How do you add $(6 - i)$ $(6-i)$ and $(- 7 + 2 i)$ $(-7+2i)$ in trigonometric form?

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How do you add $(6 - i)$ $(6-i)$ and $(- 7 + 2 i)$ $(-7+2i)$ in trigonometric form?

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Answer 1 · 2016-04-02T01:28:20

$(6 - i) + (- 7 + 2 i) =$ $(6-i)+(-7+2i)=$

Explanation:

First we add algebraically
$(6 - i) + (- 7 + 2 i) = (6 - 7) + (- 1 + 2) i = - 1 + i = \sqrt{{(- 1)}^{2} + 1^{2}} [cos (arctan (\frac{1}{- 1})) + i sin (arctan (\frac{1}{- 1}))]$ $(6-i)+(-7+2i)=(6-7)+(-1+2)i=-1+i=sqrt((-1)^2+1^2)[cos(arctan(1/-1))+isin(arctan(1/-1))]$

$= \sqrt{2} [cos (arctan (\frac{1}{- 1})) + i sin (arctan (\frac{1}{- 1}))]$ $=sqrt2[cos(arctan(1/-1))+isin(arctan(1/-1))]$

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How do you add $(6 - i)$ $(6-i)$ and $(- 7 + 2 i)$ $(-7+2i)$ in trigonometric form?

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Explanation:

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How do you add $(6 - i)$ $(6-i)$ and $(- 7 + 2 i)$ $(-7+2i)$ in trigonometric form?