Question

In: Math

If 3/(2+ cos θ+i sin θ) = a+ib, then [(a-2)2+b2] is

If 3/(2+ cos θ+i sin θ) = a+ib, then [(a-2)2+b2] is

Solutions

Expert Solution

Given - 3/(2+ cos θ+ i sin θ) = a+ib

 

3 (2+ cos θ) – i sin θ)/(2+ cos θ + i sin θ)(2+cos θ)- i sin θ)

= ((6+ 3 cos θ) – i 3 sin θ)/((2+cos θ)²+ sin2 θ)

= ((6+ 3 cos θ) – i(3 sin θ))/(5+4 cos θ)

 

Comparing with a+ib, we get

a = (6+ 3 sin θ)/(5+ 4 cos θ)

b = -3 sin θ/(5+ 4 cos θ)

 

(a-2)²+b² = [(6+ 3 sin θ)/(5+ 4 cos θ)]-2)²+(-3 sin θ/(5+ 4 cos θ))²

= (-4-5 cos θ)²+9 sin²θ)/(5+ 4 cos θ)²

= ((4+5 cos θ)²+9 sin²θ)/(5+ 4 cos θ)²

= (16 + 40 cos θ +25 cos²θ+ 9 sin2θ)/(5 +4 cos θ)²

= (16 + 40 cos θ+ 16 cos²θ+ 9 (sin²θ +cos²θ))/(5+ 4 cos θ)²

= (16 + 40 cos θ+ 16 cos²θ+ 9)/(5 +4 cos θ)²

= (16 cos²θ+ 40 cos θ+25)/(5+4 cos θ)²

= (4 cos θ+5)²/(5+4 cos θ)²

= 1

 

 


Value of [(a-2)²+b²] = 1

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