Further Pure 1

Lesson 6 – Complex Numbers

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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. Re

Im

x

y

θ

r

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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)

Re

Im

x

y

θ

r

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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) = 5arg z = inv tan (4/3)

= 0.927

2) 5 - 5j |z| = √(52+52) = 5√2arg z = inv tan (5/-5)

= -π/4

3) -2√3 + 2j |z| = √((2√3)2+22) = 4arg z = inv tan (2/-2√3)

= 5π/6

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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.

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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)

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Modulus-Argument form of a complex number (x,y) (r, θ)

cosθ = x/r, sinθ = y/r x = r cosθ, y = r sinθ,

Im

r

Re

Im

xθ

y

Re

WiltshireConverting

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

x

y

Re

r

θ

WiltshireConverting

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

x

y

Re

r

θ

WiltshireUsing 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 the Cartesian 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 )

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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.

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Modulus-Argument form of a complex number Write the following numbers in Modulus-

Argument form:i) 5 + 12j ii) √3 – j iii) -4 –

4j

Solutionsi) 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))

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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

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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

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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 Re

Im

(x1,y1)

(x2,y2)

y2- y1

x2- x1

α

WiltshireQuestions

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 68Im

Re

π/3

Im

Re

π/3

Im

Re

π/3

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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.

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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.

WiltshireComplex 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

WiltshireComplex numbers and equations

You can now try to spot the remaining factors by comparing coefficients or you can use polynomial division.

4z – 3z2 – 2z + 5 4z3 – 11z2 + 26z – 15 = 0

4z3 – 8z2 + 20z – 3z2 + 6z – 15 – 3z2 + 6z – 15

0Hence 4z3 – 11z2 + 26z – 15 = (z2 – 2z + 5)(4z – 3) The roots are, 1 + 2j, 1 – 2j and 3/4

WiltshireComplex 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