How to solve this to calculate the input impedance?

[putrinh]

How did you find PF?: Google search

I got a problem, how to calculate this formula from euler to imajiner? or how calculate this case?

 

[berkeman]

Welcome to the PF.

It looks pretty straightforward, but it would help if you could define a few of the terms for us. It looks like a transmission line calculation for a reflection at an impedance boundary? Are you mainly asking for help understanding the rectangular/polar forms of the equations?

https://en.wikipedia.org/wiki/Reflection_coefficient

 

[Fred Wright]

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,$$
Z_L=50\frac{(1-\alpha^2 + 2j\sin (\beta))}{(1+\alpha^2 - 2\cos (\beta))}$$