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Further Pure 1

Further Pure 1. Lesson 6 – Complex Numbers. Modulus/Argument of a complex number. The position of a complex number (z) can be represented by the distance that z is from the origin (r) and the angle made with the positive real axis ( θ ). Distance is given by r = |z| r = |z| = √(x 2 +y 2 )

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Further Pure 1

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  1. Further Pure 1 Lesson 6 – Complex Numbers

  2. Modulus/Argument of a complex number • The position of a complex number (z) can be represented by the distance that z is from the origin (r) and the angle made with the positive real axis (θ). • Distance is given by r = |z| • r = |z| = √(x2+y2) • r is known as the modulus of a complex number. Im r y θ x Re

  3. Argument of a complex number • Now lets look at the angle (θ) that the line makes with the positive real axis. • NOTE: θ is measured in radians in an anticlockwise direction from the positive real axis. • θ is measured from –π to π and is known as the principal argument of z. • Argument z = arg z = θ • y = r sinθ, x = r cosθ • tan θ = y/x • θ = inv tan (y/x) Im r y θ x Re

  4. Modulus/Argument of a complex number • Calculate the modulus and argument of the following complex numbers. (Hint, it helps to draw a diagram) 1) 3 + 4j |z| = √(32+42) = 5 arg z = inv tan (4/3) = 0.927 2) 5 - 5j |z| = √(52+52) = 5√2 arg z = inv tan (5/-5) = -π/4 3) -2√3 + 2j |z| = √((2√3)2+22) = 4 arg z = inv tan (2/-2√3) = 5π/6

  5. Modulus/Argument of a complex number • Note, inv tan (y/x) will return answer in first/fourth quadrant. • Last example on previous slide inv tan (2/-2√3) on your calculator will return, -π/6, however the answer we want is 5π/6. • In some circumstances you way need to add or subtract π. • This is why a diagram is useful.

  6. Modulus-Argument form of a complex number • So far we have plotted the position of a complex number on the Argand diagram by going horizontally on the real axis and vertically on the imaginary. • (This is just like plotting co-ordinates on an x,y axis) • However it is also possible to locate the position of a complex number by the distance travelled from the origin (pole), and the angle turned through from the positive x-axis. • (This is sometimes known as Polar co-ordinates and can be studied in another course)

  7. (x,y) (r, θ) cosθ = x/r, sinθ = y/r x = r cosθ, y = r sinθ, Modulus-Argument form of a complex number y Im Im θ y x r y θ x Re Re

  8. Converting • Converting from Cartesian to Polar • (x,y) = [√(x2+y2),(inv tan (y/x))] = (r, θ) • Convert the following from Cartesian to Polar i) (1,1) = (√2,π/4) ii) (-√3,1) = (2,5π/6) iii) (-4,-4√3) = (8,-2π/3) Im r y θ x Re

  9. Converting • Converting from Polar to Cartesian • (r, θ) = (r cosθ, r sin θ) • Convert the following from Polar to Cartesian i) (4,π/3) = (2,2√3) ii) (3√2,-π/4) = (3,-3) iii) (6√2,3π/4) = (-6,6) Im r y θ x Re

  10. Using the Calculator • It is possible to convert from Cartesian to Polar and back using your calculator. • (Casio fx-83MS – other calculators may be different) • Make sure that your calculator is in Radians. • The polar coordinates (√2,π/4) equal theCartesian co-ordinates (1,1) • On the calculator press • The answer shown is the first Cartesian co-ordinate x. To view the y press RCL F. To view the x again press RCL E. • Now try and convert back from the Cartesian to the Polar. Shift Rec( √ 2 , π ÷ 4 )

  11. Modulus-Argument form of a complex number • Now we can define the Modulus-Argument form of a complex number to be: z = r (cosθ + j sinθ) • Here r = |z| and θ = arg z • When writing a complex number in Modulus-Argument form it can be helpful to draw a diagram. • Remember that θ will take values between -π and π.

  12. Modulus-Argument form of a complex number • Write the following numbers in Modulus-Argument form: i) 5 + 12j ii) √3 – j iii) -4 – 4j Solutions i) 13(cos 1.176 + j sin 1.176) ii) 2(cos (-π/6) + j sin (-π/6)) iii) 4√2(cos(-3π/4) + j sin(-3π/4))

  13. Modulus-Argument form of a complex number • When writing a complex number in Modulus-Argument form it is important that it is written in the exact format as above, not: z = -r (cosθ- j sinθ) • (r must be positive) • You can use the following trig identities to ensure that z is written in the correct form. cos(π-θ) = -cosθ sin(π-θ) = sinθ cos(θ-π) = -cosθ sin(θ-π) = -sinθ cos(-θ) = cosθ sin(-θ) = -sinθ

  14. Modulus-Argument form of a complex number • Example: Re-write -4(cosθ + j sinθ) in Modulus-Argument form: -4(cosθ + j sinθ) = 4(-cosθ - j sinθ) = 4(cos(θ-π) + j sin(θ-π)) • This is now in Modulus-Argument form. • The Modulus is 4 and the Argument is (θ-π). • This may seem a bit confusing, however if you draw a diagram and tackle the problems like we did a few slides ago it makes more sense. • Now do Ex 2E pg 66

  15. Sets of points using the Modulus-Argument form • What do you think arg(z1-z2) represents? • If z1 = x1 +y1j & z2 = x2 +y2j • Then z2 – z1 = (x2 – x1) + (y2 – y1)j • Now arg(z2 – z1) = inv tan ((y2 – y1)/(x2 – x1)) • This represents the angle between the line from z1 to z2 and a line parallel to the real axis. • So arg(z2 – z1) = α Im (x2,y2) y2- y1 α (x1,y1) x2- x1 Re

  16. Im π/3 Re Im Im π/3 π/3 Re Re Questions • Draw Argand diagrams showing the sets of points z for which i) arg z = π/3 ii) arg (z - 2) = π/3 iii) 0 ≤ arg (z – 2) ≤ π/3 Now do Ex 2F page 68

  17. History of complex numbers • In the quadratic formula (b2 – 4ac) is known as the discriminant. • If this is greater than zero the quadratic will have 2 distinct solutions. • If it is equal to zero then the quadratic will have 1 repeated root. • If it is less than zero then there are no solutions. • Complex numbers where invented so that we could solve quadratic equations whose discriminant is negative. • This can be extended to solve equations of higher order like cubic and quartic. • In fact complex numbers can be used to find the roots of any polyniomial of degree n.

  18. History of complex numbers • In 1629 Albert Girard stated that an nth degree polynomial will have n roots, complex or real. Taking into account repeated roots. • For example the fifth order equation (z – 2)(z-4)2(z2+9) = 0 has five roots. 2, 4(twice) 3j and -3j. • For any polynomial with real coefficients solutions will always have complex conjugate pairs. • Many great mathematicians have tried to prove the above. • Gauss achieved it in 1799.

  19. Complex numbers and equations • Given that 1 + 2j is a root of 4z3 – 11z2 + 26z – 15 = 0 • Find the other roots. • Since real coefficients, 1 – 2j must also be a root. • Now [z – (1 + 2j)] and [z – (1 – 2j)] will be factors. • So [z – (1 + 2j)][z – (1 – 2j)] = [(z – 1) + 2j)][(z – 1) – 2j)] = (z – 1)2 – (2j)2 = z2 – 2z + 1 – (-4) = z2 – 2z + 5

  20. Complex numbers and equations • You can now try to spot the remaining factors by comparing coefficients or you can use polynomial division. 4z – 3 z2 – 2z + 5 4z3 – 11z2 + 26z – 15 = 0 4z3 – 8z2 + 20z – 3z2 + 6z – 15 – 3z2 + 6z – 15 0 Hence 4z3 – 11z2 + 26z – 15 = (z2 – 2z + 5)(4z – 3) The roots are, 1 + 2j, 1 – 2j and 3/4

  21. Complex numbers and equations • Given that -2 + j is a root of the equation z4 + az3 + bz2 + 10z + 25 = 0 • Find the values of a and b and solve the equation. • First you need to find the values of z4,z3 & z2 so that you can sub them in to the equation above. • z2 = (-2 + j)(-2 + j) = 3 – 4j • z3 = (3 – 4j)(-2 + j) = -2 + 11j • Z4 = (-2 + 11j)(-2 + j) = -7 - 24j • Now (-7 -24j) + a(-2 + 11j) + b(3 – 4j) + 10(-2 + j) +25 = 0

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