Question

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

पर्याय

• \[\frac{\left| z \right|}{2}\] 

• \[\left| z \right|\]

• \[2\left| z \right|\]

• none of these

Answer 1

\[f\left( z \right) = \frac{7 - z}{1 - z^2}\]

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

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

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

\[ = \frac{6 - 2i}{4 - 4i}\]

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

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

\[ = \frac{24 + 16i + 8}{16 + 16}\]

\[ = \frac{32 + 16i}{32}\]

\[ = 1 + \frac{1}{2}i\]

Since 

\[z = 1 + 2i\],

\[\therefore \left| z \right| = \sqrt{\left( 1 \right)^2 + \left( 2 \right)^2}\]

\[ = \sqrt{1 + 4}\]

\[ = \sqrt{5}\]

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

\[ = \sqrt{1 + \frac{1}{4}}\]

\[ = \frac{\sqrt{5}}{2}\]

\[ = \frac{\left| z \right|}{2}\]

Hence, the correct answer is option (a).