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
\(\seteqnumber{0}{14.}{7}\)\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
\(\seteqnumber{0}{14.}{8}\)\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
\(\seteqnumber{0}{14.}{8}\)\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:
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
\(\seteqnumber{0}{14.}{9}\)\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
\(\seteqnumber{0}{14.}{12}\)\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
\(\seteqnumber{0}{14.}{16}\)\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
\(\seteqnumber{0}{14.}{17}\)\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
\(\seteqnumber{0}{14.}{18}\)\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 \(z\)
\(\seteqnumber{0}{14.}{19}\)\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 \(30^{\circ } = \frac {\pi }{6} \text { rad}\)