Find the modulus of each of the following complex numbers and hence express each of
them in polar form: $\sqrt{\frac{1+\mathrm{i}}{1-\mathrm{i}}}$
$=\sqrt{\frac{1+i}{1-i}} \times \sqrt{\frac{1+i}{1+i}}$
$=\sqrt{\frac{(1+i)^{2}}{1-i^{2}}}$
$=\frac{1+i}{\sqrt{2}}$
$=\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}$
Let $Z=\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}=r(\cos \theta+i \sin \theta)$
Now, separating real and complex part , we get
$\frac{1}{\sqrt{2}}=r \cos \theta$ ……….eq.1
$\frac{1}{\sqrt{2}}=r \sin \theta$ …………eq.2
Squaring and adding eq.1 and eq.2, we get
$1=r^{2}$
Since r is always a positive no., therefore,
r = 1,
hence its modulus is 1.
now , dividing eq.2 by eq.1 , we get,
$\frac{r \sin \theta}{r \cos \theta}=\frac{\frac{i}{\sqrt{2}}}{\frac{i}{\sqrt{2}}}$
$\tan \theta=1$
Since $\cos \theta=\frac{1}{\sqrt{2}} \sin \theta=\frac{1}{\sqrt{2}}$ $\operatorname{and} \tan \theta=1$. therefore the $\theta$ lies in first quadrant.
$\operatorname{Tan} \theta=1$, therefore $\theta=\frac{\pi}{4}$
Representing the complex no. in its polar form will be
$\mathrm{Z}=1\left\{\cos \left(\frac{\pi}{4}\right)+\mathrm{i} \sin \left(\frac{\pi}{4}\right)\right\}$
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