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

Question

If $f(z)=\frac{7-z}{1-z^2}$, where $z=1+2 i$, then $|f(z)|$ is equal to :, $\frac{|z|}{2}$, $|z|$, $2|z|$, None

MARKS App · Accepted Answer

$\begin{aligned} & \mathrm{z}=1+2 \mathrm{i} \Rightarrow|\mathrm{z}|=\sqrt{1+4}=\sqrt{5} \ & \therefore \mathrm{f}(\mathrm{z})=\frac{7-\mathrm{z}}{1-\mathrm{z}^2}=\frac{7-1-2 \mathrm{i}}{1-(1+2 \mathrm{i})^2} \ & =\frac{6-2 \mathrm{i}}{1-(1-4+4 \mathrm{i})}=\frac{6-2 \mathrm{i}}{4-4 \mathrm{i}}=\frac{3-\mathrm{i}}{2-2 \mathrm{i}} \ & \Rightarrow|\mathrm{f}(\mathrm{z})|=\left|\frac{3-\mathrm{i}}{2-2 \mathrm{i}}\right|=\frac{|3-\mathrm{i}|}{|2-2 \mathrm{i}|} \ & =\frac{\sqrt{9+1}}{\sqrt{4+4}}=\frac{\sqrt{5}}{2}=\frac{|\mathrm{z}|}{2}\end{aligned}$

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