How do you add $(2 + i) + (5 + 3 i)$ $(2+i)+(5+3i)$ in trigonometric form?

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How do you add $(2 + i) + (5 + 3 i)$ $(2+i)+(5+3i)$ in trigonometric form?

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Answer 1 · 2016-03-31T02:32:03

$7 + 4 i = \sqrt{65} [cos ({tan}^{- 1} (\frac{4}{7})) + i sin ({tan}^{- 1} (\frac{4}{7}))]$ $7+4i=sqrt65[cos(tan^-1 (4/7))+i sin (tan^-1 (4/7))]$

Explanation:

From the given
$(2 + i) + (5 + 3 i)$ $(2+i)+(5+3i)$

$(2 + 5) + (i + 3 i)$ $(2+5)+(i+3i)$

$7 + 4 i$ $7+4i$

in Trigonometric Form

$a + b i = \sqrt{a^{2} + b^{2}} [cos ({tan}^{- 1} (\frac{b}{a})) + i sin ({tan}^{- 1} (\frac{b}{a}))]$ $a+bi=sqrt(a^2+b^2)[cos (tan^-1 (b/a))+i sin (tan^-1 (b/a))]$

$7 + 4 i = \sqrt{65} [cos ({tan}^{- 1} (\frac{4}{7})) + i sin ({tan}^{- 1} (\frac{4}{7}))]$ $7+4i=sqrt65[cos(tan^-1 (4/7))+i sin (tan^-1 (4/7))]$

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How do you add $(2 + i) + (5 + 3 i)$ $(2+i)+(5+3i)$ in trigonometric form?

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Explanation:

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How do you add $(2 + i) + (5 + 3 i)$ $(2+i)+(5+3i)$ in trigonometric form?