Unit 1: Trigonometry

Polar Representation of Complex Numbers

A complex number z = x + iy can be represented in the Cartesian plane (Argand diagram) by the point (x, y).

Modulus and Argument

We can also represent this point using polar coordinates (r, θ).

• Modulus (r): The distance from the origin to the point (x, y). r = |z| = √(x² + y²)
• Argument (θ): The angle made by the line joining the origin to (x, y) with the positive real (x) axis. θ = arg(z) = tan⁻¹(y/x)
  Important: The argument `θ` must be adjusted based on the quadrant of (x, y). For example, if z = -1 - i, (x, y) is in Quadrant III. tan⁻¹(-1/-1) = tan⁻¹(1) = π/4, but the correct angle is 5π/4 or -3π/4.

Polar Form

From the diagram, we can see that `x = r cos(θ)` and `y = r sin(θ)`. Substituting this into `z = x + iy` gives the polar form:

z = r(cos θ + i sin θ)

Euler's Form

Using Euler's formula, `eⁱᶿ = cos θ + i sin θ`, we get the exponential (or Euler's) form:

z = r eⁱᶿ

De Moivre's Theorem for Rational Indices

Statement (for integer index n)

For any real number θ and any integer n: (cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ)

Extension to Rational Indices (p/q)

The theorem also holds for rational indices. If `n = p/q` (where p, q are integers, q ≠ 0), then `(cos θ + i sin θ)ᵖ/q` is one of the values of `cos(pθ/q) + i sin(pθ/q)`.

More precisely, to find all q roots, we must use the general form of the argument:

θ = θ + 2kπ (where k is any integer)

So, `(cos θ + i sin θ)¹/q` has `q` distinct values, found by:

[cos(θ + 2kπ) + i sin(θ + 2kπ)]¹/q = cos((θ + 2kπ)/q) + i sin((θ + 2kπ)/q)
for k = 0, 1, 2, ..., q-1

n-th Roots of Unity

This is a direct application of De Moivre's theorem to solve the equation zⁿ = 1.

Solving zⁿ = 1

1. Write 1 in its polar form: `1 = 1 + i(0)`
   Modulus `r = 1`. Argument `θ = 0`.
2. Write 1 in its general polar form:
   `1 = cos(0 + 2kπ) + i sin(0 + 2kπ)`
3. Solve for z: `z = (1)¹/n` z = [cos(2kπ) + i sin(2kπ)]¹/n
4. Apply De Moivre's Theorem: z = cos(2kπ/n) + i sin(2kπ/n)
5. Find the `n` distinct roots by using `k = 0, 1, 2, ..., n-1`.

The Roots

The n-th roots of unity are: 1, eⁱ(²π/n), eⁱ(⁴π/n), ..., eⁱ(²(n-1)π/n)

If we let `α = eⁱ(²π/n)`, the roots are `1, α, α², ..., αⁿ⁻¹`.

Geometric Interpretation

The n-th roots of unity are the vertices of a regular n-sided polygon inscribed in the unit circle `|z| = 1` in the Argand plane, with one vertex at the point (1, 0).

Expansions of sin(nθ), cos(nθ), tan(nθ)

We can use De Moivre's Theorem and the Binomial Theorem to expand `cos(nθ)` and `sin(nθ)` in terms of powers of `cos θ` and `sin θ`.

Method

1. Start with De Moivre's Theorem:
   `cos(nθ) + i sin(nθ) = (cos θ + i sin θ)ⁿ`
2. Expand the right side using the Binomial Theorem:
   `(cos θ)ⁿ + nC₁(cos θ)ⁿ⁻¹(i sin θ)¹ + nC₂(cos θ)ⁿ⁻²(i sin θ)² + ...`
3. Simplify the powers of `i` (remember `i² = -1`, `i³ = -i`, `i⁴ = 1`, etc.).
   `= (cosⁿθ - nC₂cosⁿ⁻²θ sin²θ + nC₄cosⁿ⁻⁴θ sin⁴θ - ...) + i (nC₁cosⁿ⁻¹θ sinθ - nC₃cosⁿ⁻³θ sin³θ + ...)`
4. Equate the Real and Imaginary parts:

   Real Part: cos(nθ) = cosⁿθ - nC₂cosⁿ⁻²θ sin²θ + nC₄cosⁿ⁻⁴θ sin⁴θ - ...
   Imaginary Part: sin(nθ) = nC₁cosⁿ⁻¹θ sinθ - nC₃cosⁿ⁻³θ sin³θ + nC₅cosⁿ⁻⁵θ sin⁵θ - ...

5. For tan(nθ):
   Use `tan(nθ) = sin(nθ) / cos(nθ)`. Divide the expansions. A common trick is to divide the numerator and denominator by `cosⁿθ` to get an expansion in terms of `tan θ`.

Expansions of sinⁿ(θ) and cosⁿ(θ)

This is the reverse problem: expressing powers of `sin θ` or `cos θ` in terms of sines or cosines of multiples of θ (e.g., `cos 2θ`, `sin 3θ`).

Method (Euler's Identities)

1. Let `z = cos θ + i sin θ`.
2. From this, we know: z⁻¹ = cos(-θ) + i sin(-θ) = cos θ - i sin θ
   z + z⁻¹ = 2 cos θ
   z - z⁻¹ = 2i sin θ

   And for multiples of θ:

   zⁿ = cos(nθ) + i sin(nθ)
   z⁻ⁿ = cos(nθ) - i sin(nθ)
   zⁿ + z⁻ⁿ = 2 cos(nθ)
   zⁿ - z⁻ⁿ = 2i sin(nθ)
3. Use these to write the power (e.g., `cosⁿθ`) in terms of `z`.
   Example: `cosⁿθ = [(z + z⁻¹)/2]ⁿ`
4. Expand the right side using the Binomial Theorem.
5. Group the terms in pairs: `(zⁿ + z⁻ⁿ)`, `(zⁿ⁻² + z⁻(n-²))`, etc.
6. Substitute back: `zⁿ + z⁻ⁿ = 2 cos(nθ)`, etc.

Case: cosⁿ(θ) (n is even)

The expansion will be of the form: `A cos(nθ) + B cos((n-2)θ) + ... + K` (all cosine terms).

Case: cosⁿ(θ) (n is odd)

The expansion will be of the form: `A cos(nθ) + B cos((n-2)θ) + ... + K cos(θ)` (all cosine terms).

Case: sinⁿ(θ) (n is even)

The expansion will be of the form: `A cos(nθ) + B cos((n-2)θ) + ... + K` (all cosine terms, and signs will alternate).

Case: sinⁿ(θ) (n is odd)

The expansion will be of the form: `A sin(nθ) + B sin((n-2)θ) + ... + K sin(θ)` (all sine terms, and signs will alternate).