How do you add $(7 + 9 i) + (2 + i)$ $(7+9i)+(2+i)$ in trigonometric form?

Question

1662 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-24T11:49:06

$= \sqrt{181} \cdot (cos (0.83798) + i sin (0.83798))$ $= sqrt(181) *(cos(0.83798)+i sin(0.83798))$

It is usually easier to simplify cartesian form and then convert to trigonometric afterwards,

$(7 + 9 i) + (2 + i)$ $(7+9i)+(2+i)$

$9 + 10 i$ $9+10i$

now we convert to trigonometric.

$9 + 10 i = r \cdot c i s (θ)$ $9+10i = r*cis(theta)$

$r = \sqrt{9^{2} + 10^{2}}$ $r=sqrt(9^2+10^2)$

$r = \sqrt{181}$ $r=sqrt(181)$

$θ = {tan}^{- 1} (\frac{10}{9})$ $theta=tan^-1(10/9)$

$θ = 0.83798$ $theta=0.83798$

giving us,

$(7 + 9 i) + (2 + i) = \sqrt{181} \cdot c i s (0.83798)$ $(7+9i)+(2+i) = sqrt(181) *cis(0.83798)$

or in expanded form,

$= \sqrt{181} \cdot (cos (0.83798) + i sin (0.83798))$ $= sqrt(181) *(cos(0.83798)+i sin(0.83798))$

How do you add (7+9i)+(2+i)(7+9i)+(2+i) in trigonometric form?