# What is the trigonometric form of  (-3+12i) ?

Feb 4, 2016

$\sqrt{153} \left[\cos \left(1.81\right) + i \sin \left(1.81\right)\right]$

#### Explanation:

To convert to trig. form , require r , the modulus and $\theta ,$
the argument.

• r =sqrt(x^2 + y^2)

• theta = tan^-1 (y/x)

Here x = -3 and y = 12

$\Rightarrow r = \sqrt{{\left(- 3\right)}^{2} + {12}^{2}} = \sqrt{9 + 144} = \sqrt{153}$

[ -3 + 12i is a point in the 2nd quadrant and care must be taken to ensure that $\theta \textcolor{b l a c k}{\text{ is in this quadrant}}$]

$\theta = {\tan}^{-} 1 \left(\frac{12}{-} 3\right) = {\tan}^{-} 1 \left(- 4\right) = - 1.33 \textcolor{b l a c k}{\text{ radians}}$

and so $\theta = \left(\pi - 1.33\right) = 1.81 \textcolor{b l a c k}{\text{ radians}}$

$\Rightarrow \left(- 3 + 12 i\right) = \sqrt{153} \left[\cos \left(1.81\right) + i \sin \left(1.81\right)\right]$