Question:
If $f(z)=\frac{7-z}{1-z^{2}}$, where $z=1+2 i$, then $|f(z)|$ is
(a) $\frac{|z|}{2}$
(b) $|z|$
(c) $2|z|$
(d) none of these
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).