find the relation between real numbers a and b which satisfy the equation 1-ix by 1+ ix= a -ib for some real value of x
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(1-ix)/(1+ix) = a - ib
Multiplying the numerator and denominator of LHS by (1-ix)
(1-ix)*(1-ix) / (1+ix)*(1-ix) = a - ib
(1-ix)*(1-ix) / (1² - (ix)²)= a - ib
(1-ix)² / (1 - (ix)²)= a - ib
(1+(ix)²+2ix) / (1 - (ix)²) = a - ib
(1 - x²+2ix) / (1 + x²) = a - ib
[(1 - x²)/ (1 + x²)] - [- 2ix/ (1 + x²)]= a - ib
So,
a = (1 - x²)/ (1 + x²)
b = -2ix/ (1 + x²)
Multiplying the numerator and denominator of LHS by (1-ix)
(1-ix)*(1-ix) / (1+ix)*(1-ix) = a - ib
(1-ix)*(1-ix) / (1² - (ix)²)= a - ib
(1-ix)² / (1 - (ix)²)= a - ib
(1+(ix)²+2ix) / (1 - (ix)²) = a - ib
(1 - x²+2ix) / (1 + x²) = a - ib
[(1 - x²)/ (1 + x²)] - [- 2ix/ (1 + x²)]= a - ib
So,
a = (1 - x²)/ (1 + x²)
b = -2ix/ (1 + x²)
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