Solve the following


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



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

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

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

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

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

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

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

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

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

Since $z=1+2 i$








Hence, the correct answer is option (a).

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