How do you write $2 + 4 i$ $2 +4i$ in trigonometric form?

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How do you write $2 + 4 i$ $2 +4i$ in trigonometric form?

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Answer 1 · 2018-04-12T08:06:55

$2 \sqrt{5} (cos (1.107) + i sin (1.107))$ $2sqrt5(cos(1.107)+isin(1.107))$

Explanation:

$to convert to trigonometric form$ $"to convert to trigonometric form"$

$that is x + y i \to r (cos θ + i sin θ) using$ $"that is "x+yitor(costheta+isintheta)" using"$

$∙ x r = \sqrt{x^{2} + y^{2}}$ $•color(white)(x)r=sqrt(x^2+y^2)$

$∙ x θ = {tan}^{- 1} (\frac{y}{x}); - π < θ \leq π$ $•color(white)(x)theta=tan^-1(y/x);-pi < theta <=pi$

$here x = 2 and y = 4$ $"here "x=2" and "y=4$

$\Rightarrow r = \sqrt{2^{2} + 4^{2}} = \sqrt{20} = 2 \sqrt{5}$ $rArrr=sqrt(2^2+4^2)=sqrt20=2sqrt5$

$θ = {tan}^{- 1} (2) = 1.107 radians to 3 dec. places$ $theta=tan^-1(2)=1.107" radians to 3 dec. places"$

$\Rightarrow 2 + 4 i = 2 \sqrt{5} (cos (1.107) + i sin (1.107))$ $rArr2+4i=2sqrt5(cos(1.107)+isin(1.107))$

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How do you write $2 + 4 i$ $2 +4i$ in trigonometric form?

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How do you write $2 + 4 i$ $2 +4i$ in trigonometric form?