2. Formula b If z = a + ib then Arg (z) = tan1 ( ) + k a where k = 0, 1,1 Note that the values of k can be chosen as follows: if P lies on the rst or on the fourth quadrant, 0 k= 1 if P lies on the second quadrant, 1 if P lies on the third quadrant, y

k=1 -1 O

k=0 x

-2 P k = 1 k=0b Arg (z) = tan1 ( a ) + k

3. Examples: (a) Find the modulus and argument of (i) i, (ii) i (iii) 4 (iv) 2. What is the argument of a purely real number? What is the argument of a purely imaginary complex number? (b) Find all the the arguments of z = 1 i 3 given that 2 arg(z) 2. What is the principle one? (c) Find the Arg (z) if (ii) z = 3i, (i) z = 3+i, (iii) z = 3i, (iv) z = 3+i.

(d) (i) If arg(z) = and |z| = r, nd arg() and ||. z z 1 1 (ii) If z = cos + i sin , nd arg and prove that z = . z z Solution:

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Complex numbers

2.2 The modulus argument form of a complex numberSuppose that r and are modulus and an argument of z = a + ib respectively. From the diagram opposite, the real part and imaginary part of z can be expressed in term of and r to get the modulusargument form of a complex number.

y

b r O a

P

x

a = r cos and b = r sin . Therefore z = r(cos + i sin ) 2.2.1 Multiplication, division and powers in modulus argument form

Suppose that z = r(cos + i sin ), z1 = r1 (cos 1 + i sin 1 ) and z2 = r2 (cos 2 + i sin 2 ): z1 If z = z1 z2 , we nd the modulus argument form of z1 z2 and . z2 1. The conjugate and the reciprocal of z: z = r cos() + i sin() = r cos() i sin() and 1 1 1 = cos() + i sin() = cos() i sin() z r r Note that if |z| = 1 then z = 1 . z

2. Product z of z1 and z2 : Let r and be the modulus and argument of the product z = z1 z2 respectively, then arg(z) = = 1 + 2 , r = r1 r2 that is z = r1 r2 cos(1 + 2 ) + i sin(1 + 2 ) 3. The quotient z1 . z2 z1 is z2

The modulus argument form of the quotient

r1 z1 cos(1 2 ) + i sin(1 2 ) = z2 r2 That is |z1 | z1 z1 = and arg = arg(z1 ) arg(z2 ) z2 |z2 | z2 Petrus Ky College 8 Y12 Mathematics Extension 2

Year 12

Complex numbers

4. Powers of a complex number. De Moivres theorem: (cos + i sin )n = cos n + i sin n That is If z = cos + i sin then z n = cos n + i sin n. The power z n of z = r(cos + i sin ) If z = r(cos + i sin ) then z = r n (cos n + i sin n) That is z n = |z|n and arg(z n ) = n arg(z) Examples: 1. Express z1 = 2 + i and z2 = 4 + 4i in modulus-argument form. Hence nd the modulus-argument form of 2+i . (a) ( 2 + i)(4 + 4i), (b) 4 + 4i 2. If r and are modulus and argument of z respectively, nd, in terms of r and theta the modulus and one argument of 1 (a) z 2 , (b) , (c) iz. z 3. Show that z n = ()n . Hence nd ( 3 + i)8 + ( 3 i)8 . z Solution:

2.2.2

Roots of a complex number

1. Denition: w is an n-th root of z if wn = z. The n-th roots of z are denoted by n z. Examples

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Complex numbers

i is a square root of 1 because i2 = 1. (what is the second square root of i?) 8i is a square root of 64 2. Formula If z = r(cos + i sin ), then + k2 + k2 n + i sin z = n r cos n n where k = 0, 1, 2, (n 1)

In general, w1 , w2 , ...wn are used to denote the n-th roots of z. Note that using the values of k in the above formula we cannot get principle argument of some w1 , w2 , ...wn . Values of k can be expressed as k = 0, 1, 2, ... the sequence stops when we get n consecutive integers (negative and positive). These values provide the principle argument of all w1 , w2 , ...wn . 3. Properties: Each complex number has exactly n dierent n-th roots w1 , w2 , ...wn . The moduli of all w1 , w2 , ...wn are equal. The arguments of w1 , w2 , ...wn form an arithmetic progression. If W1 , W2 , ..., Wn are points representing w1 , w2 , ..., wn on an Argand diagram, then W1 , W2 , ..., Wn are vertices of a regular polygon inscribed in the circle centred at the origin and having radius of n r. Examples: 1. Find the complex 5th roots of unity and plot them in an Argand diagram. 2. Find the square roots of z = 2 2 + i2 2. (Note that the word unity means 1.) 3. If z is one of the three cube roots of 1, nd the two possible values of the expression z 2 z + 1. y Solution: i

-1

O

1

x

i Petrus Ky College 10 Y12 Mathematics Extension 2

Year 12

Complex numbers

Exercises 7. If z1 = 4 cos + i sin 3 3 and arguments of3 (a) z1 ,

and z2 = 2 cos3 z1 . z2

, write down the moduli + i sin 6 6

(b)

1 , z2

(c)

8. (a) Write (1 +

1 3i) in modulus-argument form.

(b) State the modulus and argument of 1 + i. Hence write (1 + i)18 in the form a + ib. 9. Write 3 + i and 3 i in modulus argument form. Hence write down ( 3 + i)10 + ( 3 i)10 in the form a + ib. 10. Find the modulus of 7i . 3 4i

11. Use the Moivre theorem to express cos(4) and sin(4) in terms of cos and sin , show that 4 tan 4 tan3 tan(4) = 1 6 tan2 + tan4 12. Find (a) the square roots of 3+i (b) the cube roots of 2 2i.

13. Solve the equation z 5 = 1. By grouping the roots in complex conjugate pairs, show that z 5 + 1 = (z + 1)(z 2 2z cos 3 + 1)(z 2 2z cos + 1) 5 5

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Complex numbers

3 Geometrical representation of a complex number as a vector3.1 Vectors 1. A vector AB can be considered as a displacement from a point A to a point B. A is called the tail and B is called the head. The direction from the tail to the head is direction of AB. Note that with two dierent points A and B we have two vectors AB and BA, and they are called opposite vectors denoted by AB = BA. 2. The magnitude of AB is the length of AB. 3. Two vectors AB and CD are parallel if AB CD. AB . AB and CD are parallel AB = kCD where |k| = CD The vectors have the same direction if k > 0, opppsite direction if k < 0. 4. Two vectors AB and CD are equal if: A they are parallel, they have the same direction, their magnitudes are equal. C Q B D P p q

= k AB = CD = P Q, p q Note that two equal vectors can replace each other, therefore we can use the notation , q to denote vectors. q 5. Addition and subtraction of vectors

q p q Figure 1 + p q p q Figure 2 + p q

p q q

p

p q q

Figure 3

There are two ways to of forming the vector sum + . p q q Method 1 (Head to tail) The vector is translated (moved parallel to itself) is at the head of . The vector sum + is constructed until the tail of q p p q to the head of as shown on Figure 1. from the tail of p q Petrus Ky College 12 Y12 Mathematics Extension 2

Year 12

Complex numbers Method 2 (Parallelogram) The vector is translated until the tail of . A q p and . The vector sum parallelogram is constructed with adjacent sides p q + is drawn along the diagonal from the common tail of and to the p q p q opposite vertex as shown on Figure 2.

To perform the vector subtraction we have two ways: p q

adding the vectors and , where is a vector with the same magnitude p q q as but in the opposite direction. q The vector is translated until the tail of is at the head of . The vector q q p is drawn from the head of to the head of as shown subtraction p q q q on 3.

3.2 Complex numbers and vectors3.2.1 Representing a complex number as a vector 1. The principle vector representing a complex number z = a + ib is the vector OP where O is the origin and P (a, b) is the point representing z. With modulus argument form z = r(cos + i sin , the magnitude of OP is equal to the modulus of z and the angle formed by the positive x-axis and OP is the principle argument of z. 2. In general, z = a + ib = r(cos + i sin is represented by any vector provided p = OP . For convenience, it is denoted by . In this case, the magnitude that p z | | is equal to the modulus of z and Arg(z) is equal to the angle formed by and z z parallel to the positive x-axis as shown on the diagram. the ray from the tail of z y y z1 + z2 z1 P (1, 1) z z1 z2 1 2 O z2 x 1 z x O Figure 4: the vectors represent z = 1 + i 3. Examples (a) If z = 2 cos( ) + i sin( ) , draw the vectors representing 6 6 1 (4i) 2z. (i)z, (ii) z , (iii) , z 1 (b) If z = 3 i, draw the vectors representing (i) z, (ii) z 3 , (iii) 2 . z Petrus Ky College 13 Y12 Mathematics Extension 2 Figure 5

Year 12

Complex numbers

3.2.2

Operations with complex numbers represented as vectors

1. Addition and subtraction z = z1 z2 The sum z1 + z2 and the vector subtraction z1 z2 are drawn on Figure 5.

Note that the point representing the complex number z1 + z2 (or z1 z2 ) must be the head of the principle vector representing that complex number. Examples (a) If z1 = 3 2i and z2 = 1 + 4i. Show on an Argand diagram vector OP , OQ, OC representing z1 , z2 and z1 + z2 respectively. Name a vector which represents z1 z2 and plot the point representing z1 z2 (b) Let A(1, 1), B(5, 2), P (1, 2 3) and Q(0, 3) be points in an Argand dia gram, nd in form a + ib the complex number represented by AB, and in modulus-argument form the complex number represented by P Q.

2. The product z = z1 z2 Let z1 = r1 (cos 1 + i sin 1 ) and z2 = r2 (cos 2 + i sin 2 ) be represented by OA and OB respectively. The vector representing z1 z2 is drawn as follows rotate the ray OA by an angle 2 , counter-clockwise if 2 > 0 and clock-wise if 2 < 0. On the new ray (OA rotated) choose a point M such that OM = r1 r2 . The vector OM represents z1 z2 .(See gure 6) y M A 2 O Figure 6 3. The quotient z = Since z1 z2 B x O Figure 7 OM = z1 z2 y A N ON = Bz1 z2

x

z1 1 1 1 can be considered as z = z1 , and = cos(2 ) + i sin(2 ) , z2 z2 z2 r2 1 z1 is performed as the product of z1 : the quotient z2 z2 rotate the ray OA by an angle (2 ), counter-clockwise if 2 > 0 and clockwise if 2 < 0.

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Complex numbers r1 On the new ray (OA rotated) choose a point N such that OM = . The r2 z1 vector ON represents .(See Figure 7) z2

Example: (a) If z1 = (i) z1 z2 Solution: 3 i and z2 = 1 + i, sketch the vectors representing z1 (ii) (iii) (z2 )3 , z2 4

(4i)

z2 .

4. The triangle inequalities On the opposite diagram (Figure 8), OA, OB rep resent z1 , z2 . OACB is a parallelogram and OC represents z1 + z2 . OC OA + OC, with equality if and only if O, A, C are collinear. OC OA + OC, since opposite sides of OACB are equal, |z1 + z2 | |z1 | + |z2 | with equality if and only if OA, OB, OC are parallel. Perform similarly we have the second triangle inequality. The triangle inequalities: |z1 + z2 | |z1 z2 | |z1 | + |z2 | with equality if and only if z1 = kz2 . |z1 | |z2 | with equality if and only if z1 = kz2 . O y C

B z2 z1 + z2 A z1

x

Figure 8

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Year 12

Complex numbers

Exercises 14. On an Argand diagram the points A and B represent the complex numbers z1 = i 1 and z2 = (1 + i) respectively. Sketch a vector representing z1 + z2 and z1 z2 . 2 What is the Argument of z1 + z2 . 15. Use the vector representation of z1 and z2 to prove that z1 + z2 is imaginary. z1 z2 and arg(z1 z2 ) arg(z1 + z2 ) = , then (b) If 0 < Arg(z2 ) < Arg(z1 ) < 2 2 |z1 | = |z2 |. (a) If |z1 | = |z2 |, then 16. Sketch OP and OQ representing z and iz. Show that OP Q is a right triangle. 17. On an Argand diagram the points P and Q represent the numbers z1 and z2 respectively. If OP Q is an equilateral triangle. Show that (z1 )2 + (z2 )2 = z1 z2 . 18. Show that |z1 | |z2 | |z1 z2 |. State the condition for equality to hold. z1 z2

19. If |z1 + z2 | = |z1 z2 |, nd the possible values of arg

20. If z1 = 3 + 4i and |z2 | = 13, nd the greatest value of |z1 + z2 |. When |z1 + z2 | reaches its greatest value, express z2 in the form a + ib.

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Year 12

Complex numbers

4 Curves and regions in the Argand diagramSince a complex z = x + iy is represented by a point P (x, y) on an Argand diagram, when z varies with some constraints, the point P moves on a xed curve (or a straight line) or on a xed region. Finding the curve (or line) or region is nding the locus of P . Note that, with help from geometric locus, the curve can be found easily.

4.1 Problems involving the real part x and the imaginary part y4.1.1 Some basic problems:

1. Problem 1: The locus is the perpendicular bisector of two given points If |z z1 | = |z z2 | where z1 and z2 are given (xed complex numbers), nd the locus of P representing z. Explanation Way 1: Let A and B be the points representing z1 and z2 respectively. The vectors AP and BP represent the complex numbers z z1 and z z2 respectively. Therefore |z z1 | = AP and |z z2 | = P B the triangle AP B is an isosceles. Hence P moves on the perpendicular bisector of AB. We can nd the equation of by noting that the line passes through the midpoint M of AB and is perpendicular to AB. Way 2: Step 1: Let A(a1 , b1 ) and B(a2 , b2 ) be the points representing z1 = (a1 + ib1 ) and z2 = a2 + ib2 respectively. A and B are xed. Evaluate z z1 and z z2 z z1 = x a1 + i(y b1 ) and z z2 = x a2 + i(y b2 ). Step 2: Evaluate |z z1 |2 = (x a1 )2 + (y b1 )2 and |z z2 |2 = (x a2 )2 + (y b2 )2 Step 3: Equate the two right hand sides of the previous equations, then expand and simply we get an equation of a straight line which is the locus of P . Note that this line is the perpendicular bisector of the interval AB. Example Find and sketch the locus of the point P (x, y) representing the complex number z = x + iy, given that |z 5 + i| = |z + 1 3i|. Solution: Way 1: y

O Way 2: Petrus Ky College 17

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Year 12

Complex numbers

2. Problem 2: The locus is a circle or regions dened by a circle Given z0 and a real numer r > 0, nd the locus of P (x, y) representing the complex number z satisfying (a) |z z0 | = r. (c) |z z0 | r.

(b) |z z0 | < r. Explanation: (a)

Way 1: Let I be the point representing z0 . Since the vector IP represents z z0 , |z z0 | = IP . On the other hand |z z0 | = r, therefore IP = r. This implies the point P moves on the circle C centred at I and its radius is r. The equation of the circle is (x a)2 + (y b)2 = r 2 . Way 2: Let I(a, b) be the point representing the complex z0 = a + ib. Step 1: Evaluate |z z0 |2 = (x a)2 + (y b)2 . Step 2: From data |z z0 |2 = r 2 , hence (x a)2 + (y b)2 = r 2 . This is the equation of the circle C, centred at I and the radius is r.

(c) The locus is the region outside the circle C, including the circle. Example (a) Find and sketch the locus of the point P (x, y) representing the complex num ber z = x + iy, given that |z 2 2i| = 2.

(b) The locus is the region inside the circle C, excluding the circle.

(b) Find the maximum and minimum values of |z| and the values of z for which these extremes are attained. (c) Find the range of possible values of Argz. (d) Sketch the region satisfying 2 |z 2 2i| < 4. Solution: (a) Way 1: y

O Way 2:

x

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Complex numbers

4.1.2

General problem:

If a complex z satises some conditions, nd the locus of the point P representing z. General solution: Let z = x + iy and P (x, y) be the point representing z. Use the constraint to nd: an equation (relation) of x and y. This is the locus of P . Or an inequation of x and y: P moves on the region dened by the inequation. Examples: 1. Sketch the points z satisfying Re(z 2 + 2i) = 4. Solution: y P z 2 + 2i = (x 2) + 2i = Re(z 2 + 2i) = (x 2) Therefore (x 2) = 4 This implies x = 6 = the point P moves on the vertical line : x = 6 Inversely, if P lies on then P represents a complex number z with Re(z) = 6, therefore Re(z z0 ) = 4 where z0 = 2 2i represented by A(2, 2) 2. On an Argand diagram, sketch the region dened by 6 Im(z) 0). Solution: O z0 A(2, 2) Im z(2 3i) < 12 and z 6 z z0 x Let z = x + iy Evaluate

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Year 12 Evaluate Im z(2 3i) : Let z = x + iy then z(23i) = (x+iy)(23i) = (2x+3y)+i(2y3x) Im z(2 3i) = 2y 3x The condition 6 Im z(2 3i) < 12 becomes 6 2y 3x < 12 This means the point P (x, y) representing z lies on the region between two parallel lines 1 :3x 2y + 6 = 0 and 2 :3x 2y + 12 = 0. The second condition Im(z) 0 shows that the point P is above the x-axis. The region satisfying both conditions is the the region above the x-axis, between the two line 1 and 2 .

Complex numbers

y

O

x

4.2 Problem involving the argument 4.2.1 The locus is a ray or a region between two rays

1. Problem 1 (a) Find the locus of z such that arg(z) = 0 . (b) Sketch the region on an Argand diagram dened by 1 Explanation: (a) Since arg(z) = , the point P representing z lies on the ray Ot, formed with the positive x-axis an angle of 0 , excluding the origin O. (b) The region is between the two rays Ot1 and Ot2 formed with the positive x-axis angles 1 and 2 respectively. 2. Problem 2: Given z0 = a + ib and angles , and , (a) Find the locus of z such that arg(z z0 ) = Petrus Ky College 20 Y12 Mathematics Extension 2 O x y arg(z) < 2

Year 12

Complex numbers

(b) Sketch the region of z satisfying < arg(z z0 ) Explanation:

(a) Let A(a, b) be the point representing z0 . From A draw the ray Ax parallel to the positive x-axis, then draw the ray At formed with Ax an angle of . The locus is the ray At excluding A as shown on the left diagram below. Indeed, any point P on the ray At represents a complex number z = x + iy. Since the vectors OP and OA represent z and z0 respectively, the vector AP represents z z0 . On the other hand Arg(z z0 ) = x At = (co-interior angles). y y t t t

A O

x x

A

x O x

(b) The region is between two rays At and At which formed with the ray Ax (parallel to the xaxis) angles and respectively, excluding At and inncluding Ax as shown on the right diagram above. Examples 1. Sketch the locus of z given that arg(z) = . 6 2. Sketch the locus of z if arg(z + 1 i) = . 6 3. Shade the region satisfying the inequalities 6 Solution

arg(z 1 + i)

complex numbers

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