Question

If $f\left(z\right)=\frac{7-z}{1-z^2}$, where $z=1+2i$, then $\left|f\left(z\right)\right|$ is

(a) $\frac{\left|z\right|}{2}$
(b) $\left|z\right|$
(c) $2\left|z\right|$
(d) none of these

Answer 1

$$\begin{gathered} f\left(z\right)=\frac{7-z}{1-z^2} \\ =\frac{7-\left(1+2i\right)}{1-{\left(1+2i\right)}^2} \\ =\frac{7-1-2i}{1-\left(1^2+2^2{i}^2+4i\right)} \\ =\frac{6-2i}{1-1+4-4i} \\ =\frac{6-2i}{4-4i} \\ =\frac{6-2i}{4-4i}\times \frac{4+4i}{4+4i} \\ =\frac{24+24i-8i-8i^2}{4^2-4^2{i}^2} \\ =\frac{24+16i+8}{16+16} \\ =\frac{32+16i}{32} \\ =1+\frac{1}{2}i \end{gathered}$$

Since $z=1+2i$,
$$\begin{gathered} \therefore \left|z\right|=\sqrt{{\left(1\right)}^2+{\left(2\right)}^2} \\ =\sqrt{1+4} \\ =\sqrt{5} \end{gathered}$$

$$\begin{gathered} \therefore \left|f\left(z\right)\right|=\sqrt{{\left(1\right)}^2+{\left(\frac{1}{2}\right)}^2} \\ =\sqrt{1+\frac{1}{4}} \\ =\frac{\sqrt{5}}{2} \\ =\frac{\left|z\right|}{2} \end{gathered}$$

Hence, the correct answer is option (a).