Question

WILL MARK BRAINLIEST!!
I. Write 2-7i in polar coordinates on the complex plane.
II. Find (1+i)^8 using DeMoivre's theorem.

Answer 1

I. Polar form = (r,θ) = (7.28,254.05°)
z= -2-i7
r= √-2^2+-7^2= √53= 7.28
θ=tan^-1 (-7/-2) =3.5=74.05
Point lies in the third quadrant
Hence θ=180+74.05=254.05°
Polar form = (r,θ) = (7.28,254.05°)

II. Z=16
De Moivre's theorem states if
z= r(cos(θ) + i sin (θ)) , then z^2=r^2(cos(nθ)+
i sin (nθ))
r= √1^2+1^2= √2
θ= arctan (1/1)= π / 4
z= √2 (cos (π / 4)+i sin (π / 4))
z^8 = (√2 (cos (π / 4)+i sin (π / 4)))^8
= (√2 (cos (8π / 4)+i sin (8π / 4)))^8
=16 (cos(2π) + i sin (2π)
=16

Answer 2

(r,θ) = (7.28,254.05°) are 2-7i in polar coordinates on the complex plane.

What are polar coordinates?

The polar coordinates system is a two-dimensional coordinate system in which each point is determined by distance from the point and an angle from the reference direction.

I. Polar form = (r,θ) = (7.28,254.05°)

z = 2 - i7

r= $\sqrt{ 2^2+-7^2}$

= √53

 = 7.28

θ = tan^-1 (-7/-2)

  =3.5

  = 74.05

Point lies in the third quadrant

Hence θ = 180+74.05

              = 254.05°

Polar form = (r,θ) = (7.28,254.05°)

II. Z=16

De Moivre's theorem states if

z= r(cos(θ) + i sin (θ)) , then

$z^2=r^2(cos(n\theta)+ i sin (n\theta))$

$r= \sqrt{1^2+1^2}$

= √2

θ= arc tan (1/1)

  = π / 4

z= √2 (cos (π / 4)+i sin (π / 4))

z^8 = (√2 (cos (π / 4)+i sin (π / 4)))^8

= (√2 (cos (8π / 4)+i sin (8π / 4)))^8

=16 (cos(2π) + i sin (2π)

=16

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