How do you write $2 + 3 i$ $2+3i$ in trigonometric form?

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How do you write $2 + 3 i$ $2+3i$ in trigonometric form?

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Answer 1 · 2017-04-09T18:11:10

To convert, $a + b i \to r (cos (θ) + i sin (θ))$ $a+bitor(cos(theta)+isin(theta))$ , use:
$r = \sqrt{a^{2} + b^{2}}$ $r=sqrt(a^2+b^2)$
$θ = {tan}^{- 1} (\frac{b}{a}) + 0, π, π, 2 π$ $theta=tan^-1(b/a)+0,pi,pi,2pi$
Depending on whether the signs of "a" and "b" indicate the, 1st, 2nd, 3rd or 4th quadrant.

Explanation:

Given: $2 + 3 i$ $2+3i$

$r = \sqrt{2^{2} + 3^{2}}$ $r = sqrt(2^2+3^2)$

$r = \sqrt{4 + 9}$ $r = sqrt(4+9)$

$r = \sqrt{13}$ $r = sqrt(13)$

The signs of "a" and "b" indicate the 1st quadrant:

$θ = {tan}^{- 1} (\frac{b}{a}) + 0$ $theta = tan^-1(b/a)+0$

$θ = {tan}^{- 1} (\frac{3}{2})$ $theta = tan^-1(3/2)$

$θ \approx {56.3}^{\circ} or 0.983 radians$ $theta ~~ 56.3^@ or 0.983" radians"$

$2 + 3 i \to \sqrt{13} (cos ({56.3}^{\circ}) + i sin ({56.3}^{\circ}))$ $2+3i to sqrt(13)(cos(56.3^@)+isin(56.3^@))$

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How do you write $2 + 3 i$ $2+3i$ in trigonometric form?

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