Complex Numbers

Question 1

What is the equation linking modulus argument form and exponential form?

Answer 1

r(cos(nθ) + isin(nθ)) = re^iθ

Question 2

How do you multiply and divide complex numbers in exponential form?

Answer 2

If z₁ = r₁e^iθ₁

If z₂ = r₂e^iθ₂

then

z₁z₂ = r₁r₂e^{i(θ₁ + θ₂)}

and

z₁/z₂ = (r₁/₂)e^{i(θ₁ - θ₂)}

Question 3

What is demoivre’s theorem?

Answer 3

(r(cosθ + isinθ))ⁿ = rⁿ(cos(nθ) + isin(nθ))

Note this is useful when putting complex numbers to powers

Question 4

How do you prove trig identities in the form cos(nθ) or sin(nθ) ?

Answer 4

• Use demoivre’s theorem to expan (cosθ +isinθ)ⁿ
• Use binomial expansions to expand (cosθ +isinθ)ⁿ
• If you are proving cos equate the real parts of the two expressions if you are proving sin equate the imaginary parts
• If you are proving cos remove all the sin terms using cos² + sin² = 1 if you are proving sin remove all the cos terms

Question 5

How do you prove trig identities in the form tan(nθ) ?

Answer 5

• Use demoivre’s theorem to expan (cosθ +isinθ)ⁿ
• Use binomial expansions to expand (cosθ +isinθ)ⁿ
• Equate the real and imaginary parts of the two expressions
• Now you have expressions equal to sin(nθ) and cos(nθ) divide them to get tan(nθ)
• Divide the top and bottom of your fraction by cosⁿθ

Question 6

How do you prove trig identities in the form cosⁿθ or sinⁿθ ?

Answer 6

• For cos state (2cosθ)ⁿ = (z + 1/z)ⁿ, for sin state (2sinθ)ⁿ = (z - 1/z)ⁿ
• Expand the right hand side using the binomial expansion
• Group together your z^k and 1/z^k terms
• Use the fact that zⁿ +1/zⁿ = 2cos(nθ) or zⁿ - 1/zⁿ = 2isin(nθ) to sub into your expression
• Simplify and divide

Question 7

How do you use demoivre’s theorem to prove that if z = e^iθ then zⁿ + 1/zⁿ = 2cos(nθ)

Answer 7

• State zⁿ + z^-n
• Sub in e^iθ for z
• Change from exponential form to modulus argument form
• Use the fact that cos(-θ) = cos(θ) and sin(-θ) = -sin(θ) to simplify and solve

Question 8

How do you use demoivre’s theorem to prove that if z = e^iθ then zⁿ - 1/zⁿ = 2isin(nθ)

Answer 8

• State zⁿ - z^-n
• Sub in e^iθ for z
• Change from exponential form to modulus argument form
• Use the fact that cos(-θ) = cos(θ) and sin(-θ) = -sin(θ) to simplify and solve

Question 9

How can you use demoivre’s tirg identity proof to solve trig equations?

Answer 9

You can simplify trig terms with powers in to trig terms without powers in and then solve

Question 10

How do you find the nth roots of a complex number?

Answer 10

• State that zⁿ = a + bi
• Convert a + bi into e^iθ to get zⁿ = e^iθ
• Work out n terms by adding 2π to the argument of the exponential
• Work out z by doing the nth root of all terms sides to get z = e^iθ/n…
• Now convert these terms into whatever form the question asks for

Question 11

What are important facts about roots of unity?

Answer 11

• |w|² = 1
• The nth roots of a complex number form a regular n-gon around the origin of an argand diagram where the radius of the circle connecting these points is ⁿ√r
• The sum of the nth roots of a complex number = 0
• A root of a complex number can be rotated to other roots by multiplying by the nth roots of unity (in a+bi form)

Question 12

What are the nth roots of unity?

Answer 12

The solutions to the equation zⁿ = 1

Note these can be the cube roots of unity or etc. if you swap n for 3 or etc

Question 13

How do you solve a sums of series question in this form?
z = cos(π/n) + isin(π/n) show that 1 + z + z² + … + z^n-1 = 1 +icot(π/2n)

Answer 13

• This is a geometric sequence where a = 1 and r = z therefore
• 1 + z + z²… = 1(1-zⁿ) / (1-z)
• Swap the zⁿ for e^i*π/n
• Simplify then multiply the top and bottom by e^i*-π/2n
  *

Question 14

Given the convergent and infinite series C and S are defined by

C = cos2θ + 1/2 cos6θ + 1/4 cos10θ
S = sin2θ + 1/2 sin6θ + 1/4 sin10θ

show that C + iS = 2e^2iθ/ 2-2e^4iθ

Answer 14

• Add the two expressions and combine terms with the same coefficeints
• Simplify these terms by converting from modulus argument form to exponential form
• Now you have a geometric sequence so state what ‘a’ and ‘r’ are
• Now you can find C+iS by using the sun of an infinite series using your values of ‘a’ and ‘r’

Now if you are asked to find an expression for ‘C’ or ‘S’
* Multiply the top and bottom of your fraction for C + iS by the bottom of the fraction but the argument is negative
* Simplify then convert the exponential form into modulus argument form
* Keep simpifying using the results cos(-θ) = cos(θ) and sin(-θ) = -sin(θ)
* Then seperate your fraction into two fractions where one is real fraction and one is an imaginary fraction
* The real fraction equals ‘C’ and the coefficient of the imaginary fraction equals ‘S’

Question 15

How do you work out the modulus and argument of a complex number? e.g a+bi

Answer 15

Modulus: (a²+b²)^1/2

Argument: tan^-1(imaginary/real)
Note this is always true for a complex number and includes any negatives. Also you can add and subtract π to find solutions in the range

Question 16

What is the modulus and argument of a complex number?

Answer 16

Modulus: The length of the line from the origin to the complex number

Argument: The angle (radians) from the positive real axis to the complex number

Question 17

What is modulus argument form?

Answer 17

z = r(cosθ + isinθ)

Where:
r = |z|
θ = arg(z)

Question 18

How do you multiply and divide modulus argument form?

Answer 18

arg(z₁z₂) = arg(z₁) + arg(z₂)

arg(z₁/z₂) = arg(z₁) - arg(z₂)

|z₁z₂| = |z₂||z₂|

|z₁/z₂| = |z ₁|/|z₂|

Question 19

What is the equation of a circle in the argand plane?

Answer 19

|z - (z₁)| = k

Centre: z₁
Radius = k

Note is a strict innequality is used then we do not include the points along the circle and draw it as a dotted line instead

Question 20

What is the equation of a half line in the argand plane?

Answer 20

arg(z - (z₁)) = θ

Starting point: z₁
θ = angle from the positive horizontal to z₁

If we have an innequality instead e.g θ₁< arg(z) < θ₂ then this is the region between the two lines from the origin between the angles z₁ and a z₂, exclusively (dotted line)

Question 21

What is the equation of a perpendicular bisector in the argand plane?

Answer 21

|z - (z₁)| = |z - (z₂)|

This forms a perpendicular bisector of the line joining z₁ and a₂

Note if there is an innequality which ever side the closed face of the innequality symbol faces (pointy bit) then the perpendicular bisector is closer to that point

Question 22

How do you convert from argand form to cartesian form?

Answer 22

Using common sense

Question 23

What is the complex conjugate of a complex number? e.g z = a + bi

Answer 23

z* = a - bi

Question 24

How do you divide complex numbers? e.g
z₁/z₂

Answer 24

You rationalise the denominator

Multiply the top and bottom by z*₂