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 · 2017-06-13T15:44:11

I got: $z = r [cos (θ) + i sin (θ)] = 8.54 [cos ({20.56}^{\circ}) + i sin ({20.56}^{\circ})]$ $z=r[cos(theta)+isin(theta)]=8.54[cos(20.56^@)+isin(20.56^@)]$

Explanation:

Consider the diagram:

enter image source here

From Pythagoras:

$r = \sqrt{8^{2} + 3^{2}} = 8.54$ $r=sqrt(8^2+3^2)=8.54$

From Trigonometry:

$θ = arctan (\frac{3}{8}) = {20.56}^{\circ}$ $theta=arctan(3/8)=20.56^@$

So:

$z = 8 + 3 i = r [cos (θ) + i sin (θ)] = 8.54 [cos ({20.56}^{\circ}) + i sin ({20.56}^{\circ})]$ $z=8+3i=r[cos(theta)+isin(theta)]=8.54[cos(20.56^@)+isin(20.56^@)]$

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

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Science

Math

Humanities

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How do you write the complex number in trigonometric form 8+3i8+3i?

1 Answer

Explanation:

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