complex operations in trigonometric form

Complex operations in trigonometric form

2024-04-22

Complex numbers are just vectors, with an x-axis of the real numbers and a y-axis of the imaginary $\sqrt{-1}$ numbers. Of course, instead of a basic $(x,y)$ or $(r, \theta)$ , we just need to include the basis vectors (the default units) of our axes when we talk complex. In the x-axis, this is $1$ , and in the y-axis, this is $i = \sqrt{-1}$ .

”Rectangular form” is obvious: $a+bi$ comes from $(x,y)$ .

”Trigonometric form” is a bit funkier: $r(cos\theta + isin\theta)$ comes from…?

Actually, it comes from a drawing of a triangle!

First, let’s draw out some complex number with magnitude $r$ and angle $\theta$ . Connect that to the origin, and we can immediately SOHCAHTOA.

Of course, we can’t just leave things at $rsin\theta$ and $rcos\theta$ ; we need to attach the basis vectors on. Now, the hypotenuse’s absolute value is $r$ , but Pythagorean Theoreming will give us an $a+bi$ .

Trigonometric form isn’t really used to add complex numbers, but it sure is used to multiply them. Why? Well, trig identities give us the opportunity to add angles, letting us escape a soup of expanded $a+bi$ binomials when we multiply. Let’s see where they come from!

$Let\ z_1 = r_1(cos\alpha+isin\alpha)$
$Let\ z_2 = r_2(cos\beta+isin\beta)$

1. Multiplication

\begin{align} z_1 z_2 &= r_1(cos\alpha + isin\alpha) \bullet r_2(cos\beta + isin\beta) \notag \\ &= r_1r_2(cos\alpha cos\beta + icos\alpha sin\beta + isin\alpha cos\beta - sin\alpha sin\beta) \notag \\ &= r_1r_2(cos\alpha cos\beta - sin\alpha sin\beta + i(cos\alpha sin\beta + sin\alpha cos\beta))\notag \\ z_1 z_2 &= r_1r_2(cos(\alpha + \beta) + isin(\alpha + \beta)) \end{align}

Between lines 3 and 4, we used the angle addition identities.

2. Division

\begin{align} \frac{z_1}{z_2} &= \frac{r_1(cos\alpha + isin\alpha)}{r_2(cos\beta + isin\beta)} \quad \text{ Rationalize the denominator!}\notag \\ &= \frac{r_1(cos\alpha + isin\alpha)}{r_2(cos\beta + isin\beta)} \bullet \frac{cos\beta - isin\beta}{cos\beta - isin\beta}\notag \\ &= \frac{r_1}{r_2} \left(\frac{ cos\alpha cos\beta - icos\alpha sin\beta + isin\alpha cos\beta + sin\alpha sin\beta }{ (cos\beta)^2 - icos\beta sin\beta + isin\beta cos\beta + (sin\beta)^2 }\right) \notag \\ &= \frac{r_1}{r_2} \left(\frac{ cos\alpha cos\beta + sin\alpha sin\beta + i(sin\alpha cos\beta - cos\alpha sin\beta) }{ (cos\beta)^2 + (sin\beta)^2 } \right) \notag \\ &= \frac{r_1}{r_2} \left(\frac{ cos(\alpha - \beta) + isin(\alpha - \beta) }{ 1 }\right) \notag \\ \frac{z_1}{z_2} &= \frac{r_1}{r_2}(cos(\alpha - \beta) + isin(\alpha - \beta)) \end{align}

Between lines 4 and 5, we used the angle subtraction identities.

3. Exponentiation (DeMoivre’s Theorem)

When we have multiplication, exponentiation comes by easily.

\begin{align} z_1 \bullet z_1 &= r_1r_1(cos(\alpha + \alpha) + isin(\alpha + \alpha)) \notag \\ (z_1 \bullet z_1) \bullet z_1 &= r_1r_1r_1(cos(\alpha + \alpha + \alpha) + isin(\alpha + \alpha + \alpha)) \notag \\ z_1^n &= r_1^n(cos(n \bullet \alpha) + isin(n \bullet \alpha)) \end{align}

4. Rooting

Rooting is a bit interesting. While we can simply supply a fractional exponent to a complex number for the initial part of the rule,

\begin{align} z_1^{\frac{1}{n}} &= r_1^{\frac{1}{n}}(cos(\frac{1}{n}\alpha) + isin(\frac{1}{n}\alpha)) \notag \\ \sqrt[n]{z_1} &= \sqrt[n]{r_1}(cos(\frac{1}{n}\alpha) + isin(\frac{1}{n}\alpha)) \end{align}

there is something else we have to consider: multiple solutions.

Usually, with some caveats, we prefer the “principal solution” when solving for a real number’s root, i.e.
$\sqrt{4} = 2$ rather than $\sqrt{4} = -2$

However, with complex numbers we have to consider every root.
$\sqrt{-4} = \pm 2i$

Let’s think about what this means, based on our exponentiation rule. Imagine a complex number of some magnitude $r_0$ , that has an angle of $60\degree$ .
$z_0 = r_0(cos(60\degree) + isin(60\degree))$

If we want to find its cubic roots, we’ll need a complex number
$z = r(cos(\theta) + isin(\theta))$ where
$r^3 = r_0$ and
$3 \bullet \theta = 60\degree$ .

But angles are modular! They wrap around a circle!
$20\degree \times 3 = 60\degree$
$140\degree \times 3 = 420\degree \equiv 60\degree$
$260\degree \times 3 = 780\degree \equiv 60\degree$

Because of the circle’s modulus of 360 degrees, we therefore discover these properties for the solution angles of $\sqrt[n]{z}$ :

They’re spaced equally apart by $\frac{360}{n}$
The smallest angle is $\frac{\theta}{n}$ , and the largest angle must be below $360\degree$ (because otherwise we’d just subtract $360\degree$ anyway).

Hence, unless you prefer to use $2\pi$ over $360\degree$ , our final formula will look like this:

\sqrt[n]{z} = \sqrt[n]{r} \left( cos\left(\frac{\theta}{n} + k\frac{360}{n} \right) + isin\left(\frac{\theta}{n} + k\frac{360}{n} \right) \right)

where integer $k \geq 0$ and $k < n$ (so that $k\frac{360}{n}$ doesn’t surpass 360).