How do you write the complex number in trigonometric form $- 8 + 3 i$ $-8+3i$ ?

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How do you write the complex number in trigonometric form $- 8 + 3 i$ $-8+3i$ ?

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Answer 1 · 2017-08-14T09:30:16

The trigonometric form is $= 2.92 (cos ({159.4}^{\circ}) + i sin ({159.4}^{\circ})) = 2.92 e^{{159.4}^{\circ} i}$ $=2.92 (cos(159.4^@)+isin(159.4^@))=2.92e^(159.4^@i)$

Explanation:

Our complex number is

$z = - 8 + 3 i$ $z=-8+3i$

The trigonometric form is

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

If our complex number is $z = a + i b$ $z=a+ib$

$r = | z | = \sqrt{a^{2} + b^{2}}$ $r=|z|=sqrt(a^2+b^2)$

And

$cos θ = \frac{a}{| z |}$ $costheta=a/|z|$ and

$sin θ = \frac{b}{| z |}$ $sintheta=b/|z|$

Therefore,

$| z | = \sqrt{{(- 8)}^{2} + 3^{2}} = \sqrt{64 + 9} = \sqrt{73} = 2.92$ $|z|=sqrt((-8)^2+3^2)=sqrt(64+9)=sqrt73=2.92$

$cos θ = - \frac{8}{\sqrt{73}}$ $costheta=-8/sqrt73$

$sin θ = \frac{3}{\sqrt{73}}$ $sintheta=3/sqrt73$

We are in the Quadrant $I I$ $II$

$Theta=159.4^@$

The trigonometric form is

$z=2.92 (cos(159.4^@)+isin(159.4^@))=2.92e^(159.4^@i)$

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How do you write the complex number in trigonometric form $- 8 + 3 i$ $-8+3i$ ?

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