Further Pure 1 Lesson 6 Complex Numbers ModulusArgument

Further Pure 1 Lesson 6 – Complex Numbers

Modulus/Argument of a complex number n 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| n r = |z| = √(x 2+y 2) n r is known as the modulus of a complex number. n Wiltshire Im r y θ x Re

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

Modulus/Argument of a complex number Wiltshire n Calculate the modulus and argument of the following complex numbers. (Hint, it helps to draw a diagram) 1) 3 + 4 j |z| = √(32+42) = 5 arg z = inv tan (4/3) = 0. 927 2) 5 - 5 j |z| = √(52+52) = 5√ 2 arg z = inv tan (5/-5) = -π/4 3) -2√ 3 + 2 j |z| = √((2√ 3)2+22) = 4 arg z = inv tan (2/-2√ 3) = 5π/6

Modulus/Argument of a complex number Wiltshire n Note, inv tan (y/x) will return answer in first/fourth quadrant. n 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. n In some circumstances you way need to add or subtract π. n This is why a diagram is useful.

Modulus-Argument form of a complex number Wiltshire n 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. n (This is just like plotting co-ordinates on an x, y axis) n 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. n (This is sometimes known as Polar co-ordinates and can be studied in another course)

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

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

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

Using the Calculator Wiltshire n It is possible to convert from Cartesian to Polar and n n back using your calculator. (Casio fx-83 MS – other calculators may be different) Make sure that your calculator is in Radians. The polar coordinates (√ 2, π/4) equal the Cartesian coordinates (1, 1) On the calculator press Shift Rec( √ 2 , π ÷ 4 n 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. n Now try and convert back from the Cartesian to the Polar. )

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

Modulus-Argument form of a complex number n i) Wiltshire Write the following numbers in Modulus. Argument form: 5 + 12 j ii) √ 3 – j iii) -4 – 4 j 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))

Modulus-Argument form of a complex number Wiltshire n 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) n (r must be positive) n 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

Modulus-Argument form of a complex number Wiltshire n Example: Re-write -4(cosθ + j sinθ) in Modulus- Argument form: -4(cosθ + j sinθ) n n = 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 2 E pg 66

Sets of points using the Modulus-Argument form n What do you think n n n arg(z 1 -z 2) represents? If z 1 = x 1 +y 1 j & z 2 = x 2 +y 2 j Then z 2 – z 1 = (x 2 – x 1) + (y 2 – y 1)j Now arg(z 2 – z 1) = inv tan ((y 2 – y 1)/(x 2 – x 1)) This represents the angle between the line from z 1 to z 2 and a line parallel to the real axis. So arg(z 2 – z 1) = α Wiltshire Im (x 2, y 2) y 2 - y 1 α (x 1, y 1) x 2 - x 1 Re

Questions n Wiltshire 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 Im Now do Ex 2 F page 68 π/3 Re

History of complex numbers Wiltshire n In the quadratic formula (b 2 – 4 ac) is known as the n n n 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.

History of complex numbers Wiltshire n In 1629 Albert Girard stated that an nth degree n n polynomial will have n roots, complex or real. Taking into account repeated roots. For example the fifth order equation (z – 2)(z-4)2(z 2+9) = 0 has five roots. 2, 4(twice) 3 j and -3 j. 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.

Complex numbers and equations Wiltshire n Given that 1 + 2 j is a root of 4 z 3 – 11 z 2 + 26 z – 15 n n =0 Find the other roots. Since real coefficients, 1 – 2 j must also be a root. Now [z – (1 + 2 j)] and [z – (1 – 2 j)] will be factors. So [z – (1 + 2 j)][z – (1 – 2 j)] = [(z – 1) + 2 j)][(z – 1) – 2 j)] = (z – 1)2 – (2 j)2 = z 2 – 2 z + 1 – (-4) = z 2 – 2 z + 5

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

Complex numbers and equations Wiltshire n Given that -2 + j is a root of the equation n n n z 4 + az 3 + bz 2 + 10 z + 25 = 0 Find the values of a and b and solve the equation. First you need to find the values of z 4, z 3 & z 2 so that you can sub them in to the equation above. z 2 = (-2 + j) = 3 – 4 j z 3 = (3 – 4 j)(-2 + j) = -2 + 11 j Z 4 = (-2 + 11 j)(-2 + j) = -7 - 24 j Now (-7 -24 j) + a(-2 + 11 j) + b(3 – 4 j) + 10(-2 + j) +25 = 0