Question

Express ${\displaystyle \frac{2+\text{i}}{(1+\text{i})(1-2\text{i})}}$  in the form of a + ib. Find its modulus and argument

Answer 1

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

 

${\displaystyle =\frac{2+\text{i}}{1-2\text{i}+\text{i}-2{\text{i}}^2}=\frac{2+\text{i}}{3-\text{i}}}$

 

${\displaystyle \text{z}=\frac{2+\text{i}}{3-\text{i}}\times \frac{3+\text{i}}{3+\text{i}}=\frac{(2+\text{i})(3+\text{i})}{9-{\text{i}}^2}}$

 

${\displaystyle =\frac{6+5\text{i}+{\text{i}}^2}{10} =\frac{5+5\text{i}}{10} =\frac{1+\text{i}}{2}}$

 

Now,    ${\displaystyle \text{Z}=\frac{1}{2}+\frac{\text{i}}{2}=\text{a}+\text{i}\text{b}}$

 

⇒     a = ${\displaystyle \frac{1}{2}}$  , b =${\displaystyle \frac{1}{2}}$

 

|Z|  = ${\displaystyle \sqrt{{\text{a}}^2+{\text{b}}^2}=\sqrt{\frac{1}{4}+\frac{1}{4}}=\frac{1}{\sqrt{2}}}$

 

${\displaystyle \text{arg Z}={\tan }^{-1}\left(\frac{\text{b}}{\text{a}}\right)={\tan }^{-1}\left(1\right)=\left(\frac{\pi }{4}\right)}$

 

Adding (i) and (ii), we get :

 

${\displaystyle 2\text{I}={\int }_0^{\frac{\pi }{2}}\frac{\sin \text{x}-\cos \text{x}+\cos \text{x}-\sin \text{x}}{1+\sin \text{x}\cos \text{x}}\text{dx}}$

 

${\displaystyle 2\text{I}=0 \Rightarrow \text{I}=0}$

 

⇒  ${\displaystyle {\int }_0^{\frac{\pi }{2}}\frac{\sin \text{x}-\cos \text{x}}{1+\sin \text{x}\cos \text{x}}\text{dx}=0}$