Chapter 5: Complex Numbers and Quadratic Equations Exercise 5.1

Ans:

Let \[z  =  4  -  3i\text{ }\]

Then,

\[\overline{z}  =  4  +  3i  \]and \[\left| z \right|  =  {{4}^{2}}  +  {{\left( -  3 \right)}^{2}}  =  16  +  9  =  25\]

The multiplicative inverse of \[4  -  3i\text{ }\]is given by 

$\Rightarrow $${{z}^{-  1}}  =  \dfrac{\overline{z}}{{{\left| z \right|}^{2}}}$

\[=  \dfrac{4  +  3i}{25}  \]

\[=  \dfrac{4}{25}  +  \dfrac{3}{25}i\]

Ans:

Let $z  =  \sqrt{5}  +  3i$

Then, $\overline{z}  =  \sqrt{5}  -  3i$

$\left| z \right|  =  {{\left( \sqrt{5} \right)}^{2}}  +{{\left( 3 \right)}^{2}}  =  5  +  9  =  14$

The multiplicative inverse of the complex number $\sqrt{5}  +  3i$ is given by 

${{z}^{-  1}}  =  \dfrac{\overline{z}}{{{\left| z \right|}^{2}}}$

$ =   \dfrac{\sqrt{5}  -  3i}{14}   $

$=  \dfrac{\sqrt{5}}{14}  -  \dfrac{3}{14}i$

Ans:

Let $z  =  -  i  $

Then, $\overline{z}  =  i$ 

${{\left| z \right|}^{2}} =   1$

The multiplicative inverse of the complex number $-  i  $

$\Rightarrow $${{z}^{-  1}}  =  \dfrac{\overline{z}}{{{\left| z \right|}^{2}}}$

$  =  \dfrac{i}{1}  $

$=   i$

Ans:

$\Rightarrow $$\dfrac{\left( 3  +  i\sqrt{5} \right)\left( 3  -  i\sqrt{5} \right)}{\left( \sqrt{3}  +  \sqrt{2 }i \right)  -  \left( \sqrt{3}  -  i \sqrt{2} \right)}$

$=  \dfrac{\left( {{3}^{2}}  -  {{\left( i\sqrt{5} \right)}^{2}} \right)}{\left( \sqrt{3}  +  \sqrt{2 }i \right)  -  \left( \sqrt{3}  -  i \sqrt{2} \right)}  $            $\left[ \left( a  +  b \right)\left( a  -  b \right)  =  {{a}^{2}}  -  {{b}^{2}} \right]$

$=\,\dfrac{\left( 9  -  5{{i}^{2}} \right)}{2\sqrt{2}i}$

$=  \dfrac{9  -  5\left( -  1 \right)}{2\sqrt{2}i}          \left[ {{i}^{2}}  =  -  1 \right]  $

$=  \dfrac{14i}{2\sqrt{2}{{i}^{2}}}$

$=  \dfrac{14i}{2\sqrt{2}\left( -  1 \right)}$

$=  \dfrac{-  7i}{\sqrt{2}}\times   \dfrac{\sqrt{2}}{\sqrt{2}}$

$=  \dfrac{-  7\sqrt{2}i}{2}$