Question

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

Answer 1

Answer:

D) 3 i

    $\frac{Z_{1} }{Z_{2} } = 3 i$

Step-by-step 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 = $\frac{-\sqrt{3} }{2}$

           Sin (5π/6) = Sin (150°) = Sin(90°+60°)   = Cos 60 = $\frac{1 }{2}$

   

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

          $= 9(\frac{-\sqrt{3} }{2} + i\frac{1}{2} )$

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

       $= 3(\frac{1 }{2} + i\frac{\sqrt{3} }{2} )$

Step(ii):-

$\frac{Z_{1} }{Z_{2} } = \frac{ 9(\frac{-\sqrt{3} }{2} + i\frac{1}{2} )}{3(\frac{1 }{2} + i\frac{\sqrt{3} }{2} )}$

$\frac{Z_{1} }{Z_{2} } =\frac{3 (-\sqrt{3}+i) }{1+i\sqrt{3}) }$

Rationalize with 1 - i √3 and we get

$\frac{Z_{1} }{Z_{2} } =\frac{3 (-\sqrt{3}+i) }{1+i\sqrt{3}) } X\frac{1-i\sqrt{3} }{1-i\sqrt{3} }$

on simplification , we will use formulas

i² = -1 and    (a+b)(a-b) = a² - b²

$\frac{Z_{1} }{Z_{2} } =\frac{3(-\sqrt{3} + 3 i + i +\sqrt{3} )}{1 - i^{2} (\sqrt{3} )}$

$\frac{Z_{1} }{Z_{2} } =\frac{3(4 i )}{1 - i^{2} (\sqrt{3} )^{2} }$

$\frac{Z_{1} }{Z_{2} } =\frac{3(4 i )}{1 - i^{2} (\sqrt{3} )^{2} } = \frac{3(4 i)}{1-(-3)}= \frac{3(4 i)}{4}$

$\frac{Z_{1} }{Z_{2} } = 3 i$

Final answer:-

$\frac{Z_{1} }{Z_{2} } = 3 i$