How do you add $(5 + 4 i) + (- 2 - i)$ $(5+4i)+(-2-i)$ in trigonometric form?

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How do you add $(5 + 4 i) + (- 2 - i)$ $(5+4i)+(-2-i)$ in trigonometric form?

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Answer 1 · 2016-05-29T01:15:56

$- \sqrt{\frac{41}{5}} (cos θ + i sin θ)$ $-sqrt(41/5) (cos theta+isin theta)$ , where $θ = {tan}^{- 1} (\frac{3}{14}) or arctan (\frac{3}{14})$ $theta=tan^(-1) (3/14)or arctan(3/14)$

Explanation:

$\frac{5 + 4 i}{- (2 + i)} = \frac{(5 + 4 i) (2 - i)}{- (2 + i) (2 - i)}$ $(5+4i)/(-(2+i))= ((5+4i)(2-i))/(-(2+i)(2-i))$

= $\frac{14 + 3 i}{- 5} = - \frac{1}{5} (14 + 3 i)$ $(14+3i)/(-5) = -1/5 (14+3i)$

= $- \frac{1}{5} \cdot \sqrt{205} (\frac{14}{\sqrt{205}} + \frac{3}{\sqrt{205}} i)$ $-1/5 *sqrt205 ( 14/sqrt205 +3/sqrt205 i)$

= $- \sqrt{\frac{41}{5}} (cos θ + i sin θ)$ $-sqrt(41/5) (cos theta+isin theta)$ , where $θ = {tan}^{- 1} (\frac{3}{14}) or arctan (\frac{3}{14})$ $theta=tan^(-1) (3/14)or arctan(3/14)$

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How do you add $(5 + 4 i) + (- 2 - i)$ $(5+4i)+(-2-i)$ in trigonometric form?

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How do you add $(5 + 4 i) + (- 2 - i)$ $(5+4i)+(-2-i)$ in trigonometric form?