How do you add $(8 - 8 i) + (- 7 - i)$ $(8-8i)+(-7-i)$ in trigonometric form?

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Answer 1 · 2017-02-15T17:41:32

$1 - 9 i = \sqrt{82} (cos (arctan (- 9)) + i sin ((arctan (- 9)))$ $1-9i = sqrt(82)(cos(arctan(-9)) + i sin ((arctan (-9)))$

Add them in rectangular form $(a + b i)$ $(a + bi)$ and then convert to trigonometric form:

Combine the like terms: $(8 + - 7) + (- 8 i + - i) = 1 - 9 i$ $(8 + -7) + (-8i + -i) = 1-9i$
$r = \sqrt{1^{2} + {(- 9)}^{2}} = \sqrt{82}$ $r = sqrt (1^2 + (-9)^2) = sqrt (82)$
$θ = arctan (\frac{b}{a}) = arctan (- \frac{9}{1}) = arctan (- 9)$ $theta = arctan(b/a) = arctan (-9/1) = arctan(-9)$
$z = r (cos θ + i sin θ)$ $z = r(cos theta + i sin theta)$
$= \sqrt{82} (cos (arctan (- 9)) + i sin ((arctan (- 9)))$ $= sqrt(82)(cos(arctan(-9)) + i sin ((arctan (-9)))$

How do you add (8−8i)+(−7−i)(8-8i)+(-7-i) in trigonometric form?