Solve the following

Solution:

$f(z)=\frac{7-z}{1-z^{2}}$

$=\frac{7-(1+2 i)}{1-(1+2 i)^{2}}$

$=\frac{7-1-2 i}{1-\left(1^{2}+2^{2} i^{2}+4 i\right)}$

$=\frac{6-2 i}{1-1+4-4 i}$

$=\frac{6-2 i}{4-4 i}$

$=\frac{6-2 i}{4-4 i} \times \frac{4+4 i}{4+4 i}$

$=\frac{24+24 i-8 i-8 i^{2}}{4^{2}-4^{2} i^{2}}$

$=\frac{24+16 i+8}{16+16}$

$=\frac{32+16 i}{32}$

$=1+\frac{1}{2} i$

Since $z=1+2 i$

$\therefore|z|=\sqrt{(1)^{2}+(2)^{2}}$

$=\sqrt{1+4}$

$=\sqrt{5}$

$\therefore|f(z)|=\sqrt{(1)^{2}+\left(\frac{1}{2}\right)^{2}}$

$=\sqrt{1+\frac{1}{4}}$

$=\frac{\sqrt{5}}{2}$

$=\frac{|z|}{2}$

Hence, the correct answer is option (a).