For complex numbers z satisfying | z - 8 - 16 i | = 2 5 , find the least possible value of | z | and the greatest possible value of | z | .

Given | z - 8 - 16 i | = 2 5 Let the complex number z = x + i y , where x and y are real, the locus of the point M ( x , y ) that satisfies If z - z 1 = r , z - z 1 < r , z - z 1 ≤ r where r is a constant and z 1 is the fixed point a + b i where a and b are real. z - z 1 = r is the circumference of a circle with centre a , b and radius r . Now comparing the given | z - 8 - 16 i | = 2 5 with z - z 1 = r we get z 1 = 8 + 16 i Therefore, the centre is 8 , 16 ∴ | z - 8 - 16 i | = 2 5 is a circle with centre 8 , 16 and radius 2 5 . It is only the circle, not the points within the circle. The circle is drawn as a solid line as it is included. Draw a straight line starting from origin and passing through the centre of circle. Distance between two points, x 1 , y 1 and x 2 , y 2 is x 2 - x 1 2 + y 2 - y 1 2 . The distance between origin O ( 0 , 0 ) and centre C ( 8 , 16 ) is O C = 8 - 0 2 + 16 - 0 2 = 320 = 8 5 The least possible value of z is distance between O A . ⇒ O A = O C - A C ⇒ O A = 8 5 - 2 5 ⇒ O A = 6 5 . The greatest possible value of z is distance between O B ⇒ O B = O C + C B ⇒ O B = 8 5 + 2 5 ⇒ O B = 10 5 .

On a single Argand diagram, sketch the loci | z + 3 - 2 i | = 4 and arg ( z + 1 ) = π 4 . Hence determine the value of z that satisfies both loci, giving your answer in Cartesian form.

Given | z + 3 - 2 i | = 4 Let the complex number z = x + i y , where x and y are real, the locus of the point M ( x , y ) that satisfies If z - z 1 = r , where r is a constant and z 1 is the fixed point a + b i where a and b are real, then z - z 1 = r is the circumference of a circle with centre a , b and radius r . Now comparing the given | z + 3 - 2 i | = 4 with z - z 1 = r we get z 1 = - 3 + 2 i Therefore, the centre is - 3 , 2 ∴ | z + 3 - 2 i | = 4 is a circle with centre - 3 , 2 and radius 4 . The Cartesian equation is x + 3 2 + ( y - 2 ) 2 = 4 2 ⇒ x 2 + y 2 + 9 + 4 + 6 x - 4 y = 16 ⇒ x 2 + y 2 + 6 x - 4 y = 3 It is only the circle, not the points within the circle. The circle is drawn as a solid line as it is included. Given arg ( z + 1 ) = π 4 If a r g ( z - z 1 ) = α , where α is a fixed angle and z 1 is the fixed point a + b i , then this represents a half line from ( a , b ) with an angle α . a r g ( z - z 1 ) is the angle made between the direction parallel to the real axis and the direction of the vector which has point Q ( a , b ) and an unknown point P ( x , y ) . Points below Q ( a , b ) are not part of the locus as the angle made between these points and the relevant horizontal line is π - α , not α . ⇒ arg ( z + 1 ) = π 4 Comparing with a r g ( z - z 1 ) = α we get z - z 1 = z + 1 ⇒ z - z 1 = z - ( - 1 + 0 i ) ⇒ z 1 = - 1 + 0 i Therefore, the fixed point is - 1 , 0 Therefore, the locus is half-line from ( - 1 , 0 ) at an angle of π 4 radians. The equation of a line from a point ( x 1 , y 1 ) with slope m is y - y 1 = m x - x 1 Here x 1 , y 1 = - 1 , 0 and s l o p e ( m ) = tan π 4 = 1 ⇒ y - 0 = 1 × x + 1 ⇒ y = x + 1 From above information we can draw below Argand diagram The loci intersect when x 2 + y 2 + 6 x - 4 y = 3 and y = x + 1 intersects ⇒ x 2 + x + 1 2 + 6 x - 4 x + 1 = 3 ⇒ x 2 + x 2 + 1 + 2 x + 6 x - 4 x - 4 = 3 ⇒ 2 x 2 + 4 x = 6 ⇒ x 2 + 2 x - 3 = 0 ⇒ x 2 + 3 x - x - 3 = 0 ⇒ x x + 3 - 1 ( x + 3 ) = 0 ⇒ x - 1 x + 3 = 0 ⇒ x = 1 , - 3 x = - 3 does not satisfy because to satisfy both loci should intersect from graph we can say it is not intersecting at x = - 3 ⇒ y = 1 + 1 ⇒ y = 2 Hence, the complex number which satisfies the both is z = 1 + 2 i .

Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Complex Numbers, Exercise 8: EXERCISE 11E

Author:Sue Pemberton, Julianne Hughes & Julian Gilbey

Sue Pemberton Mathematics Solutions for Exercise - Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Complex Numbers, Exercise 8: EXERCISE 11E

Attempt the practice questions on Chapter 11: Complex Numbers, Exercise 8: EXERCISE 11E with hints and solutions to strengthen your understanding. Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book solutions are prepared by Experienced Embibe Experts.

Questions from Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Complex Numbers, Exercise 8: EXERCISE 11E with Hints & Solutions

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 11 - Complex Numbers>EXERCISE 11E>Q 5

Sketch the locus $| z - (3 - 6 i) | = 3$ on an Argand diagram. Write down the Cartesian equation of this locus.

HARD

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 11 - Complex Numbers>EXERCISE 11E>Q 6

Sketch the loci $\arg (z + 4 + 2 i) = \frac{5 π}{6}$ and $| z - 5 i | = 4$ on an Argand diagram. Determine whether or not there is a complex number, $z$ , that satisfies both loci.

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 11 - Complex Numbers>EXERCISE 11E>Q 7

For complex numbers $z$ satisfying $| z - 8 - 16 i | = 2 \sqrt{5},$ find the least possible value of $| z |$ and the greatest possible value of $| z | .$

HARD

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 11 - Complex Numbers>EXERCISE 11E>Q 8

By sketching the appropriate loci on an Argand diagram, find the value of $z$ that satisfies $| z | = 13$ and $\arg (z - 12) = \frac{π}{2} .$

HARD

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 11 - Complex Numbers>EXERCISE 11E>Q 9

On a single Argand diagram, sketch the loci $| z + 3 - 2 i | = 4$ and $\arg (z + 1) = \frac{π}{4} .$ Hence determine the value of $z$ that satisfies both loci, giving your answer in Cartesian form.

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 11 - Complex Numbers>EXERCISE 11E>Q 10

The complex number $z$ is represented by the point $P (x, y)$ on an Argand diagram.
It is given that the locus of $P$ is $| z - 5 - 5 i | = 5 .$
Write down the Cartesian equation of the locus of $P$ .

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 11 - Complex Numbers>EXERCISE 11E>Q 10

The complex number $z$ is represented by the point $P (x, y)$ on an Argand diagram.
It is given that the locus of $P$ is $| z - 5 - 5 i | = 5 .$
Sketch an Argand diagram to show the locus of $P$ .

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 11 - Complex Numbers>EXERCISE 11E>Q 10

The complex number $z$ is represented by the point $P (x, y)$ on an Argand diagram.
It is given that the locus of $P$ is $| z - 5 - 5 i | = 5 .$
Write down the greatest possible value of $\arg z$ and the least possible value of $\arg z .$

Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Complex Numbers, Exercise 8: EXERCISE 11E

Sue Pemberton Mathematics Solutions for Exercise - Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Complex Numbers, Exercise 8: EXERCISE 11E

Questions from Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Complex Numbers, Exercise 8: EXERCISE 11E with Hints & Solutions

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