## Trigonometry With Complex Numbers

Take a pencil and place it along the edge of a ruler, so that the ruler and the pencil are parallel, and so that the end of the eraser is at zero, and the point of the pencil points towards increasing numbers on the ruler.

Recall that a polynomial of degree two is called a **Quadratic Polynomial**.

A general quadratic polynomial (over the real numbers) is of the form:

\[ az^{2} + bz + c = 0. \]

The formula for solving for the roots of this equation is:

\[ z = \frac{-b \pm \sqrt{b^{2} -4ac}}{2a} .\]

Two complex numbers of the form:

\[ z_+ = x + iy ,\]

and,

\[ z_- = x - iy ,\]

are called **Complex Conjugates**.

For any complex number \[ z = x + iy ,\]

\[ z \bar{z} = (x + iy)(x -iy) \]

\[ = x^{2} + ixy - ixy + y^{2} = x^{2} + y^{2} .\]

That is, any complex number times its complex conjugate equals the square of its real part plus the square of its imaginary part.

Because both the real and imaginary parts of any complex number are real:

\[ z \bar{z} = x^{2} + y^{2} \geq 0, \forall z \in \mathbb{C}.\]

We define the positive square root of \[ z \bar{z}: \]

\[ +\sqrt{z\bar{z}} : = |z| ,\]

to be the **Modulus **of the complex number \[ z .\]

(Note: this is a quick and dirty introduction to exponentials. I will not bother with any epsilons or deltas, but for those of you who are concerned about such things, be assured that all convergence is uniform, all sequences are Cauchy, etc. The series that defines the exponential is about as well behaved as an infinite series can be.)

The function \[ e^{x} \]is defined to be the limit as \( n \) approaches infinity of \[ ( 1 + \frac{x}{n})^{n}: \]

exp(x) = \[ e^{x} = \lim_{n \to \infty} (1 + \frac{x}{n})^{n} .\]

Consider first the exponential of a constant, a:

\[ e^{a} = \lim_{n \to \infty} (1 + \frac{a}{n})^{n} = \sum_{k=0}^{\infty} \frac{a^{k}}{k!}. \]

If b is another constant, what is \[ e^{a}e^{b} ? \]

Let's go back to the definition of exponential:

\[ e^{a}e^{b} \ = \lim_{n \to \infty} f(n) .\]

Where:

\[ f(n) = (1 + \frac{a}{n})^{n} (1 + \frac{b}{n})^{n} = ( 1 + \frac{ a + b}{n} + \frac{ab}{n^{2}})^{n} \]

\[ = \sum_{k=0}^{n} \binom{n}{k} c^{k}, \]

Consider the exponential of a purely imaginary number:

\[ z = i\alpha \]

with \[ \alpha \in \mathbb{R}\ .\]

\[ e^{i\alpha} = \lim_{n \to \infty } (1 + \frac{i\alpha}{n})^{n} \]

\[ = \sum_{k=0}^{\infty }\frac{i\alpha^{k}}{k!} . \]

From the definition of cosine:

\[ \cos{\theta} =\frac{e^{i\theta} + e^{-i\theta}}{2} \]

we find:

\[ 2\cos{(\alpha + \beta)} = e^{i(\alpha + \beta)} + e^{-i(\alpha + \beta)} .\]

The real part of \[ e^{i\alpha} \ \]

is:

\[ \text{Re}(e^{i \alpha}) = \frac{ e^{i\alpha} + e^{-i\alpha}}{2} \]

\[ =\sum_{k=0}^{\infty} (-1)^{k}\frac{\alpha^{2k}}{(2k)!} . \]

The imaginary part of \[ e^{i\alpha} \ \]

is:

\[ \text{Im}(e^{i\alpha}) = \frac{ e^{i\alpha} - e^{-i\alpha}}{2i} \]

\[ = \sum_{k=1}^{\infty}(-1)^{k -1}\frac{\alpha^{2k -1}}{(2k -1)!} . \]

We define:

\[ \text{Re}(e^{i\alpha}) = \frac{ e^{i\alpha} + e^{-i\alpha}}{2} := \cos{\alpha} ;\]

\[ \text{Im}(e^{i\alpha}) = \frac{ e^{i\alpha} - e^{-i\alpha}}{2i} := \sin{\alpha}.\]

Therefore:

\[ e^{i\alpha} = \text{Re} (e^{i\alpha}) + i\text{Im}(e^{i\alpha}) =\cos{\alpha} + i \sin{\alpha}. \]

This is Euler's Formula.

Euler's formula makes life much easier when it comes to deriving trigonometric identities.

Note, first of all, that \[ |e^{i\alpha}| = 1, \]gives us:

\[ \left( \text{Re} (e^{i\alpha}) + i\text{Im}(e^{i\alpha}) \right) \left( \text{Re}(e^{i\alpha}) – i\text{Im}(e^{i\alpha}) \right) \]

\[ = \left( \cos{\alpha} + i\sin{\alpha} \right) \left( \cos{\alpha} - i\sin{\alpha} \right) \]

\[ = \cos^{2}{\alpha} + \sin^{2}{\alpha} = 1. \]

Consider any complex number \[ z, \ \]with modulus

\[ \sqrt{z\bar{z} } = |z| = r \ ,\]

where \[ r \in \mathbb{R^{+}}\ . \]

Note that \[ \frac{z}{r} \ \]would be a complex number with modulus one:

\[ | \frac{z}{r} | = | \frac{z}{|z|} | = 1. \]

We can find the half angle formula for cosine by starting with the double angle formula:

\[ \cos{(2\alpha)} = \cos^{2}{\alpha} - \sin^{2}{\alpha} \ . \]

Making the substitution:

\[ \gamma = 2\alpha ,\]

leads to the equation:

\[ \cos{\gamma} = \cos^{2}{(\frac{\gamma}{2})} - \sin^{2}{(\frac{\gamma}{2})} . \]