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

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

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Answer 1 · 2018-07-15T19:08:34

Hint is given below

Explanation:

HINT:

$- 7 + 7 i = 7 \sqrt{2} (cos (\frac{3 π}{4}) + i sin (\frac{3 π}{4}))$ $-7+7i=7\sqrt2(\cos ({3\pi}/4)+i\sin({3\pi}/4))$

$6 - 2 i = 2 \sqrt{10} (cos (- {tan}^{- 1} (\frac{1}{3})) + i sin (- {tan}^{- 1} (\frac{1}{3})))$ $6-2i=2\sqrt10(\cos (-\tan^{-1}(1/3))+i\sin(-\tan^{-1}(1/3)))$

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

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

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