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Complex Numbers and Quadratic Equations — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsComplex Numbers and Quadratic Equations2 Marks
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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