The trick to solving a problem with a ratio of complex numbers is to multiply the numerator and denominator by the complex conjugate of the denominator (this makes the denominator real). Let ##\alpha = 0.1029## and ##\beta=120.94##. We have,$$
Z_L=50\frac{(1+\alpha e^{j\beta}) }{(1-\alpha e^{j\beta})}$$I hope you can see that the complex conjugate of the denominator is ##1-\alpha e^{-j\beta}## Therefore,$$
Z_L=50\frac{(1+\alpha e^{j\beta} )}{(1-\alpha e^{j\beta})}\frac{(1-\alpha e^{-j\beta} )}{(1-\alpha e^{-j\beta})}\\
=50\frac{(1+\alpha (e^{j\beta}-e^{-j\beta})- \alpha^2)}{(1-\alpha (e^{j\beta}+e^{-j\beta})+ \alpha^2)}$$Recall that$$
\cos(\beta)=\frac{e^{j\beta}+e^{-j\beta}}{2}\\
\sin(\beta)=\frac{e^{j\beta}-e^{-j\beta}}{2j}$$
Thus we have,$$
Z_L=50\frac{(1-\alpha^2 + 2j\sin (\beta))}{(1+\alpha^2 - 2\cos (\beta))}$$
