What is the trigonometric form of $(5 - 2 i) (2 + 3 i)$ $(5-2i)(2+3i)$ ?

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What is the trigonometric form of $(5 - 2 i) (2 + 3 i)$ $(5-2i)(2+3i)$ ?

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Answer 1 · 2016-01-06T09:46:11

$\sqrt{377} {cos ({tan}^{- 1} (\frac{11}{16})) + i sin ({tan}^{- 1} (\frac{11}{16}))}$ $sqrt(377){cos(tan^-1(11/16))+isin(tan^-1(11/16))}$

Explanation:

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

Multiply using FOIL

$(5) (2) + 5 (3 i) + (- 2 i) (2) + (- 2 i) (3 i)$ $(5)(2)+5(3i)+(-2i)(2)+(-2i)(3i)$
$10 + 15 i - 4 i - 6 i^{2}$ $10+15i-4i-6i^2$
$= 10 + 11 i - 6 (- 1)$ $=10+11i-6(-1)$ since $i^{2} = - 1$ $i^2=-1$
$= 10 + 6 + 11 i$ $=10+6+11i$
$= 16 + 11 i$ $=16+11i$

Trigonometric form is $r (cos (θ) + i sin (θ))$ $r(cos(theta)+isin(theta))$
if $z = x + i y$ $z=x+iy$

$r = \sqrt{x^{2} + y^{2}}$ $r=sqrt(x^2+y^2)$ and $θ = {tan}^{- 1} (\frac{y}{x})$ $theta = tan^-1(y/x)$

$r = \sqrt{16^{2} + 11^{2}}$ $r=sqrt(16^2+11^2)$ and $θ = {tan}^{- 1} (\frac{11}{16})$ $theta = tan^-1(11/16)$
$r = \sqrt{256 + 121}$ $r=sqrt(256+121)$
$r = \sqrt{377}$ $r=sqrt(377)$
Trigonometric form

$\sqrt{377} {cos ({tan}^{- 1} (\frac{11}{16})) + i sin ({tan}^{- 1} (\frac{11}{16}))}$ $sqrt(377){cos(tan^-1(11/16))+isin(tan^-1(11/16))}$

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What is the trigonometric form of $(5 - 2 i) (2 + 3 i)$ $(5-2i)(2+3i)$ ?

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Explanation:

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Math

Humanities

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What is the trigonometric form of (5−2i)(2+3i) (5-2i)(2+3i) ?

1 Answer

Explanation:

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What is the trigonometric form of $(5 - 2 i) (2 + 3 i)$ $(5-2i)(2+3i)$ ?