# A calculus of the absurd

#### 15.3 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

$$z := r \rbrackets {\cos (\theta ) + i \sin (\theta )}$$

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

$$1 + i = \sqrt {2} e ^{\frac {\sqrt {2}}{2}i}$$

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).