# How do you write the trigonometric form of 2-2i?

Mar 28, 2017

$z = 2 \sqrt{2} \left[\cos 7 \frac{\pi}{4} + i \sin 7 \frac{\pi}{4}\right]$

#### Explanation:

We are trying to get it to the form; $z = r \left(\cos \theta + i \sin \theta\right)$, usually called $z = r c i s \theta$

Where;
The complex number is $z = a + b i$

The modulus (or absolute value) of the complex number $z$ is $r = | z | = \sqrt{{a}^{2} + {b}^{2}}$

And $\tan \theta = \frac{b}{a}$

Now to the question, first we find the modulus of $| z |$ or $r$

$z = 2 - 2 i$

$| z | = r = \sqrt{{2}^{2} + \left(- {2}^{2}\right)} = \sqrt{4 + 4} = \sqrt{8} = 2 \sqrt{2}$

Let's find $\theta$

$\tan \theta = \frac{- 2}{2} = - 1$

We know that $\tan \theta = \frac{o p p o s i t e}{a \mathrm{dj} a c e n t} = \sin \frac{\theta}{\cos} \theta$,

when $r = 1$ for the latter.

If you sketch a right angled triangle and put $- 2$ opposite $\theta$ and $2$ adjacent to $\theta$, the hypotenuse will be $- \sqrt{2}$.

If you are familiar with the unit circle, you will know that the reference angle for the triangle is 45° or pi/4.

If you are not familiar with the unit circle, using the triangle drawn,

we know this because $\cos = \frac{a \mathrm{dj} a c e n t}{h y p o t e n u s e} = \frac{\sqrt{2}}{2}$ and

$\sin = \frac{o p p o s i t e}{h y p o t e n u s e} = \frac{\sqrt{2}}{2}$ which are the values for 45°

Here though the tangent of 45° or pi/4 is $1$ and not $- 1$.

How do we figure out what gives us a tangent of $- 1$?

Easy! We know that $a = 2$ and $b = - 2$.

If $a$, the abscissa (what we normally refer to as the $x$- coordinate) is positive, then it means we are looking at Quadrant $I$ or $I V$.

If $b$, the ordinate (what we normally refer to as the $y$-coordinate) is negative, then it means we are looking at Quadrant $I V$.

Thus, we know that the complex number $z$ lies in Quadrant $I V$.

If it lies in $I V$, then the angle that makes tangent $- 1$ is;

theta=315° or 7pi/4 because that is the 45° or pi/4 angle in $I V$

Now, we have the modulus and $\theta$ so we can go ahead and substitute it into our trig form of $z$.

$z = 2 \sqrt{2} \left[\cos \left(7 \frac{\pi}{4}\right) + i \sin \left(7 \frac{\pi}{4}\right)\right]$

I hope this was well explained...