Question

If $z=\frac{(1+i)(1-i)}{\left(2\right)+i}$

find $\log \mid z\mid$

Answer 1

Given,

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

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

on rationalizing, we get,

$z=\frac{2(2-i)}{5}$

$\therefore z=\frac{4}{5}-\frac{2}{5}i$

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

$\therefore \log z=\log \left(\frac{2}{\sqrt{5}}\right)=\log 2-\log \sqrt{5}$

$=\log 2-\frac{1}{2}\log 5$

$=\log 2-\frac{1}{2}\log \left(\frac{10}{2}\right)$

$=\log 2-\frac{1}{2}(\log 10-\log 2)$

$=\log 2-\frac{1}{2}(1-\log 2)$

$=\log 2-\frac{1}{2}+\frac{1}{2}\log 2$

$=\frac{3}{2}\log 2-\frac{1}{2}$

$=\frac{3}{2}\left(0.3010\right)-\frac{1}{2}$

$=-0.048$