Browse Questions

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

$(a)\;1 , \large\frac{\pi}{2}\qquad(b)\;3 , \large\frac{\pi}{2}\qquad(c)\;1 , \large\frac{\pi}{4}\qquad(d)\;5 , \large\frac{\pi}{2}$

Answer : $\;1 , \large\frac{\pi}{2}$
Explanation :
We have , $\; \large\frac{1+i}{1-i}$
$\large\frac{1+i}{1-i}= \large\frac{1+i}{1-i} \times \large\frac{1+i}{1+i}$
$= \large\frac{(1+i)^{2}}{1^{2}-i^{2}} = \large\frac{1^{2} +i^{2} +2i}{1+1}$
$= \large\frac{2i}{2} = i = 0+ i$
Now let us put $\;0 = r cos \theta \;$ and $\;1=r sin \theta$
On squaring and adding , we obtain
$r^{2}(cos^{2} \theta + sin^{2} \theta)^{2}= 0^{2}+ 1^{2}$
$r^{2} = 1\;$ so that
$\; cos \theta =0\;$ and $sin \theta =1$
Therefore , $\; \theta = \large\frac{\pi}{2}$
Hence , the modulus of $\; \large\frac{1+i}{1-i}$ is 1 and the argument is $\;\large\frac{\pi}{2}$