# How do you add (8+8i)+(-4+6i) in trigonometric form?

Oct 7, 2017

See below.

#### Explanation:

To convert complex numbers to trigonometric form, find $r$, the distance of the point away from the origin, and $\theta$, the angle.

$\left(8 + 8 i\right) + \left(- 4 + 6 i\right) = 8 - 4 + 8 i + 6 i = 4 + 14 i$

$4 + 14 i$ is in the form $a + b i$. First, find $r$:

${r}^{2} = {a}^{2} + {b}^{2}$

${r}^{2} = {4}^{2} + {14}^{2}$

$r = \sqrt{252} = 2 \sqrt{53}$

Find $\theta$:

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

$\tan \theta = \frac{14}{4}$

$\theta = {\tan}^{-} 1 \left(\frac{14}{4}\right) \approx 1.29$

In trigonometric form, this is $r \left(\cos \theta + i \sin \theta\right)$ or in shorthand,
$r$ $c i s$ $\theta$.

Thus the answer is $2 \sqrt{53} \left(\cos 1.29 + i \sin 1.29\right)$ or $2 \sqrt{53}$ $c i s$ $1.29$.