Figure 5 has S-parameters for ADF4113 RF input. You can calculate input impedance from these.
\$R=\dfrac{Z_0(1-\rm{MAG}^2)}{1+\rm{MAG}^2-2\cdot\rm{MAG}\cdot\cos(\frac{\rm{ANG}}{360}\cdot 2\pi)}\$
\$X=\dfrac{2\cdot\rm{MAG}\cdot\sin(\frac{\rm{ANG}}{360}\cdot 2\pi)\cdot 50}{1+\rm{MAG}^2-2\cdot\rm{MAG}\cdot\cos(\frac{\rm{ANG}}{360}\cdot 2\pi)}\$
\$Z = \sqrt{R^2 + X^2}\$
Where \$Z_0 = 50\$
I hope I latexed the formulas correctly, just in case they are from this page.
It is a good thing for a voltage input, as if the input impedance is high compared to the source impedance then the voltage level will not drop too much due to the divider effect.
For example, say we have a \$10V\$ signal with \$1k\Omega\$ impedance.
We connect this to a \$1M\Omega\$ input, the input voltage will be \$ 10V\cdot\frac{1M\Omega}{1M\Omega+1k\Omega} = 9.99V \$.
If we reduce the input impedance to \$10k\Omega\$, we get \$10V \cdot \frac{10k\Omega}{10k\Omega + 1k\Omega} = 9.09V\$
Reduce it to 1k and we get \$ 10V \cdot \frac{1k\Omega}{1k\Omega + 1k\Omega} = 5V\$
Hopefully you get the picture - generally an input impedance of at least 10 times the source impedance is a good idea to prevent significant loading.
High input impedance is not always a good thing though, for example if you want to transfer as much power as possible then the source and load impedance should be equal. So in the above example the 1k input impedance would be the best choice.
For a current input a low input impedance (ideally zero) is desired, for example in a transimpedance (current to voltage) amplifier.
Best Answer
It's one of the rules. For an ideal opamp goes that
If these requirements aren't met several basic opamp circuits wouldn't work. Take for instance the inverting amplifier.
The transfer function is
\$ \mathrm{V_{OUT} = - \dfrac{R_f}{R_{IN}} \cdot V_{IN}} \$
as derived in this answer. The proof relies on the infinite input impedance, but you can't explain the transfer function based on both inputs equal, because that's not a property of the opamp! So-called proofs that start from the fact that the inverting input is at ground are invalid.
Note that FET input opamps do better than their BJT input counterparts. The former will have pA input current, whereas for the latter this may be several \$\mu\$A.
Further reading:
Opamps for everyone