Find the modulus and argument of the complex numbers:
$\frac{1}{1+i}$
We have $\frac{1}{1+i}=\frac{1-i}{(1+i)(1-i)}=\frac{1-i}{1+1}=\frac{1}{2}-\frac{i}{2}$
Let $\frac{1}{2}=r \cos \theta,-\frac{1}{2}=r \sin \theta$
Proceeding as in part $(i)$ above, we get $r=\frac{1}{\sqrt{2}} ; \cos \theta=\frac{1}{\sqrt{2}}, \sin \theta=\frac{-1}{\sqrt{2}}$
Therefore $\theta=\frac{-\pi}{4}$
Hence, the modulus of $\frac{1}{1+i}$ is $\frac{1}{\sqrt{2}},$ argument is $\frac{-\pi}{4}$.
For any two complex numbers ${z_1},{z_2}$we have $|{z_1} + {z_2}{|^2} = $ $|{z_1}{|^2} + |{z_2}{|^2}$ then
Find the modulus and argument of the complex number $\frac{1+2 i}{1-3 i}$
The set of all $\alpha \in R$, for which $w = \frac{{1 + \left( {1 - 8\alpha } \right)z}}{{1 - z}}$ is a purely imaginary number, for all $z \in C$ satisfying $\left| z \right| = 1$ and ${\mathop{\rm Re}\nolimits} \,z \ne 1$, is
If $\frac{{z - i}}{{z + i}}(z \ne - i)$ is a purely imaginary number, then $z.\bar z$ is equal to
Find the modulus of $\frac{1+i}{1-i}-\frac{1-i}{1+i}$