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FP1: Chapter 1 Complex Numbers Dr J Frost (jfrost@tiffin.kingston.sch.uk) Last modified: 2 nd January 2014

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i i captain! Heres something that someone has almost certainly spoiled for you already… ?

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Complex versus imaginary numbers ? ? Imaginary partReal part ??

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Putting complex numbers in the preferred form We tend to ensure real and imaginary components are grouped together. (a, b and c are real constants) a + 3i – 4 + bi = a – 4 + (3 + b)i 2a – 3bi + 3 – 6ci = 2a+3 – 3(b + 2c)i Just like wed write 6 instead of 6, the i appears after any real constants, so we might write 5ki or i. An exception is when we involve a function. e.g. i sin and i 3 Why? This avoids ambiguity over whether the function is being applied to the i. Convention 1 ? ? ? Convention 2 In the same way that we often initially use x and y as real-valued variables, we often use z first to represent complex values, then w.

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Why complex numbers? Complex numbers were originally introduced by the Italian mathematician Cardano in the 1500s to allow him to represent the roots of polynomials which werent real. They can also be used to represent outputs of functions for inputs not in the usual valid domain, e.g. Logs of negative numbers, or even the factorial of negative numbers! Some other major applications of Complex Numbers: Fractals1 A Mandelbrot Set is the most popular fractal. For each possible complex number c, we see if z n+1 = z n 2 + c is not divergent (using z 0 = 0), leading to the diagram on the right. Coloured diagrams can be obtained by seeing how quickly divergence occurs for each complex c (if divergent). This is called an Argand Diagram. Well use it later. Analytic Number Theory2 Number Theory is the study of integers. Analytic Number Theory treats integers as reals/complex numbers to use other (analytic) methods to study them. For example, the Riemann Zeta Function allows complex numbers as inputs, and is closely related to the distribution of prime numbers. Physics and Engineering3 Used in Signal Analysis, Quantum Mechanics, Fluid Dynamics, Relativity, Control Theory...

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Manipulation and application Simplify (8 + i)(3 – 2i) = 24 – 16i + 3i – 2i 2 = 24 – 16i + 3i + 2 = 26 – 13i Heres a flavour of some of things youll be initially expected to do... ? ? ? ? ?

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Multiplying Example:Let z 1 = 1 + 3i and z 2 = 5 – 2i, find z 1 z 2. (1 + 3i)(5 – 2i) = 5 – 2i + 5i – 6i 2 = 5 – 2i + 5i + 6 = 11 + 3i Calculate z 2 for the following z. The quick way to think about this is that i 2 reverses the sign. a)z = 1 + i z 2 = 2i b)z = 1 – i z 2 = -2i c)z = 3 + 2i z 2 = 5 + 12i d)z = 7 – 4iz 2 = 33 – 56i e)z = -3 + 3iz 2 = -18i f)z = a + biz 2 = a2 – b 2 + 2abi Calculate z 1 z 2. a)z 1 = 1 + i z 2 = 1 – iz 1 z 2 = 2 b)z 1 = 2 + i z 2 = 2 – 2iz 1 z 2 = 6 – 2i c)z 1 = 3 + 2i z 2 = 4 – 3iz 1 z 2 = 18 – i d)z 1 = -3 + i z 2 = 5 + 7iz 1 z 2 = -22 – 16i e)z 1 = 7 – 3i z 2 = 3 – 7i z 1 z 2 = -58i f)z 1 = -1 – i z 2 = 9 + 8iz 1 z 2 = -1 – 17i 2 3 4 By using a Binomial Expansion, determine (1 + i) 5. = 1 + 5i + 10i 2 +10i 3 + 5i 4 + i 5 = 1 + 5i – 10 – 10i + 5+ i = -4 – 4i What is the value of: a)i 100 = (i 4 ) 25 = 1 25 = 1 b)i 2003 = i 2000 x i 3 = 1 x i 3 = -i 5 Evaluate the following: a) (-16)= 4i b) (-25)= 5i c) (-3)= i 3 d) (-7)= i 7 e) (-8)=2i 2 1 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?

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Complex Conjugates The same trick works with complex numbers. If z = x + yi, then we define z* = x – yi. z* is known as the complex conjugate of z. z z* = x 2 + y 2 z + z* = 2x which are both real. The fact the first is real will help us with dividing complex numbers. Well see the significance of the second later. ? ? ? ?

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Quickfire Questions Given z, determine z z*. z = 3 + 2iz z* = 13 z = 3 – 2iz z* = 13 z = 5 + 4iz z* = 41 z = 1 + iz z* = 2 z = 4 – 2i z z* = 20 ? ? ? ? ? 1 2 3 4 5

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Dividing __26__ 2 + 3i (2 – 3i) ( )(2 – 3i) = 52 – 78i 13 =4 – 6i Click to Brosolve

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Exercises 1 - i 1 + i Put all the following in the form a + bi. _10_ 3 + i _1_ 1 + i 8 – 4i 1 – 3i = 3 - i = -i Note that mark schemes permit the single fraction. 1 - i 2 = 10 + 3i 1 + 2i 16 – 17i 5 = _i_ 1 – i -1 + i 2 = 14 – 5i 3 – 2i = 4 + i = 2 + 2i a + i a – i a 2 – 1 + 2ai a 2 + 1 = ? ? ? ? ? ? ? ? Edexcel June 2013 (Retracted) Given that (2 + i)(z + 3i) = 10 – 5i, find z, giving your answer in the form a + bi. z = 3 – 7i (either by dividing 10 – 5i by 2 + i and subtracting 3i, or replacing z with a + bi before expanding and comparing real and imaginary parts). 1 2 3 4 5 6 7 8 9 ?

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Argand Diagrams Argand Diagrams are a well of geometrically representing complex numbers. z = x + yi The x-axis is the real component. The y-axis is the imaginary component. Im[z] Re[z] -3 -2 -1 1 2 3 3 2 1 -2 -3 1 + i2 – 3i-1 + 2i-3 – i Click to move.

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Argument and Modulus In FP2, youll encounter something called polar coordinates. This is an alternative way of representing coordinate, which instead of using the (x,y) position (known as a Cartesian coordinate), uses the distance from the origin and the angle. Im[z] Re[z] -3 -2 -1 1 2 3 3 2 1 -2 -3 z = 2 + 3i Distance from origin: (Dont write anything down yet!) (2 2 + 3 2 ) = 13 This is known as the modulus of z, and we write |2 + 3i| = 13 Angle: (anticlockwise from the real axis) Using trigonometry, we have (in radians): tan -1 (3/2) = 0.983 This is known as the argument of z, and we write: arg(2 + 3i) = 0.983 ? ?

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Argument and Modulus Im[z] Re[z] arg z |z|

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Check your Understanding z|z|arg(z) 110 -i1 - /2 -1 + i 2 3 /4 -5 – 2i 29 -2.761 ?? ?? ?? ??

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Exercises z|z|arg(z) 1 i1 /2 1 + i 2 p/4 1 + 2i 5 1.107 1 – 2i 5 -1.107 -1 + 2i 5 2.034 -1 – 2i 5 -2.034 3 + 4i50.927 -5 + 12i131.966 1 – i 3 7 - /3 Give exact answers where possible, otherwise to 3dp. z|z|arg(z) - 3 - i 7 -5 /6 3 + i 10 0.322 2 – 5i 29 -1.190 -4 + 3i52.498 -1 – 4i 17 -1.816 Given that arg(3 + a + 4i) = p/3 and that a is real, determine a. 4/(3 + a) = 3. Thus a = (4/ 3) – 3 Give that arg(5 + i + ai) = p/4 and that a is real, determine a. (a + 1)/5 = 1. So a = 4 Given that arg(santa + 3 + 2i) = /2 and that santa is real, determine santa. santa = -3 12 3 4 5 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?

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Argument-modulus form Im[z] Re[z] arg z |z| Suppose that r = |z| and = arg(z). How could we express z in terms of r and o? z = x + yi = r cos + r i sin = r(cos + i sin ) This is known as the modulus- argument form (or the polar form) of z.

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z|z|arg(z)Mod-arg form 1 z = cos + i sin i1 /2z = cos( /2) + i sin( /2) 1 + i 2 p/4 z = 2(cos( /4) + i sin( /4)) 1 + 2i 5 1.107 z = 5(cos 1.107 + i sin 1.107) 1 – 2i 5 -1.107 z = 5(cos(-1.107) + i sin(-1.107)) -1 + 2i 5 2.034 z = 5(cos 2.034 + i sin 2.034) -1 – 2i 5 -2.034 z = 5(cos(-2.034) + i sin(-2.034)) 3 + 4i50.927z = 5(cos 0.927 + i sin 0.927) -5 + 12i131.966z = 13(cos 1.966+ i sin 1.966) 1 – i 3 7 - /3 z = 7(cos(- /3) + i sin(- /3)) 1 Question 1 from earlier... Argument-modulus form ? ? ? ? ? ? ? ? ? ?

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|z 1 z 2 | and arg(z 1 z 2 ) ?

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Comparing coefficients ? ? Q Q

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Exercise 1G ? 1 4 7 9 10 11 ? ? ? ? ?

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Roots of polynomials Any cubic (with coefficient of x 3 of 1) can be expressed as: y = (x – α)(x – )(x – ) where α, and are the roots. However, these 3 roots may not necessarily all be real... α α All 3 roots are real.1 real root, 2 complex roots. Are there any other possibilities? No: since cubics have a range of – to +, it must cross the x axis. And it cant cross an even number of times, otherwise the cubic would start and end in the same vertical direction. ? α = 3 real roots (one repeated) ??

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Roots of polynomials Classic Question: x = 2 is one of the roots of the polynomial x 3 – x – 6 Find the other two roots. Classic Question: x = 2 is one of the roots of the polynomial x 3 – x – 6 Find the other two roots. Use polynomial division (as per C2) to divide x 3 – x – 6 by (x – 2). This gives x 2 + 2x + 3. Using the quadratic formula, we obtain the roots: x = -1 i 2 Your Go: x = -1/2 is one of the roots of the polynomial 2x 3 – 5x 2 + 5x + 4. Find the other two roots. Your Go: x = -1/2 is one of the roots of the polynomial 2x 3 – 5x 2 + 5x + 4. Find the other two roots. We could divide by (x + ½), but it would be cleaner to divide by (2x + 1). This gives x^2 – 3x + 4. Using the quadratic formula, we obtain the roots: x = - ½ (3 i 7) ? ?

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Relationship between complex roots For the first cubic, we found the two complex roots were: x = -1 i 2 What is the relationship between these two roots? Theyre complex conjugates. Can we prove this will always be the case for cubics with real coefficients? Suppose a and b are the two complex roots. Then (x – α)(x – β) in the expansion must give real coefficients. (x – α)(x – β) = x 2 – (a + β)x + αβ Therefore α + β and αβ must both be real. We saw earlier that this is satisfied when α and β are complex conjugates. (Although technically our proof isnt complete, because weve shown that the roots being complex conjugates is a sufficient condition for the coefficients of the cubic being real, not a necessary one. However, we know the imaginary components of α and β must be the same but negative of each other, so that in α + β they cancel out, and we can proceed from there to show the real components of α and β must be the same). Q Q ? ? In summary, complex roots of polynomials always come in complex conjugate pairs.

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Finding other roots Another classic type of exam question: -1 + 2i is one of the roots of the cubic x 3 – x 2 – x – 15. Find the other two roots. Another classic type of exam question: -1 + 2i is one of the roots of the cubic x 3 – x 2 – x – 15. Find the other two roots. Other complex root is the complex conjugate: -1 – 2i. Now expand (x – (-1 + 2i))(x – (-1 – 2i)) = x 2 – (-1 – 2i)x – (-1 + 2i)x + (-1 + 2i)(-1 – 2i) = x 2 + 2x + 5 We can now either use polynomial division or Factor Theorem with trial and error to establish the real root as 3. ? Your turn: 2 – i is one of the roots of the cubic x 3 – 11x + 20. Find the other two roots. Your turn: 2 – i is one of the roots of the cubic x 3 – 11x + 20. Find the other two roots. 2 + i is other complex root. (x – (2 + i))(x – (2 – i)) = x 2 – 4x + 5. Dividing we get (x + 4), so real root is -4. ?

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Quartics Any quartic (with coefficient of x 4 of 1) can be expressed as: y = (x – α)(x – )(x – )(x – δ) where α,, and δ are the roots. α δ α Possibility 1: 4 real roots (some potentially repeated) Possibility 2: 2 real roots, pair of complex conjugate roots. Possibility 3: No real roots. Two pairs of complex conjugate roots. ?? ? Any other possibilities? No. On y-axis, both ends of line must either be both positive infinity or both negative. x- axis must therefore be crossed an even number of times, so an even number of real roots. ?

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Quartics Edexcel Jan 2013 (Retracted) a)4x 2 + 9 = 0 x 2 = -9/4,x = i (3/2) x 2 – 2x + 5 = 0 Using the quadratic formula, x = 1 2i Im[z] Re[z] b) Since real roots appear on the x-axis and complex roots come in conjugate pairs, we know well have symmetry in the line y = 0 ? ?

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Exercise 1H 1 3 5 7 9 10 ? ? ? ? ? ?

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