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# Find the modulus and argument of the complex number $\; \large\frac{1}{1+i}$

$(a)\;\large\frac{1}{\sqrt{2}} , -\large\frac{\pi}{4}\qquad(b)\;3 , \large\frac{\pi}{2}\qquad(c)\;\large\frac{1}{\sqrt{3}} , \large\frac{\pi}{4}\qquad(d)\;\large\frac{1}{\sqrt{2}} , \large\frac{\pi}{4}$

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Answer : $\;\large\frac{1}{\sqrt{2}} , -\large\frac{\pi}{4}$
Explanation :
We have , $\; \large\frac{1}{1+i}$
$\; \large\frac{1}{1+i} = \large\frac{1}{1+i} \times \large\frac{1-i}{1-i}$
$=\large\frac{1-i}{1^{2}-i^{2}}$
$=\large\frac{1-i}{2}$
$= \large\frac{1}{2} - \large\frac{i}{2}$
Now let us put $\;\large\frac{1}{2} = r cos \theta \;$ and $\;- \large\frac{1}{2}=r sin \theta$
On squaring and adding , we obtain
$r^{2}(cos^{2} \theta + sin^{2} \theta)^{2}= (\large\frac{1}{2})^{2}+ (-\large\frac{1}{2})^{2}$
$r^{2}= (\large\frac{1}{4})+ (\large\frac{1}{4})$
$r^{2}= \large\frac{1}{2}$
$r= \large\frac{1}{\sqrt{2}}$
$\; cos \theta = \large\frac{1}{\sqrt{2}}\;$ and $sin \theta = -\large\frac{1}{\sqrt{2}}$
Therefore , $\; \theta = -\large\frac{\pi}{4}$
Hence , the modulus of $\; \large\frac{1}{1+i}$ is $\;\large\frac{1}{\sqrt{2}}\;$ and the argument is $\; - \large\frac{\pi}{4}$
answered Apr 16, 2014 by

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