How do you add $(2 - 3 i)$ $(2-3i)$ and $(12 - 2 i)$ $(12-2i)$ in trigonometric form?

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How do you add $(2 - 3 i)$ $(2-3i)$ and $(12 - 2 i)$ $(12-2i)$ in trigonometric form?

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Answer 1 · 2016-05-17T11:10:23

$\sqrt{221} (cos (0.343) - i sin (0.343))$ $sqrt221(cos(0.343)-isin(0.343))$

Explanation:

A complex number z = x +iy can be expressed in trig. form as shown.

$z = x + i y = r (cos θ + i sin θ) where$ $z=x+iy=r(costheta+isintheta)" where"$

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

Now to get this sum in trig form we have to add the numbers together and then convert to trig.

$\Rightarrow (2 - 3 i) + (12 - 2 i) = 14 - 5 i$ $rArr(2-3i)+(12-2i)=14-5i$

Using x = 14 and y = -5 , convert to trig form.

$\Rightarrow r = \sqrt{14^{2} + {(- 5)}^{2}} = \sqrt{221} does not simplify further$ $rArrr=sqrt(14^2+(-5)^2)=sqrt221" does not simplify further"$

and $θ = {tan}^{- 1} (- \frac{5}{14}) \approx - 0.343 radians$ $theta=tan^-1(-5/14)≈-0.343" radians"$

$\Rightarrow 14 - 5 i = \sqrt{221} (cos (- 0.343) + i sin (- 0.343))$ $rArr14-5i=sqrt221(cos(-0.343)+isin(-0.343))$

using $cos (- θ) = cos θ and sin (- θ) = - sin θ$ $cos(-theta)=costheta" and "sin(-theta)=-sintheta$

we can also express in trig form as

$14 - 5 i = \sqrt{221} (cos (0.343) - i sin (0.343))$ $14-5i=sqrt221(cos(0.343)-isin(0.343))$

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How do you add $(2 - 3 i)$ $(2-3i)$ and $(12 - 2 i)$ $(12-2i)$ in trigonometric form?

1 Answer

Explanation:

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How do you add $(2 - 3 i)$ $(2-3i)$ and $(12 - 2 i)$ $(12-2i)$ in trigonometric form?