# A calculus of the absurd

#### 14.3 The trigonometric form of a complex number

When we say $$z := x + iy$$ (also known as “z is defined as $$x + iy$$”), $$x$$ and $$y$$ can be anything! For example, $$x$$ and $$y$$ could equal $$\cos (\theta )$$ and $$\sin (\theta )$$ respectively. By setting different values of theta, we can obtain any point on a circle with radius 1, and centre $$(0, 0)$$. For example, if we have $$z := \cos (\theta ) + i \sin (\theta )$$, then to obtain the number $$1$$, we can just set $$\theta = 0$$. To be able to obtain every complex number, however, we need to introduce a second variable (which we can call $$r$$). This “scales” the circle, so that for each value of $$r$$, the values of $$\theta$$ between $$0$$ and $$2\pi$$ ($$2\pi$$ is not inclusive) we can obtain a circle with that radius. Overall, we can write any complex number in the form

\begin{equation} z := r \rbrackets {\cos (\theta ) + i \sin (\theta )}. \end{equation}

Additionally, we can do this with $$r \geqq 0$$ and $$-\pi \leqq \theta \leqq \pi$$.

To find the trigonometric form of a given complex number there are two ways.

To convert a complex number (e.g. $$1 + 3i$$) into trigonometric form, the first way algebra is to use algebra (in particular “comparing coefficients”). By comparing $$r\cos (\theta ) + ri\sin (\theta )$$ with $$1 + 3i$$, we obtain that $$r \cos (\theta ) = 1$$ and that $$r\sin (\theta ) = 3$$. Thus

\begin{align} r^2 \cos ^2(\theta ) + r^2 \sin ^2(\theta ) &= r^2(\cos ^2(\theta ) + \sin ^2(\theta ) \notag \\ &= r^2 \notag \\ &= 2 \text { (from $1+3i$)} \notag \end{align}

from which we can deduce that $$r=\pm \sqrt {2}$$. The other thing we can do is divide through, thus obtaining that $$\frac {r \sin (\theta )}{r \cos (\theta )} = \frac {1}{1}$$, and thus that $$\theta = \arctan (1)$$ (which is $$\frac {\sqrt {2}}{2}$$). Overall, we can then write that

\begin{equation} 1 + i = \sqrt {2} e ^{\frac {\sqrt {2}}{2}i} \end{equation}

which we can also do for any complex number.

The second way involves geometry (I still need to find where I originally wrote my notes on this, but the method is to draw the complex number - not necessary, but usually helpful - and to then find the modulus and argument of the complex number). First we can apply this useful fact:

• Theorem 14.3.1 For a complex number $$z = r (\cos (\theta ) + i\sin (\theta ))$$, $$\arg (z) = \theta$$ and $$|z| = r$$

The proof is definitely not relevant for any A Level examination, but it’s also not too difficult.

• • First we prove that $$\arg (z) = \theta$$. We know that for a complex number $$x + iy$$, there are several cases for the argument (rather unfortunately we need to consider each quadrant separately)

• – In the first case, $$x, y \geqq 0$$, where $$\arg (z) = \arctan \left (\frac {y}{x}\right )$$. For our $$z$$, this means

\begin{align} \arg (z) &= \arctan \left (\frac {r\sin (\theta )}{r\cos (\theta )}\right ) \\ &= \arctan (\tan (\theta )) \\ &= \theta \end{align}

• TODO: go through the other three cases

• • Second we prove that $$|z| = r$$. This proof is quite a bit more satisfying than the previous proof, we just apply the definition of the modulus, that is that

\begin{align} z^2 &= \sqrt {x^2 + y^2} \\ &= \sqrt {\Big (\big (r\cos (\theta )\big )^2 + \big (r\sin (\theta )\big )^2\Big )} \\ &= \sqrt {r^2 \big (\cos ^2(\theta ) + \sin ^2(\theta )\big )} \\ &= \sqrt {r^2} \end{align} Note that we also defined $$r \geqq 0$$ (well, hopefully we did) and thus

\begin{equation} \sqrt {r^2} = r \end{equation}

##### 14.3.1 Using the trigonometric form of a complex number
• Example 14.3.1 This is probably not the most exciting example, but we might want to consider the case of

\begin{equation} z = 2(\cos (330^{\circ }) - \sin (330^{\circ })) \end{equation}

If we can write this in trigonometric form, then by 14.3.1 we can just read the argument and modulus. Examining the expression, it looks pretty close to something in the form

\begin{equation} z = r (\cos (\theta ) + \sin (\theta )). \end{equation}

The only problem we have is that there’s a pesky negative hanging around in there, but not to worry, we can apply two useful facts here; $$\cos (\theta ) = \cos (-\theta )$$ and $$\sin (\theta ) = -\sin (-\theta )$$ (these two equations are referred to as the “even” and “odd” property of $$\cos$$ and $$\sin$$ respectively - see Section 14.7.3 for a proof of this).

Therefore, we can rewrite $$z$$

\begin{align} z &= 2(\cos (330^{\circ }) - \sin (330^{\circ })) \\ &= 2(\cos (-330^{\circ }) + \sin (-330^{\circ })) \\ &= 2(\cos (-30^{\circ }) + \sin (-30^{\circ })) \end{align}

Therefore the argument is $$-30^{\circ } = -\frac {\pi }{6} \text { rad}$$