Complex Numbers Flashcards
Exponential form
re^iθ = r(cos θ + i sin θ)
Complex to exponential
Find the mod (r) and the argument (θ) and substitute into the exponential format
Exponential to complex
Put into modulus-argument form and then into the complex form
Multiply and divide in exponential form
Multiply/divide the mods and add/subtract the args before writing in exponential form
When to simplify mod-arg
When there are simple values for θ that give exact values e.g. π/2
De Moivre’s theorem
r(cos(θ) + isin(θ))^n
r^n(cos(nθ) + isin(nθ))
Proving de Moivre’s theorem by induction
do z^k x z and simplify to r^k+1(cos(k+1)θ + isin(k+1)θ) using the addition formulae
de Moive’s theorem
re^iθ
n inθ
r^e^
(x + yi)^n
Write x + yi in mod-arg and apply de Moivre’s theorem
Express cos/sin nθ in terms of powers of cos/sin nθ
- Use de Moivre’s theorem with n as the power
- Expand (cosθ + isinθ)^n using binomial expansion
- Set the real/imaginary part of each side equal
- Simplify to be in terms of cos/sin
1
z^n + —-
z^n
2cosnθ
1
z^n - —-
z^n
2i sin(nθ)
cos/sin^nθ in terms of cos/sin nθ
- Use (2cosθ)^n = (z + 1/z)^n or (2sinθ) = (z - 1/z)^n
- Expand both sides remembering the 2/2i
- Group the RHS with z^n +/- 1/z^n
- Use the identities for (z +/- 1/z) and substitute
- Divide both sides by the coefficient on the LHS
If it has both cos and sin expand both terms and multiply
Showing z^n +/- z^-n is 2cosθ or 2isinθ
Use de Moivre’s theorem with n and -n, simplifying the negative signs
n-1
∑ w z^r
r=0
w(z^n-1)/(z - 1)
∞
∑ w z^r if |z| < 1
r=0
w/(z-1)
Simplifying sums to an-1
- Write out the sum of series rule
- Replace w with the first term and z with its exponential form, using e^πi = -1
- Multiply top and bottom by e^(-1/2 x power in denominator)
- Use the 2cosθ and 2isinθ rules in the denominator
- Multiply top and bottom by i, use I^2 = -1 and simplify
- Put numerator in mod-arg and simplify to the needed form
e^πi
-1
z^n= x + iy method
- Put the RHS in mod-arg form
- Write each θ as (θ + 2kπ)
- Use de Moivre’s theorem to raise each side by 1/n
- Substitute k=0, k=1… for n values of k and put in principal argument form
Geometric problems
No matter what the roots are the ratio between them is the same as the roots of unity for that power
ω = cos(2π/n) + isin(2π/n)
Multiply by ω to get as many points as necessary, use exponential form
Series of cos and sin from an infinite series
Find the real and imaginary parts of the series
Infinite series from (cos θ + cos 2θ + cos 3θ) + i(sin θ + sin 2θ + sin 3θ)
Write as z + z^2 + z^3 + …
Where z = e^iθ
A 1 first can be z^0
Infinite series from (cos θ + kcos 2θ + k^2cos 3θ) + i(sin θ + ksin 2θ + k^2sin 3θ)
Write as regular sum of series with e^iθ as the numerator
Multiply by the denominator with the power of e inversed
Write in mod arg and simplify with real and imaginary