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COMPLEX NUMBERS AND QUADRATIC EQUATIONS — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSCOMPLEX NUMBERS AND QUADRATIC EQUATIONS1 Mark
Question
Find the modulus and argument of complex number $\frac { 1 } { 1 + i }$.
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Answer

We have, $\frac { 1 } { 1 + i } = \frac { 1 } { 1 + i } \times \frac { 1 - i } { 1 - i } = \frac { 1 - i } { 1 ^ { 2 } - i ^ { 2 } } = \frac { 1 - i } { 1 + 1 }$
= $\frac { 1 - i } { 2 } = \frac { 1 } { 2 } - \frac { i } { 2 }$
Let r cos$\theta$ = $\frac { 1 } { 2 }$ ...(i)
and r sin$\theta$ = - $\frac { 1 } { 2 }$ ...(ii)
On squaring and adding Eqs. (i) and (ii), we get
$r^2 \cos ^2 \theta+r^2 \sin ^2 \theta$ = $\left( \frac { 1 } { 2 } \right) ^ { 2 } + \left( - \frac { 1 } { 2 } \right) ^ { 2 }$
$\Rightarrow r^2$ =$\frac { 1 } { 4 } + \frac { 1 } { 4 } = \frac { 1 } { 2 } \Rightarrow$ r = $\frac { 1 } { \sqrt { 2 } }$ [$\because$ r is positive]
On putting the value of r in Eqs. (i) and (ii), we get
cos$\theta$ = $\frac { 1 } { \sqrt { 2 } }$ and sin$\theta$ = - $\frac { 1 } { \sqrt { 2 } }$
Since, cos$\theta$ is positive and sin$\theta$ is negative.
So, $\theta$ lies in IV quadrant.
$\therefore$ $\theta$ = - $\frac { \pi } { 4 }$
Hence, modulus of $\frac { 1 } { 1 + i }$ is $\frac { 1 } { \sqrt { 2 } }$ and argument is $\frac { - \pi } { 4 }$.

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