Complex Numbers Flashcards
What is the equation linking modulus argument form and exponential form?
r(cos(nθ) + isin(nθ)) = reiθ
How do you multiply and divide complex numbers in exponential form?
If z1 = r1eiθ1
If z2 = r2eiθ2
then
z1z2 = r1r2ei(θ1 + θ2)
and
z1/z2 = (r1/2)ei(θ1 - θ2)
What is demoivre’s theorem?
(r(cosθ + isinθ))n = rn(cos(nθ) + isin(nθ))
Note this is useful when putting complex numbers to powers
How do you prove trig identities in the form cos(nθ) or sin(nθ) ?
- 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
How do you prove trig identities in the form tan(nθ) ?
- 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
How do you prove trig identities in the form cosnθ or sinnθ ?
- 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 zk and 1/zk terms
- Use the fact that zn +1/zn = 2cos(nθ) or zn - 1/zn = 2isin(nθ) to sub into your expression
- Simplify and divide
How do you use demoivre’s theorem to prove that if z = eiθ then zn + 1/zn = 2cos(nθ)
- State zn + z-n
- Sub in eiθ 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
How do you use demoivre’s theorem to prove that if z = eiθ then zn - 1/zn = 2isin(nθ)
- State zn - z-n
- Sub in eiθ 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
How can you use demoivre’s tirg identity proof to solve trig equations?
You can simplify trig terms with powers in to trig terms without powers in and then solve
How do you find the nth roots of a complex number?
- State that zn = a + bi
- Convert a + bi into eiθ to get zn = eiθ
- 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 = eiθ/n…
- Now convert these terms into whatever form the question asks for
What are important facts about roots of unity?
- |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)
What are the nth roots of unity?
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
How do you solve a sums of series question in this form?
z = cos(π/n) + isin(π/n) show that 1 + z + z2 + … + zn-1 = 1 +icot(π/2n)
- This is a geometric sequence where a = 1 and r = z therefore
- 1 + z + z2… = 1(1-zn) / (1-z)
- Swap the zn for ei*π/n
- Simplify then multiply the top and bottom by ei*-π/2n
*
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 = 2e2iθ/ 2-2e4iθ
- 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’
How do you work out the modulus and argument of a complex number? e.g a+bi
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
What is the modulus and argument of a complex number?
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
What is modulus argument form?
z = r(cosθ + isinθ)
Where:
r = |z|
θ = arg(z)
How do you multiply and divide modulus argument form?
arg(z1z2) = arg(z1) + arg(z2)
arg(z1/z2) = arg(z1) - arg(z2)
|z1z2| = |z2||z2|
|z1/z2| = |z 1|/|z2|
What is the equation of a circle in the argand plane?
|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
What is the equation of a half line in the argand plane?
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)
What is the equation of a perpendicular bisector in the argand plane?
|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
How do you convert from argand form to cartesian form?
Using common sense
What is the complex conjugate of a complex number? e.g z = a + bi
z* = a - bi
How do you divide complex numbers? e.g
z1/z2
You rationalise the denominator
Multiply the top and bottom by z*2
