# What is the trigonometric form of  (-3+15i) ?

Jun 17, 2018

 -3 + 15 i = 3 sqrt{26} \ "cis" \ ( "Arc""tan"(-5) + 180^circ )

#### Explanation:

This one is in the second quadrant, so the principal value of the inverse tangent misses it.

Trigonometric form of $a + b i$ is

$r \setminus \text{cis} \setminus \theta = r \left(\cos \theta + i \sin \theta\right)$

where

$a = r \cos \theta \mathmr{and} b = r \sin \theta$

Squaring and adding we see

${a}^{2} + {b}^{2} = {r}^{2} {\cos}^{2} \theta + {r}^{2} {\sin}^{2} \theta = {r}^{2}$

We can divide and write $\theta = \arctan \left(\frac{b}{a}\right)$ but that's inadequate as it conflates pairs of quadrants. Let's write

$\theta = \text{arc""tan2} \left(b / , a\right)$

That's deliberately the two parameter, four quadrant inverse tangent. The funky "$/ ,$" is deliberate, reminding us this is two parameters and which is which. It works for $a = 0.$ If ${r}^{2} = {a}^{2} + {b}^{2}$ the four quadrant inverse tangent assures

$a = r \cos \theta \mathmr{and} b = r \sin \theta$

In the second quadrant we have

$\text{arc""tan2"(15 //, -3 ) = "Arc""tan} \left(- 5\right) + {180}^{\circ} \approx {101.3}^{\circ}$

$r = \setminus \sqrt{{\left(- 3\right)}^{2} + {\left(15\right)}^{2}} = 3 \sqrt{26}$

 -3 + 15 i = 3 sqrt{26} \ "cis" \ ( "Arc""tan"(-5) + 180^circ )

$- 3 + 15 i \approx 3 \sqrt{26} \setminus \text{cis} \setminus \left({101.3}^{\circ}\right)$