Presentation on theme: "FP1: Chapter 1 Complex Numbers"— Presentation transcript:
1 FP1: Chapter 1 Complex Numbers Dr J FrostLast modified: 2nd January 2014
2 i i captain!Here’s something that someone has almost certainly spoiled for you already…𝑖= −1?
3 3+4𝑖 Complex versus imaginary numbers Imaginary number: of form 𝑏𝑖, 𝑏∈ℝ?Complex number: of form𝑎+𝑏𝑖, 𝑎,𝑏∈ℝ??Real part?Imaginary part3+4𝑖
4 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)i2a – 3bi + 3 – 6ci = 2a+3 – 3(b + 2c)i??Convention 1Just like we’d 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√3Why? This avoids ambiguity over whether the function is being applied to the i.?Convention 2In 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.
5 Why complex numbers? 1 Fractals 2 Analytic Number Theory 3 Complex numbers were originally introduced by the Italian mathematician Cardano in the 1500s to allow him to represent the roots of polynomials which weren’t ‘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:1FractalsA Mandelbrot Set is the most popular ‘fractal’. For each possible complex number c, we see if zn+1 = zn2 + c is not divergent (using z0 = 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. We’ll use it later.2Analytic Number TheoryNumber 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.3Physics and EngineeringUsed in Signal Analysis, Quantum Mechanics, Fluid Dynamics, Relativity, Control Theory...
6 Manipulation and application Here’s a flavour of some of things you’ll be initially expected to do...−36 = 𝟑𝟔 −𝟏 =𝟔𝒊Solve 𝑥 2 +9=0𝒙 𝟐 =−𝟗 𝒙=± −𝟗 𝒙=±𝟑𝒊Solve 𝑥 2 +6𝑥+25=0Using the quadratic formula:𝒙=−𝟑±𝟒𝒊Simplify 5+2𝑖 + 8+9𝑖=13+11𝑖??Simplify (8 + i)(3 – 2i)= 24 – 16i + 3i – 2i2= 24 – 16i + 3i + 2= 26 – 13i???
7 MultiplyingExample: Let z1 = 1 + 3i and z2 = 5 – 2i, find z1z2.(1 + 3i)(5 – 2i) = 5 – 2i + 5i – 6i2= 5 – 2i + 5i + 6= i?The quick way to think about this is that i2 reverses the sign.1Evaluate the following:√(-16) = 4i√(-25) = 5i√(-3) = i√3√(-7) = i√7√(-8) =2i√23Calculate z1z2.z1 = 1 + i z2 = 1 – i z1z2 = 2z1 = 2 + i z2 = 2 – 2i z1z2 = 6 – 2iz1 = 3 + 2i z2 = 4 – 3i z1z2 = 18 – iz1 = -3 + i z2 = 5 + 7i z1z2 = -22 – 16iz1 = 7 – 3i z2 = 3 – 7i z1z2 = -58iz1 = -1 – i z2 = 9 + 8i z1z2 = -1 – 17i???????????4By using a Binomial Expansion, determine (1 + i)5.= 1 + 5i + 10i2 +10i3 + 5i4 + i5= 1 + 5i – 10 – 10i + 5+ i= -4 – 4iWhat is the value of:i100 = (i4)25 = 125 = 1i2003 = i2000 x i3 = 1 x i3 = -i2Calculate z2 for the following z.z = 1 + i z2 = 2iz = 1 – i z2 = -2iz = 3 + 2i z2 = iz = 7 – 4i z2 = 33 – 56iz = i z2 = -18iz = a + bi z2 = a2 – b2 + 2abi?????5????
8 Complex ConjugatesRemember in C1 how we rationalised the denominator?3 4− 2Suppose x1 = a + √b and x2 = a - √bThen:x1x2 = a2 – bx1 + x2 = 2aBoth these results are rational.?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* = x2 + y2z + z* = 2xwhich are both real.The fact the first is real will help us with dividing complex numbers.We’ll see the significance of the second later.???
9 Quickfire Questions z = 3 + 2i z z* = 13 z = 3 – 2i z z* = 13 Given z, determine z z*.z = 3 + 2i z z* = 13z = 3 – 2i z z* = 13z = 5 + 4i z z* = 41z = 1 + i z z* = 2z = 4 – 2i z z* = 20?1?2?3?4?5
11 Exercises ? ? ? ? ? ? ? ? ? _1_ 1 + i 1 - i 2 10 + 3i 1 + 2i 16 – 17i Put all the following in the form a + bi._1_1 + i1 - i210 + 3i1 + 2i16 – 17i5??1==Note that mark schemes permit the single fraction.5_i_1 – i-1 + i2_10_3 + i??26== 3 - i8 – 4i1 – 3i?= 2 + 2i1 - i1 + i7?3= -ia + ia – ia2 – 1 + 2aia2 + 1?=14 – 5i3 – 2i?8= 4 + i4Edexcel 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).9?
12 Argand Diagrams z = x + yi Argand Diagrams are a well of geometrically representing complex numbers.Im[z]Click to move.321-1-2-31 + i2 – 3iRe[z]-1 + 2i-3 – iz = x + yiThe x-axis is the real component.The y-axis is the imaginary component.
13 (Don’t write anything down yet!) Argument and ModulusIn FP2, you’ll 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.(Don’t write anything down yet!)Im[z]z = 2 + 3i321-1-2-3Distance from origin:?√( ) = √13This is known as the modulus of z, and we write|2 + 3i| = √13Re[z]Angle: (anticlockwise from the real axis)?Using trigonometry, we have (in radians):tan-1(3/2) = 0.983This is known as the argument of z, and we write:arg(2 + 3i) = 0.983
14 Use trig common sense for other quadrants. Argument and ModulusIm[z]!If 𝑧=𝑥+𝑦𝑖|z|Modulus of z:𝒛 = 𝒙 𝟐 + 𝒚 𝟐arg zRe[z]Argument of z:Usual range - < arg z ≤ . Known as the principal argument if in this range.𝒂𝒓𝒈 𝒛=𝒕𝒂 𝒏 −𝟏 𝒚 𝒙when − 𝝅 𝟐 <𝒂𝒓𝒈 𝒛< 𝝅 𝟐Use trig common sense for other quadrants.
15 Check your Understanding z|z|arg(z)1-i-/2-1 + i√23/4-5 – 2i√29-2.761????????
16 ExercisesGive exact answers where possible, otherwise to 3dp.1z|z|arg(z)-11i/21 + i√2p/41 + 2i√51.1071 – 2i-1.107-1 + 2i2.034-1 – 2i-2.0343 + 4i50.927i131.9661 – i√3√7-/32z|z|arg(z)-√3 - i√7-5/63 + i√100.3222 – 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) – 3Give that arg(5 + i + ai) = p/4 and that a is real, determine a.(a + 1)/5 = 1. So a = 4Given that arg(santa i) = /2 and that santa is real, determine santa.santa = -33????4??5?
17 Argument-modulus form What’s the point of |z| and arg(z)?We’ll see in FP2 that we can express a complex number as 𝑧 𝑒 𝑖 arg 𝑧 ., where we’ll also see the infamous 𝑒 𝜋𝑖 =−1. But for now, let’s get halfway there...Im[z]Suppose that r = |z| and = arg(z).How could we express z in terms of r and o?|z|!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.arg zRe[z]
18 Argument-modulus form Question 1 from earlier...1z|z|arg(z)Mod-arg form-11z = cos + i sin i/2z = cos(/2) + i sin(/2)1 + i√2p/4z = √2(cos(/4) + i sin(/4))1 + 2i√51.107z = √5(cos i sin 1.107)1 – 2i-1.107z = √5(cos(-1.107) + i sin(-1.107))-1 + 2i2.034z = √5(cos i sin 2.034)-1 – 2i-2.034z = √5(cos(-2.034) + i sin(-2.034))3 + 4i50.927z = 5(cos i sin 0.927)i131.966z = 13(cos i sin 1.966)1 – i√3√7-/3z = √7(cos(-/3) + i sin(-/3))??????????
19 𝒛 𝟏 𝒛 𝟐 = 𝒛 𝟏 𝒛 𝟐 and 𝐚𝐫𝐠 𝒛 𝟏 𝒛 𝟐 = 𝐚𝐫𝐠 𝒛 𝟏 + 𝐚𝐫𝐠 𝒛 𝟐 |z1z2| and arg(z1z2)When we multiply two complex numbers, we multiply their moduli, and we add their arguments.So can we prove this, i.e. that:𝒛 𝟏 𝒛 𝟐 = 𝒛 𝟏 𝒛 𝟐 and 𝐚𝐫𝐠 𝒛 𝟏 𝒛 𝟐 = 𝐚𝐫𝐠 𝒛 𝟏 + 𝐚𝐫𝐠 𝒛 𝟐Let 𝒛 𝟏 = 𝒓 𝟏 𝐜𝐨𝐬 𝜽 𝟏 +𝒊 𝐬𝐢𝐧 𝜽 𝟐 and 𝒛 𝟏 = 𝒓 𝟏 𝒄𝒐𝒔 𝜽 𝟏 +𝒊 𝒔𝒊𝒏 𝜽 𝟐 .Then 𝒛 𝟏 𝒛 𝟐 = 𝒓 𝟏 𝒓 𝟐 𝐜𝐨𝐬 𝜽 𝟏 𝐜𝐨𝐬 𝜽 𝟐 − 𝐬𝐢𝐧 𝜽 𝟏 𝐬𝐢𝐧 𝜽 𝟐 +𝒊 𝐬𝐢𝐧 𝜽 𝟏 𝒄𝒐𝒔 𝜽 𝟐 + 𝒔𝒊𝒏 𝜽 𝟏 𝐜𝐨𝐬 𝜽 𝟐= 𝒓 𝟏 𝒓 𝟐 𝐜𝐨𝐬 𝜽 𝟏 + 𝜽 𝟐 +𝒊 𝐬𝐢𝐧 𝜽 𝟏 + 𝜽 𝟐So 𝒛 𝟏 𝒛 𝟐 = 𝒓 𝟏 𝒓 𝟐 = 𝒛 𝟏 𝒛 𝟐and 𝒂𝒓𝒈 𝒛 𝟏 𝒛 𝟐 = 𝜽 𝟏 + 𝜽 𝟐 = 𝐚𝐫𝐠 𝒛 𝟏 + 𝐚𝐫𝐠 𝒛 𝟐?The following C3 identities may be useful:cos 𝑎+𝑏 = cos 𝑎 cos 𝑏 − sin 𝑎 sin 𝑏sin 𝑎+𝑏 = sin 𝑎 cos 𝑏 + cos 𝑎 sin 𝑏
20 Comparing coefficients If two complex numbers are equal, then clearly both their real and imaginary parts are equal. i.e. if 𝑎 1 + 𝑏 1 𝑖= 𝑎 2 + 𝑏 2 𝑖 then 𝑎 1 = 𝑎 2 and 𝑏 1 = 𝑏 2 .QGiven that 3+5𝑖=(𝑎+𝑖𝑏)(1+𝑖) where 𝑎 and 𝑏 are real, find the value of 𝑎 and 𝑏.?Expanding: 𝒂−𝒃 +𝒊 𝒂+𝒃 =𝟑+𝟓𝒊So 𝒂−𝒃=𝟑 and 𝒂+𝒃=𝟓.Solving: 𝒂=𝟒, 𝒃=𝟏. We could have also found 𝟑+𝟓𝒊 𝟏+𝒊QCalculate 𝑖 .𝒊 =𝒂+𝒊𝒃𝒊= 𝒂+𝒊𝒃 𝟐 = 𝒂 𝟐 +𝟐𝒂𝒃𝒊− 𝒃 𝟐So 𝟎+𝟏𝒊= 𝒂 𝟐 − 𝒃 𝟐 + 𝟐𝒂𝒃 𝒊𝒂 𝟐 − 𝒃 𝟐 =𝟎 and 𝟐𝒂𝒃=𝟏𝒂=𝒃= 𝟏 𝟐So 𝒊 = 𝟏 𝟐 + 𝟏 𝟐 𝒊?
21 Exercise 1G1𝑎+2𝑏+2𝑎𝑖=4+6𝑖 where 𝑎 and 𝑏 are real.Find the value of 𝑎 and of 𝑏.𝒂=𝟑, 𝒃= 𝟏 𝟐𝑎+𝑖 3 =18+26𝑖 where 𝑎 is real. Find the value of 𝑎.𝒂=𝟑Find real 𝑥 and 𝑦 such that 1 𝑥+𝑖𝑦 =3−2𝑖𝒙= 𝟑 𝟐 , 𝒚=− 𝟏 𝟐Find the square roots of 7+24𝑖± 𝟒+𝟑𝒊Find the square roots of 11+60𝑖±(𝟔+𝟓𝒊)Find the square roots of 5−12𝑖± 𝟑−𝟐𝒊?4?7?9?10?11?
22 Roots of polynomialsAny cubic (with coefficient of x3 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...ααα=All 3 roots are real.?1 real root, 2 complex roots.?3 real roots (one repeated)Are there any other possibilities?No: since cubics have a range of – to +, it must cross the x axis. And it can’t cross an even number of times, otherwise the cubic would start and end in the same vertical direction.?
23 Roots of polynomials ? ? Classic Question: x = 2 is one of the roots of the polynomial x3 – x – 6 Find the other two roots.?Use polynomial division (as per C2) to divide x3 – x – 6 by (x – 2).This gives x2 + 2x + 3.Using the quadratic formula, we obtain the roots:x = -1 i√2Your Go:x = -1/2 is one of the roots of the polynomial 2x3 – 5x2 + 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)
24 Relationship between complex roots For the first cubic, we found the two complex roots were:x = -1 i√2What is the relationship between these two roots?They’re 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) = x2 – (a + β)x + αβTherefore 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 isn’t complete, because we’ve 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.
25 Finding other roots ? ? Another classic type of exam question: -1 + 2i is one of the roots of the cubic x3 – x2 – x – 15.Find the other two roots.?Other complex root is the complex conjugate: -1 – 2i.Now expand (x – (-1 + 2i))(x – (-1 – 2i))= x2 – (-1 – 2i)x – (-1 + 2i)x + (-1 + 2i)(-1 – 2i)= x2 + 2x + 5We 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 x3 – 11x Find the other two roots.?2 + i is other complex root.(x – (2 + i))(x – (2 – i)) = x2 – 4x + 5.Dividing we get (x + 4), so real root is -4.
26 y = (x – α)(x – )(x – )(x – δ) QuarticsAny quartic (with coefficient of x4 of 1) can be expressed as:y = (x – α)(x – )(x – )(x – δ)where α, , and δ are the roots.αδαPossibility 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.?
27 Quartics ? ? Edexcel Jan 2013 (Retracted) 4x2 + 9 = 0 x2 = -9/4, x = i√(3/2) x2 – 2x + 5 = 0 Using the quadratic formula, x = 1 2i?b)?Im[z]Since real roots appear on the x-axis and complex roots come in conjugate pairs, we know we’ll have symmetry in the line y = 0Re[z]
28 Exercise 1H1Given that 1 + 2i is one of the roots of a quadratic equation with real coefficients, find the equation.𝒙 𝟐 −𝟐𝒙+𝟓=𝟎Given that 𝑎+4𝑖, where 𝑎 is real, is one of the roots of a quadratic equation with real coefficients, find the equation.𝒙 𝟐 −𝟐𝒂𝒙+ 𝒂 𝟐 +𝟏𝟔=𝟎Show that 𝑥=3 is a root of the equation 2 𝑥 3 −4 𝑥 2 −5𝑥−3=0.Hence solve the equation completely.Roots are 3, − 𝟏 𝟐 + 𝟏 𝟐 𝒊, − 𝟏 𝟐 − 𝟏 𝟐 𝒊Given that −4+𝑖 is one of the roots of the equation 𝑥 3 +4 𝑥 2 −15𝑥−68=0, solve the equation completely.Roots are 4, -4 + i and -4 - iGiven that 2+3𝑖 is one of the roots of the equation 𝑥 4 +2 𝑥 3 − 𝑥 2 +38𝑥+130=0, solve the equation completely.Roots are 2 + 3i, 2 – 3i, -3 + i and -3 – iFind the four roots of the equation 𝑥 4 −16=0. Show these roots on an Argand Diagram.Roots are 2, -2, 2i, -2i?3?5?7?9?10?