What is the trigonometric form of $(1 - 3 i)$ $(1-3i)$ ?

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What is the trigonometric form of $(1 - 3 i)$ $(1-3i)$ ?

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Answer 1 · 2016-05-02T19:03:16

$\sqrt{10} (cos (- 1.25) + i sin (- 1.25))$ $sqrt10(cos(-1.25)+isin(-1.25) )$

Explanation:

Given a complex number z = x + iy , then in trig.form it is written

z = $r (cos θ + i sin θ)$ $r(costheta + isintheta)$

where $| z | = | x + i y | = r = \sqrt{x^{2} + y^{2}}$ $|z|=|x+iy|=r=sqrt(x^2+y^2)$

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

here x = 1 and y = - 3

$\Rightarrow r = \sqrt{1^{2} + {(- 3)}^{2}} = \sqrt{10}$ $rArr r=sqrt(1^2+(-3)^2)=sqrt10$

and $θ = {tan}^{- 1} (- 3) = - 1.25 radians$ $theta=tan^-1(-3)=-1.25" radians "$

$\Rightarrow (1 - 3 i) = \sqrt{10} (cos (- 1.25) + i sin (- 1.25))$ $rArr(1-3i)=sqrt10(cos(-1.25)+isin(-1.25))$

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What is the trigonometric form of $(1 - 3 i)$ $(1-3i)$ ?

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

What is the trigonometric form of (1−3i) (1-3i) ?

1 Answer

Explanation:

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Impact of this question

What is the trigonometric form of $(1 - 3 i)$ $(1-3i)$ ?