Answer:
D) 3 i
[tex]\frac{Z_{1} }{Z_{2} } = 3 i[/tex]
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 = [tex]\frac{-\sqrt{3} }{2}[/tex]
Sin (5π/6) = Sin (150°) = Sin(90°+60°) = Cos 60 = [tex]\frac{1 }{2}[/tex]
Z₁ = 9 ( Cos (5π/6) +i Sin (5π/6))
[tex]= 9(\frac{-\sqrt{3} }{2} + i\frac{1}{2} )[/tex]
Z₂ = 3 ( Cos (π/3) +i Sin (π/3))
[tex]= 3(\frac{1 }{2} + i\frac{\sqrt{3} }{2} )[/tex]
Step(ii):-
[tex]\frac{Z_{1} }{Z_{2} } = \frac{ 9(\frac{-\sqrt{3} }{2} + i\frac{1}{2} )}{3(\frac{1 }{2} + i\frac{\sqrt{3} }{2} )}[/tex]
[tex]\frac{Z_{1} }{Z_{2} } =\frac{3 (-\sqrt{3}+i) }{1+i\sqrt{3}) }[/tex]
Rationalize with 1 - i √3 and we get
[tex]\frac{Z_{1} }{Z_{2} } =\frac{3 (-\sqrt{3}+i) }{1+i\sqrt{3}) } X\frac{1-i\sqrt{3} }{1-i\sqrt{3} }[/tex]
on simplification , we will use formulas
i² = -1 and (a+b)(a-b) = a² - b²
[tex]\frac{Z_{1} }{Z_{2} } =\frac{3(-\sqrt{3} + 3 i + i +\sqrt{3} )}{1 - i^{2} (\sqrt{3} )}[/tex]
[tex]\frac{Z_{1} }{Z_{2} } =\frac{3(4 i )}{1 - i^{2} (\sqrt{3} )^{2} }[/tex]
[tex]\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}[/tex]
[tex]\frac{Z_{1} }{Z_{2} } = 3 i[/tex]
Final answer:-
[tex]\frac{Z_{1} }{Z_{2} } = 3 i[/tex]
