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COMPLEX NUMBERS Assumed Knowledge from Extension 1 Make sure you are familiar with the following before proceeding • Solving Quadratic Equations • Absolute Values • Preliminary Trigonometry • Locus • Circle Geometry Where are we up to? • Complex Numbers and Quadratic Equations • The Argand Diagram • Modulus-Argument Form • Vector Representation • Locus Problems • Roots of Unity 1. Introduction to complex numbers 1.1 Number Systems Throughout your progression in primary to high school mathematics you were introduced to different number systems which eventually led to the commonly used real number system: • Natural Numbers N = {0, 1, 2, 3, ....} - the problem with this system is that the operation of say 1 − 2 gives no solution in the natural number set, so we introduce the concept of negative numbers • Integers Z = {...., −2, −1, 0, 1, 2, ....} - the problem with this system is that the operation of say 1 ÷ 2 gives no solution in the integer set, so we introduce the concept of fractions o nm : m ∈ Z, n ∈ Z - the problem with this system is that some • Rational Numbers Q = n numbers cannot be written as a fraction (e.g. π), so we introduce the concept of irrational numbers • Real Numbers R = Q ∪ Q - the problem with this system is that the operation of say √ −1 has no solution in the real number set, so we introduce the concept of IMAGINARY NUMBERS • Complex Numbers C - is the union of real numbers and imaginary numbers. We will now look at this number system in detail Copyright - The information and materials on this worksheet are copyright NQ Corp Pty Ltd 1.2 Imaginary numbers These are based on the following definition of the imaginary unit i where: i2 = −1 In other words we are proposing there exists this unusual ’thing’ called i such that when it is squared we get negative one. All imaginary numbers exist in the form iy where y is a real number. In other words, any real number multiplied by i becomes an imaginary number e.g. 2i, 2i , 7i. They have no real physical meaning, as they are ’imaginary’. 1.3 Complex numbers These are the UNION (combination) of real numbers and imaginary numbers. Therefore, all complex numbers (commonly called z) can be written in the form z = x + iy where: x is the REAL PART of z or Re(z) y is the IMAGINARY PART z or Im(z) If Re(z) = 0, then we say that z is purely imaginary (real part ’does not exist’) If Im(z) = 0, then we say that z is purely real (imaginary part ’does not exist’) The complex number system has no ordering. In other words, we cannot say one complex number is greater than or less than another complex number. 1.4 Arithmetic involving complex numbers Addition, subtraction and multiplication on the set of complex numbers are calculated using the standard arithmetic procedures we are used to doing in real numbers and algebra. e.g. Addition: (1 + 2i) + (2 + 6i) = .............................. e.g. Subtraction: (1 + 2i) − (2 + 6i) = .............................. e.g. Multiplication: (1 + 2i)(2 + 6i) = ............................................................................................ Notice that the final answer is often written with their respective real and imaginary parts in the form x + iy. Division of complex numbers involves a bit more effort to get it in the form x + iy, so before we look at division we need to first look at complex conjugates. 1.5 Complex conjugates If z = x + iy, then we define the complex CONJUGATE of z as z = x − iy. This allows us to derive a very useful property. Note that: zz = (x + iy)(x − iy) = ............................ = ............................ Copyright - The information and materials on this worksheet are copyright NQ Corp Pty Ltd Note that this number is purely real (i.e. no imaginary parts) and is always non-negative. e.g. If z = 3 + 2i then find z, hence find the value of zz. 1.6 Division involving complex numbers The idea of complex conjugates is needed to divide two complex numbers and simplify them to the form x + iy. The first step is to make the denominator REAL (sometimes called ’realising’ the denominator). In order to do this we needed to multiply the numerator and denominator by the conjugate of the denominator, because this converts the denominator into a real number (as shown in the previous part) rather than a non-real number. It is very similar to the technique of rationalising the denominator. e.g. Simplify 3 + 2i in the form x + iy. 1 + 4i 3 + 2i 1 − 4i 3 + 2i = × 1 + 4i 1 + 4i 1 − 4i = ...................................................... = ...................................................... = ...................................................... BRIEF EXERCISES - express the following in the form x + iy 1. (3 + 2i) + (4 − 5i) 2. (2 + 3i) − (3 − 2i) Copyright - The information and materials on this worksheet are copyright NQ Corp Pty Ltd 3. (1 + i)(2 + i) 4. 3 − 4i 1 + 2i 2. Quadratic Equations Involving Complex Numbers 2.1 Equality of complex numbers Suppose z = a + ib and say ω = c + id. If z = ω then we can say that a = c and b = d. In other words Re(z) = Re(ω) (the real parts are equal) and Im(z) = Im(ω) (the imaginary parts are equal). e.g. If (2 + i)2 = x + iy, find the values of x and y. e.g. If 1+i = x + iy, find the values of x and y. 1−i Copyright - The information and materials on this worksheet are copyright NQ Corp Pty Ltd √ 3 1 i = cos θ + i sin θ for 0 < θ < π2 , find the value of θ. e.g. If + 2 2 (Note that we are using radian measures of angles rather than degrees. The conversion from degrees to radians uses π radians = 180 degrees. You will be shown the reason for this in more detail in the latter part of the Extension 1 course, but for now just accept the result.) 2.2 Square root of complex numbers √ Brief Exercise - Suppose a + ib = x + iy, find an expression for a and find another expression for b in terms of x and y. √ Suppose we want to find say 3 + 4i in the form x + iy (keep in mind that x and y are real numbers). Using the same method employed above (here a = 3 and b = 4), notice that we have a set of simultaneous equations to solve for x and y. Since one of the equations √ is quadratic, we expect TWO pairs of solutions which means there are TWO ways to write 3 + 4i in the form x + iy. We solve this problem as follows: √ 3 + 4i = x + iy ⇒ 3 + 4i = (x + iy)2 Copyright - The information and materials on this worksheet are copyright NQ Corp Pty Ltd BRIEF EXERCISES - express the following in the form x + iy √ 1. 24 + 10i Copyright - The information and materials on this worksheet are copyright NQ Corp Pty Ltd 2. √ 15 + 8i 3. √ −21 + 20i Copyright - The information and materials on this worksheet are copyright NQ Corp Pty Ltd 4. √ i 2.3 Solving quadratic equations with real coefficients Many quadratic equations you may have encountered may have had no real solution because you end up square rooting negative numbers. This does not necessarily mean that it is impossible to have a solution. It just means that the solutions that exist are not in the real number system. Using the knowledge of complex numbers, it is now possible to solve EVERY quadratic equation. In other words, whilst the solutions of a quadratic equation may not exist in the real number system R, they will always exist in the complex number system C. Copyright - The information and materials on this worksheet are copyright NQ Corp Pty Ltd e.g. Solve x2 + 2 = 0 if a) x ∈ R x2 = −2 gives no real solutions for x b) x ∈ C x2 = −2 √ x = ± −2 √ x=i± 2 Note that there are always two solutions in the complex number system. We can now solve equations which were otherwise unsolvable in the real number system. e.g. Solve x2 + 2x + 5 = 0 for x ∈ C e.g. Solve 2z 2 − 2z + 1 = 0 for z ∈ C Copyright - The information and materials on this worksheet are copyright NQ Corp Pty Ltd 2.4 Solving quadratic equations with non-real coefficients The complex number system also allows us to invent quadratic equations with complex coefficients. These are solvable using the same methods as solving any real quadratic equation. However, they often require a bit more effort to simplify to the form x + iy because we need to evaluate the square root of a complex number. e.g. Solve 3z 2 − 3z − i = 0 for z ∈ C p (−3)2 − 4(3)(−i) z= 2(3) √ 3 ± 9 + 12i = 6 3± √ The 9 + 12i needs to be simplified so that z can be written in the form x + iy. We employ the standard procedure for finding the square root of the complex number. Let √ 9 + 12i = x + iy ⇒ 9 + 12i = (x + iy)2 Copyright - The information and materials on this worksheet are copyright NQ Corp Pty Ltd BRIEF EXERCISES - solve the following quadratic equations expressing your solutions in the form x + iy 1. z 2 + 1 = 0 2. u2 − 2u + 3 = 0 3. 3y 2 + 3y + 1 = 0 4. x2 − (4 + i)x + 9 + 7i = 0 Copyright - The information and materials on this worksheet are copyright NQ Corp Pty Ltd 5. a2 − (1 − 4i)a − 5 + i = 0 Copyright - The information and materials on this worksheet are copyright NQ Corp Pty Ltd