Correct answer is (d) \(\frac{-asinω}{1-2acosω+a^2}\)
Easiest explanation: Given, X(ω)=\(\frac{1}{1-ae^{-jω}}\), |a|<1
By multiplying both the numerator and denominator of the above equation by the complex conjugate of the denominator, we obtain
X(ω)=\(\frac{1-ae^{jω}}{(1-ae^{-jω})(1-ae^{jω})} = \frac{1-acosω-jasinω}{1-2acosω+a^2}\)
This expression can be subdivided into real and imaginary parts, thus we obtain
XI(ω)=\(\frac{-asinω}{1-2acosω+a^2}\).