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Published byEnrique Mould Modified over 5 years ago

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Complex Numbers If we wish to work with , we need to extend the set of real numbers Definitions i is a number such that i2 = -1 C is the set of numbers Z, of the form where a and b are real numbers. a is called the real part of Z and we write a = R(z) of a = Re(z) b is called the imaginary part of Z and we write b = i(z) or b = Im(z)

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Given and Addition is defined by: Multiplication is defined by: We may write a + bi or a + ib, whichever we find more convenient

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a) Given find (i) (ii) (iii) (i) (ii) (iii)

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b) Solve the equation Using the quadratic formula

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Page 90 Exercise 1 Questions 1, 2, 3, 6, 7, 8

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**Complex Conjugate When , then its complex conjugate is denoted by**

This is useful when we wish to carry out a division.

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a) Calculate

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b) Calculate Let where Then Equating parts we get: also

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Since

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Page 91 Ex 2 Questions 1(a), (b), (c), 2(c), (e) 3(a), (b), (f), 5(a), (b) TJ Exercise 2. TJ Exercise 1 - if needed.

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Argand Diagrams The complex number is represented on the plane by the point P(x,y). The plane is referred to as “The Complex Plane”, and diagrams of this sort are called Argand Diagrams. r y x p Any point on the x-axis represents a purely Real Number Any point on the y-axis represents a purely imaginary number

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**The size of the rotation is called the amplitude or argument of z.**

It is often denoted Arg z. This angle could be We refer to the value of Arg z which lies in the range -< as the principal argument. It is denoted arg z, lower case ‘a’. r y x p By simple trigonometry: This is referred to as the Polar form of z.

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**a) Find the modulus and argument of the complex number**

Since (3,4) lies in the first quadrant, n = 0 b) Find the modulus and argument of the complex number Since (-3,-4) lies in the third quadrant, n = -1

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(2,2) is in Q1

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Page 94 Exercise 3 Questions 3a, b, d, e, i 6a, b, f 7a, b, c

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**Loci-Set of points on the complex plane**

This is a circle, centre the origin radius 4 4 -4 y x (i) 4 -4 y x (ii)

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y x This is a circle centre (2, 0) radius 3 units.

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**This is a straight line through the origin gradient**

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**Page 96 Exercise 4 Questions 1a, b, d, f, j**

4a, c TJ Exercise 7

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**Polar Form and Multiplication**

Note arg(z1z2) lies in the range (-, ) and adjustments have to be made by adding or subtracting 2 as appropriate if Arg(z1z2) goes outside that range during the calculation.

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Note:

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**Now turn to page 96 Exercise 5 Questions 1 and 2.**

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**Let us now look at question 3 on page 99.**

This leads to the pattern:

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De Moivre’s Theorem

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a) Given find

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b) Given find Round your answer to the nearest integer

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**Page 101 Exercise 6 questions 1 to 3, 4g, h, i, j.**

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**Roots of a complex number**

It would appear that if then

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**The solutions are radians apart,**

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By De Moivre’s theorem, when finding the nth root of a complex number we are effectively dividing the argument by n. We should therefore study arguments in the range (-n, n) so that we have all the solutions in the range (-, ) after division. The position vectors of the solution will divide the circle of radius r, centre the origin, into n equal sectors.

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For k = 0 For k = 1

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For k = 2

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**Page 106 Exercise 7: Question 2 plus a selection from 1**

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Polynomials In 1799 Gauss proved that every polynomial equation with complex coefficients, f(z) = 0, where z C, has at least one root in the set of complex numbers. He later called this theorem the fundamental theorem of algebra. In this course we restrict ourselves to real coefficients but the fundamental theorem still applies since real numbers are also complex.

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We need to find z2, z3 and z4 And substitute them into the Original equation.

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Using Division Hence the complimentary real factor is Hence all four roots are:

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**Page 108 Exercise 8 Questions 2, 3, 4, 5 and 6**

Review on Page 110

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