Complex numbers

Question 1

Solving complex number equations

Answer 1

Complex numbers can only be equal if their real and imaginary parts are equal.

1. Make real parts of the equation equal to each other
2. Make imaginary parts of the equation equal to each other
3. Solve separately to find the value of each variable. If necessary, solve simultaneously.

Question 2

Discriminant of quadratic equations

Answer 2

The discriminant of quadratic equations is △ = b^2 - 4ac

1. If b^2 - 4ac > 0
There are two distinct, real solutions
2. If b^2 - 4ac = 0
There is one repeated, real solution 
3. b^2 - 4ac < 0
There are two, imaginary solutions

Question 3

Showing addition and subtraction of points on an argand diagram

Answer 3

1. Mark the coordinates of points
2. Draw arrows out from (0,0) to the points
3. If a point needs to be subtracted, flip the direction of the arrow by 180°
4. Draw arrows from head to tail, with the first arrow tail starting at (0,0)
5. Draw the resultant vector from the tail of the first arrow to the head of the last arrow
6. The coordinate of the end point reached is the result of the addition/subtraction.

Question 4

Multiplying and dividing complex numbers

Answer 4

• To multiply complex numbers, multiply out the brackets, replace i variables with their corresponding number value, then simplify like terms in i
• To divide complex numbers, first multiply the numerator and denominator by its complex conjugate, then simplify like terms in i

Question 5

How can i variables be simplfied?

Answer 5

Substitute powers of i for their equivalent values to simplify expressions

i^0 = 1 
i^1 = i
i^2 = -1 
i^3 = -i
And so on, the cycle repeats

-i = √-1

Question 6

What is the complex conjugate of a number z?

Answer 6

If z = x + iy, then its complex conjugate is ˉz = x - iy

• If a question is ˉxˉy, solve for xy first, the find the complex conjugate of the result

Question 7

What does the complex conjugate correspond to in an argand diagram?

Answer 7

The complex conjugate corresponds to the reflection in the real (x) axis.

Question 8

What is the modulus of a number z?

Answer 8

If z = x + iy, then its modulus is |z| = √x^2+y^2

Question 9

What does the modulus correspond to in an argand diagram?

Answer 9

The modulus corresponds to the distance from the origin (length of vector).

Question 10

Finding the modulus

Answer 10

Substitute the x (real) and y (imaginary) values into the formula, then simplify. Not that ‘i’ is not included in the y (imaginary) value, as only the coefficients are used.

Question 11

What does the argument correspond to in an argand diagram?

Answer 11

The argument corresponds to the angle that a line joining a complex number z to the origin makes with the positive direction of the real (x) axis.

Question 12

Converting from degrees to radians

Answer 12

Degrees x π / 180 = radians

Question 13

What is the argument of a number z

Answer 13

If z = x + iy, then its argument is θ = tan^-1 (y/x)

• Depending on which quadrant the point lies in, the answer will need to be changed accordingly

1. Quadrant one
   θ = a
2. Quadrant two
   θ = a + 180
3. Quadrant three
   θ = a - 180
4. Quadrant four
   θ = a

Question 14

Things to remember when completing complex number calculations (rectangular form)

Answer 14

• ‘i’ (but not its coefficient) can be moved inside a root when being multiplied
• Use complex arithmetic rule (a+bi)(a-bi) = a^2 + b^2
• Use trig identities such as sin^2 θ + cos^2 θ = 1 to simplify answers
• For any complex number z = x + iy and thus substitution can be used
• For any complex number ˉz x z = |z|^2 and thus substitution can be used
• Multiplying the conjugate solutions in factor form (x - (conjugate 1))(x - (conjugate 2)) removes the i values. Remember this for if a question does not give a proper/complete equation and asks for the factors and roots to be calculated.

Question 15

What does the locus correspond to in an argand diagram?

Answer 15

The locus corresponds to the pathway between two complex numbers.

Question 16

Loci involving |z| or r

Answer 16

|z| or r is the modulus of a complex number. Therefore, the locus |z| = k can be thought of as the set of all complex numbers that are a fixed distance k from the origin, which results in a circle with the centre at the origin and radius k.

Question 17

Loci involving |z - z1|

Answer 17

z1 is a fixed complex number. Therefore, the locus |z - z1| = k can be thought of as the set of all complex numbers that are a fixed distance k from a fixed point z1 = x1 + iy1, which results in a circle, with the centre at z1 = x1 + iy1 and radius k.

Question 18

Loci involving |z - z1| = |z - z2|

Answer 18

The equation |z - z1| = |z - z2| can be thought of as the locus of a general point that moves equidistant from two fixed points, which is a perpendicular bisector of the line segment joining z1 and z2.

Question 19

Loci involving arg(z)

Answer 19

The locus arg(z) = θ is a ray from the origin, which makes an angle of θ with the positive direction for the real (x) axis. Loci involving arg (z - z1) = θ gives a ray that starts from z1 and makes an angle of θ with the positive direction of the real (x) axis.

Question 20

When is arg(z) meaningful?

Answer 20

In all cases for arg(z) loci, the starting point for the ray is represented by an open circle, which shows that the point is not included in the locus. Therefore, arg is only meaningful when a direction is defined (a single point has no direction).

Question 21

Writing loci equations

Answer 21

Rewrite loci equations from |z + x + iy| to |z - (x +iy)|

Question 22

Synthetic division

Answer 22

1. Arrange the numerator and denominator so that each is in descending powers of x.
2. If any terms with consecutive powers of x are missing, these should be included by using 0 as their coefficient.
3. Each term of the quotient is obtained by dividing the first term of the new dividend.
4. After each division, subtract and then bring down the next term.

Question 23

Remainder theorem

When p(x)/(x - b)

Answer 23

Remainder = p(b)

Where the value of a is changed, then substituted into the polynomial function

Question 24

Remainder theorem

When p(x)/(x + b)

Answer 24

Remainder = p(-b)

Where the value of a is changed, then substituted into the polynomial function

Question 25

Remainder theorem

When p(x)/(ax + b)

Answer 25

Remainder = p(-b/a)

Question 26

Remainder theorem

When p(x)/(ax - b)

Answer 26

Remainder = p(b/a)

Question 27

Factor theorem

Answer 27

(x-a) will only be a factor of any polynomial p(x) if p(a) = 0

Question 28

Using the factor theorem to factorise polynomials

Answer 28

Substitute chosen numbers into the polynomial until the value 0 is obtained (thus indicating a factor). Use guess and check (first try numbers that are factors of the constant term).

Question 29

Using the factor theorem to solve equations

Answer 29

1. Find the first factor where p(a) = 0
2. Use long division to divide the original equation by the answer to find the resultant equation.
3. Factorise the resultant equation.
4. Answer is the three brackets, all multiplied together
5. If there are powers above 3 in the equation, repeat the process twice, and the answer will be the four brackets, all multiplied together.

Question 30

Conjugate pairs

Answer 30

Roots are complex conjugate pairs if they are conjugates of each other. Any pair of complex conjugate roots gives a real polynomial.

[x - (a + ib)][x - (a - ib)] = x^2 - 2ax + (a^2 + b^2)

Question 31

Conjugate root theorem

Answer 31

The complex conjugate roots of any polynomial over the real numbers come in conjugate pairs. Thus, if p(x) is a polynomial over ℝ with a complex root α, then ˉα is also a root.

If p(α) = 0, then p(ˉα) = 0

Question 32

What does the conjugate root theorem hold for?

Answer 32

The conjugate root theorem holds only for polynomials with real coefficients.

Question 33

Using the conjugate root theorem

Answer 33

1. If (a + ib) is a root, then (a - ib) is also a root.
2. The third root will be real, and correspond to a factor in the form of (ax + b)
3. The coefficient of x^3 will be equal to ‘a’
4. The constant terms in the factors will multiply to the non-x term. Solve (a + ib)(a - ib)b = (non-x term)
   to calculate ‘b’
5. Rewrite third factor in the form (ax + b) using the ‘a’ and ‘b’ values found
6. Calculate the third root by isolating ‘x’
7. Rewrite all three factors in brackets as being multiplied together

Question 34

Polar form

Answer 34

r [cos (θ) + i sin(θ)]

= r cis (θ)

Question 35

Converting complex numbers from rectangular to polar form

Answer 35

1. Use tan^-1 (y/x) to find θ

2. Use √a^2 + b^2 to find r and determine the value of θ

Question 36

Converting complex numbers from polar to rectangular form

Answer 36

1. Use the equation a + ib = r [cos(θ) + i sin(θ)] to substitute in values

Question 37

Modulus and argument of z

Answer 37

|z| = r
arg(z) = θ

Question 38

Things to remember when completing complex number calculations (polar form)

Answer 38

• To get coordinates, solve equation for real and imaginary parts
• Use radians
• Remember to equate moduli and arguments (coefficients and angles) from the original and new equations to solve for each of the r and θ.
• Adjust arguments to their principal value (lying between -180° and 180°) by adding or subtracting 360° until it is

Question 39

Multiplication of complex numbers in polar form

Answer 39

To multiply complex numbers in polar form, multiply the moduli and add the arguments.

z1 z2 = r1 r2 cis(θ1 + θ2)

Question 40

Division of complex numbers in polar form

Answer 40

To divide complex numbers in polar form, divide the moduli and subtract the arguments.

z1/z2 = r1/r2 cis(θ1 - θ2)

Question 41

Powers of complex numbers in polar form

Answer 41

To raise complex numbers to a power in polar form, raise the moduli to the power and multiply the argument and power.

[r cis(θ)]^n = r^n cis(θn)

Question 42

Fundamental theorem of algebra

Answer 42

If
- p(x) = i=nΣi=0 ai x^i
- ai are real numbers
There will be n roots over ℂ (may be repeated roots)

Question 43

Things to remember when factorising polynomials

Answer 43

• Use the quadratic equation to factorise if needed

- Re-write x solutions (x = ± __) to equal 0 for final answer (x ± __ = 0)