How do you convert -4 + 8i−4+8i to polar form?

2 Answers
David G.
Jul 15, 2017

In polar coordinates, the point will be (r,theta)=(6.92i, -1.11i)(r,θ)=(6.92i,−1.11i)

Explanation:

For coordinates in polar form, (r, theta)(r,θ), we need to find the value of rr and the value of thetaθ.

We will use Pythagoras' theorem to find rr and the tantan function to find thetaθ.

r=sqrt((-4)^2+8i^2)=sqrt(16-64) =sqrt(-48)=6.92ir=√(−4)2+8i2=√16−64=√−48=6.92i

(since 8i^2=8ixx8i=8sqrt(-1)xx8sqrt(-1)=64xx-1=-648i2=8i×8i=8√−1×8√−1=64×−1=−64)

tantheta=(opp)/(adj)=(8i)/-4=-8/4i=-2itanθ=oppadj=8i−4=−84i=−2i

So theta=tan^-1(-2i)=-1.11iθ=tan−1(−2i)=−1.11i

Jim G.
Jul 15, 2017

(4sqrt5,2.03)(4√5,2.03)

Explanation:

"to convert from "color(blue)"cartesian to polar form"to convert from cartesian to polar form

"that is "(x,y)to(r,theta)" where"that is (x,y)→(r,θ) where

•color(white)(x)r=sqrt(x^2+y^2)∙xr=√x2+y2

•color(white)(x)theta=tan^-1(y/x)color(white)(x)-pi< theta <=pi∙xθ=tan−1(yx)x−π<θ≤π

"here " x=-4" and "y=8here x=−4 and y=8

rArrr=sqrt((-4)^2+8^2)=sqrt80=4sqrt5⇒r=√(−4)2+82=√80=4√5

-4+8i" is in the second quadrant "−4+8i is in the second quadrant

"so we must ensure that "theta" is in the second quadrant"so we must ensure that θ is in the second quadrant

theta=tan^-1(2)=1.11larrcolor(red)" related acute angle"θ=tan−1(2)=1.11← related acute angle

rArrtheta=(pi-1.11)=2.03larrcolor(red)" in second quadrant"⇒θ=(π−1.11)=2.03← in second quadrant

rArr-4+8ito(-4,8)to(4sqrt5,2.03)⇒−4+8i→(−4,8)→(4√5,2.03)

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