I thought of putting this on the Electrical Engineering Exchange but I thought since this seems more mathematical than related to engineering I thought I should place it here instead.
Question: Why for a quarter wave transformer does the equation for Zin of a lossless tranformer simplify to the following equation:
$$ Z_{in} = \frac{(Z_o)^2}{Z_L} $$
Usually the equation is:
$$ Z_{in} = Z_o \left[ \frac{Z_L + Z_o i tan(\beta l)}{Z_o + Z_L i tan(\beta l)}\right] $$
Where $i$ is the imaginary unit, $\beta = \frac{2\pi}{\lambda}$ and $l = \frac{\lambda}{4}$ So putting those values for $\beta$ and $\lambda$ we get the following equation:
$$ Z_{in} = Z_o \left[ \frac{Z_L + Z_o i tan(\frac{\pi}{2})}{Z_o + Z_L i tan(\frac{\pi}{2})}\right] $$
Now my question how do I get from the above equation to:
$$ Z_{in} = \frac{(Z_o)^2}{Z_L} $$
I know $\tan(\frac{\pi}{2}) = \infty$ so how can I simplify? My book just assumes I know how to get there. Is there some relationship that I don't know or have forgotten.
If $Z_L$ and $Z_0$ are non-zero:
$$ \begin{align} Z_{in} &= Z_0 \left( \frac{Z_L + Z_0 i \tan(\pi/2)}{Z_0 + Z_L i \tan(\pi/2)}\right)\\ &= Z_0 \left( \frac{\dfrac{Z_L}{\tan(\pi/2)} + Z_0 i }{\dfrac{Z_0}{\tan(\pi/2)} + Z_L i}\right)\\ &= Z_0 \left( \frac{0 + Z_0 i }{0 + Z_L i}\right)\\ &= \frac{(Z_0)^2 }{ Z_L } \end{align} $$
Above $\tan(\pi/2)$ is an abuse of notation in mathematics. I assume $\dfrac{c}{\tan(\pi/2)}=0$ for any non-zero $c$, since you use $\tan(\pi/2)$ in your equations freely.