Question

For z1=9cis(5π/6) and z2=3cis(π/3), find z1/z2 in rectangular form.

A. -3

B. 3

C. -3i

D. 3i

Answer 1

Answer: D. 3i

Step-by-step explanation: I got this right on Edmentum.

Answer 2

D) 3 i

`(Z_(1) )/(Z_(2) ) = 3 i`

Explanation:

Step(i):-

Given Z₁ = 9 Cis ( 5π/6) = 9 ( Cos (5π/6) +i Sin (5π/6))

Z₂ = 3 Cis ( π/3) = 3 ( Cos (π/3) +i Sin (π/3))

Now Cos (5π/6) = cos (150°) = cos(90°+60°) = -sin 60 =
`(-√(3) )/(2)`

Sin (5π/6) = Sin (150°) = Sin(90°+60°) = Cos 60 =
`(1 )/(2)`

Z₁ = 9 ( Cos (5π/6) +i Sin (5π/6))

`= 9((-√(3) )/(2) + i(1)/(2) )`

Z₂ = 3 ( Cos (π/3) +i Sin (π/3))

`= 3((1 )/(2) + i(√(3) )/(2) )`

Step(ii):-

`(Z_(1) )/(Z_(2) ) = ( 9((-√(3) )/(2) + i(1)/(2) ))/(3((1 )/(2) + i(√(3) )/(2) ))`

`(Z_(1) )/(Z_(2) ) =(3 (-√(3)+i) )/(1+i√(3)) )`

Rationalize with 1 - i √3 and we get

`(Z_(1) )/(Z_(2) ) =(3 (-√(3)+i) )/(1+i√(3)) ) X(1-i√(3) )/(1-i√(3) )`

on simplification , we will use formulas

i² = -1 and (a+b)(a-b) = a² - b²

`(Z_(1) )/(Z_(2) ) =(3(-√(3) + 3 i + i +√(3) ))/(1 - i^(2) (√(3) ))`

`(Z_(1) )/(Z_(2) ) =(3(4 i ))/(1 - i^(2) (√(3) )^(2) )`

`(Z_(1) )/(Z_(2) ) =(3(4 i ))/(1 - i^(2) (√(3) )^(2) ) = (3(4 i))/(1-(-3))= (3(4 i))/(4)`

`(Z_(1) )/(Z_(2) ) = 3 i`

Final answer:-

`(Z_(1) )/(Z_(2) ) = 3 i`