Simplifying trigonometric products into trigonometric sums

It can be hard to differentiate products of trigonometric functions like this one: \[ \cos{y}\sin{y} \] It is easier when the functions are separated like this: \[ \cos{y}+\sin{y} \] for example. Therefore I will here describe how to transform equations of the first form into equations of the second form using the Euler's formula.

Euler's formula lets us transform the $\sin$ and $\cos$ parts into this: \[ \begin{aligned} \DeclareMathOperator{\e}{e} \cos(y) &= \frac{\e^{iy} + \e^{-iy}}{2} \\ \sin(y) &= \frac{\e^{iy} - \e^{-iy}}{2i} \end{aligned} \]

By doing this we can make use of the rules for products of $\e$, which are preserved for complex numbers: \[ \e^{z_1}\;\e^{z_2} = \e^{z_1+z_2} \] and magically doing this will end up with some parts that can be put back on the regular trigonometric form be reversing Euler's formula.

Worked example

I will now rewrite trigonometric products like this: \[ \sin{3x}\cos{5x} = \frac{1}{2} (\sin{8x} - \sin{2x}). \]

First, I will rewrite the left side using Euler's formula: \[ \left( \frac{\e^{3xi} - \e^{-3xi}}{2i} \right) \left( \frac{\e^{5xi} + \e^{-5xi}}{2} \right). \]

I pull the denominator out in front, and use the product rules of $\e$: \[ \begin{aligned} &\frac{1}{4i} (\e^{3xi + 5xi} + \e^{-3xi + 5xi} - \e^{-3xi + 5xi} - \e{-3xi - 5xi}) \\ = &\frac{1}{4i}(\e^{8xi} + \e^{-2xi} - \e{2xi} - \e^{-8xi}) \end{aligned} \]

I break it up in two parts: \[ \frac{1}{4i} (\e^{8xi} - \e^{-8xi}) + \frac{1}{4i} (\e^{-2xi} - \e^{2xi}) \]

Now what we are trying to do is to get it on the form defined by Euler's formula. Because there is a minus between the $\e$-s in both the first and the second part, we assume that we want to go back to sinus in both cases. Sinus was defined as: \[ \sin(y) = \frac{\e^{iy} - \e^{-iy}}{2i} \] and therefore part of the denominator of $\frac{1}{4i}$, is “pulled” into the equation like this: \[ \frac{1}{2} \left(\frac{\e^{8xi} - \e^{-8xi}}{2i}\right) + \frac{1}{2} \left(\frac{\e^{-2xi} - \e^{2xi}}{2i}\right) \]

Now the “equation” is on the Euler's form, and can be put back as an ordinary trigonometric function. Taking the signing into condition, it ends up like this: \[ \frac{1}{2} (\sin{8x} - \sin{2x}) \] and thereby the first statement is proved.

Solving roots in binomial complex equations

In this post I will describe how to solve binomial complex equations of the form \[ z^n = a + bi. \]

This is done by first putting the right side on polar form using the following: \[ \begin{aligned} R &= \sqrt{a^2 + b^2} \\ \theta &= \left\{ \begin{aligned} &\tan^{-1}\frac{b}{a} &\text{if}\; a > 0 \\ &\tan^{-1}\frac{b}{a} + \pi &\text{if}\; a < 0 \\ &\frac{\pi}{2} &\text{if}\; a = 0, b > 0 \\ &-\frac{\pi}{2} &\text{if}\; a = 0, b < 0 \\ &\text{random} &\text{if}\; a = 0, b = 0 \end{aligned} \right. . \end{aligned} \]

The roots are now found using the following: \[ z = \left( \sqrt[n]{R} \right)_{\frac{a}{n} + p\frac{2\pi}{n}} \qquad p = 0, 1, \ldots, n-1 \] where p is altered from $0$ to $n-1$, to find all roots.

Worked example

Given: \[ z^4 = 3+4i \]

Finding $R$: \[ \begin{aligned} R &= \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \\ R &= 5 \end{aligned} \]

Finding $\theta$ (note: $a = 3 \Leftrightarrow a > 0$): \[ \theta = \tan^{-1}\frac{4}{3} \]

The roots are therefore: \[ \begin{aligned} z_0 &= \left( \sqrt[4]{5} \right)_{\frac{3}{4}} \approx 1.495_{0.75}\\ z_1 &= \left( \sqrt[4]{5} \right)_{\frac{3}{4} + 1\frac{2\pi}{4}} \approx 1.495_{2.32}\\ z_2 &= \left( \sqrt[4]{5} \right)_{\frac{3}{4} + 2\frac{2\pi}{4}} \approx 1.495_{3.89}\\ z_4 &= \left( \sqrt[4]{5} \right)_{\frac{3}{4} + 3\frac{2\pi}{4}} \approx 1.495_{5.46} \end{aligned} \] in polar coordinates. To reverse to regular coordinates, just use the following: \[ \begin{aligned} a &= R\cos{\theta} \\ b &= R\sin{\theta} \end{aligned} \]

Looking at the complex plane the roots are distributed as follows:

Complex numbers: Roots in polynomials of degree two

Here is how to solve the roots in polynomials of degree two with complex numbers. After explaining the approach I will go through an example.

We have a polynomial of the form: \[ p(z) = az^2 + bz + c \] To find the roots of this, we need to know the discriminant $D$, to know if the roots are complex. This is the case if $D < 0$. \[ D = b^2 -4ac \] If $D \geq 0$ the root(s) are real, and may be found using the formula: \[ x = \frac{-b\pm \sqrt{D}}{2a} \] If the discriminant is negative or complex, we need replace $\sqrt{D}$ with $w$, which is the solution to the equation $z^2-D = 0$. This is found by using the formula: \[ w = \pm\left( \sqrt{\frac{r+\alpha}{2}} + i\,\text{sgn}(\beta)\sqrt{\frac{r-\alpha}{2}} \right) \] Where the discriminant is given as $D = \alpha + \beta i$ and $r = \sqrt{\alpha^2 + \beta^2}$. $\text{sgn}$ is $+1$ or $-1$ depending on the value of $\beta$: \[ \text{sgn}(\beta) = \left\{ \begin{aligned}1,\qquad \text{if}\; \beta \geq 0, \\ -1,\qquad \text{if}\; \beta < 0. \end{aligned} \right. \] Now the roots may be found using the formula: \[ z = \frac{-b \pm w}{2a} \] Using $a$, $b$, and $c$ from the original polynomial and $w$ just found.

Worked example

Here's a worked example with the polynomial \[ z^2 + 2z - (2+4i) = 0 \] First we find the discriminant: \[ D = b^2 - 4ac = 2^2 - 4\cdot 1\cdot (-(2+4i)) = 12 +16i \] As this is a complex determinant, we now need to find $w$. With the discriminant $D = 12 + 16i$, $r = \sqrt{12^2 + 16^2} = \sqrt{400} = 20$ and $\text{sqn}(\beta)=1$. \[ w = \pm\left( \sqrt{\frac{20+12}{2}} + i\cdot 1\cdot \sqrt{\frac{20-12}{2}} \right) = \pm \left( \sqrt{16} +i\sqrt{4} \right) = \pm (4-2i) \] This enables us to find the roots for the original polynomial: \[ \begin{align} z &= \frac{-b\pm w}{2a} = \frac{-2 \pm (4-2i)}{2\cdot 1} = \frac{-2}{2} \pm \frac{4-2i}{2} \\ z_1 &= -1 + 2-i = 1-i &\qquad\text{Using} + \\ z_2 &= -1 -2+i = -3+i &\qquad\text{Using} - \\ \end{align} \]