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  • If where z = 1 + 2i, then |f(z)| is A. B. |z| C. 2|z| D. none of these

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

B. |z|
C. 2|z|
D. none of these

Answers (1)

The answer is the option (a).

Given that \\ z=1+2i \\ |z|=\sqrt{(1)^2+(2)^2 }=\sqrt5

Now  f(z)=\frac{7-z}{1-z^2}

\begin{aligned} =\frac{7-(1+2 i)}{1-(1+2 i)^{2}} &=\frac{7-1-2 i}{1-1-4 i^{2}-4 i} \\ =\frac{6-2 i}{4-4 i} &=\frac{3-i}{2-2 i} \end{aligned}

\begin{aligned} =\frac{3-i}{2-2 i} * \frac{2+2 i}{2+2 i}=\frac{6+6 i-2 i-2 i^{2}}{4-4 i} \\ =\frac{6+4 i+2}{4+4} &=\frac{8+4 i}{8} \\ =1+\frac{1}{2} i & \end{aligned}

f(z)=\sqrt{(1)^{2}+\left(\frac{1}{2}\right)^{2}}=\sqrt{1+\frac{1}{4}}=\frac{\sqrt{5}}{2}=\frac{|z|}{2}

Hence, option a is correct

 

Posted by

infoexpert21

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