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lora16 [44]
3 years ago
13

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

Mathematics
1 answer:
lyudmila [28]3 years ago
4 0

Answer:

D) 3 i

    \frac{Z_{1} }{Z_{2} } = 3 i

Step-by-step explanation:

<u><em>Step(i)</em></u>:-

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} )

<u><em>Step(ii)</em></u>:-

\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

<u><em>Final answer</em></u>:-

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

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