Question 1073369: If z+1/z-1=cos(theta)+isin(theta), where theta is not equal to zero, show that z=-icot(theta/2). I tried to solve this problem but I'm stuck. I tried to get rid of the fraction by multiplying z-1 to the right hand side, but I'm not sure about the next step. Additional information is that this is one of the "mini" questions in a series of questions. The answers to the previous questions tell you that abs(z+1/z-1)=1, and that z is purely imaginary. I'm not sure if this additional info would help, but I would really appreciate it if you could teach me how to solve this question. Thank you so much for your time and effort!
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! (z+1) / (z-1) = cos(theta) + isin(theta)
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rewrite right side of equation
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(z+1) / (z-1) = cos(theta) + (0 + 1i) * sin(theta)
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multiply both sides of = by (z - 1)
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z + 1 = (z - 1) * [cos(theta) + (0 + 1i) * sin(theta)]
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z + 1 = zcos(theta) +z(0+1i)sin(theta) -cos(theta) -(0+1i)sin(theta)
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collect terms on right side
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z + 1 = -cos(theta) +z[cos(theta) +(0+1i)sin(theta)] -(0+1i)sin(theta)
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subtract -1 from both sides of =
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z = -1 -cos(theta) +z[cos(theta) +(0+1i)sin(theta)] -(0+1i)sin(theta)
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subtract z[cos(theta) +(0+1i)sin(theta)] from both sides of =
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z -z[cos(theta) +(0+1i)sin(theta)] = -1 -cos(theta) -(0+1i)sin(theta)
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z(1 -cos(theta) +(0+1i)sin(theta)) = -1 -cos(theta) -(0+1i)sin(theta)
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divide both sides of = by (1 -cos(theta) +(0+1i)sin(theta))
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1) z = (-1 -cos(theta) -(0+1i)sin(theta)) / (1 -cos(theta) +(0+1i)sin(theta))
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z = -i * cot(theta/2)
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note that i^2 = -1, half-angle identity and euler identity e^(i*theta) = cos(theta) + (i * sin(theta))
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using this information equation 1 reduces to
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z = (cos(theta) + i * sin(theta) + 1) / (cos(theta) + i * sin(theta) - 1)
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using euler identity
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z = (e^(i*theta) + 1) / (e^(i*theta) - 1) = -i * cot(theta/2)
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