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

Question

1654 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-07T15:19:27

See below.

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

$(8 + 8i) + (-4 + 6i) = 8 - 4 + 8i + 6i = 4 + 14i$

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

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

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

$r = sqrt252 = 2sqrt53$

Find $theta$ :

$tan theta = b/a$

$tan theta = 14/4$

$theta = tan^-1(14/4) ~~ 1.29$

In trigonometric form, this is $r(cos theta + i sin theta)$ or in shorthand,
$r$ $cis$ $theta$ .

Thus the answer is $2sqrt53 (cos 1.29 + i sin 1.29)$ or $2sqrt 53$ $cis$ $1.29$ .

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