Question

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

Answer 1

Consider,

$z=\frac{1}{1+i}$

$z=\frac{1}{1+i}\times \frac{1-i}{1-i}$

$z=\frac{1-i}{(1+i)(1-i)}$

$z=\frac{1-i}{1^2-i^2}$

$z=\frac{1-i}{1-(-1)}$ -------- We know that $i^2=-1$

$z=\frac{1-i}{2}$

$z=\frac{1}{2}-\frac{1}{2}i$

This is the complex number of the form $z=x+iy$

$x=\frac{1}{2}$ and $y=-\frac{1}{2}$

We know that, $Modulus=\sqrt{x^2+y^2}$

$=\sqrt{\frac{1}{4}+\frac{1}{4}}$

$=\sqrt{\frac{2}{4}}$

$=\frac{1}{\sqrt{2}}$

Argument:

$\frac{1}{2}+(-\frac{1}{2})i=r\cos \theta +ir\sin \theta$

Equating the real parts,

$\frac{1}{2}=r\cos \theta$

$\frac{1}{2}=\frac{1}{\sqrt{2}}\cos \theta$ ------ Above, $modulus=r=\frac{1}{\sqrt{2}}$

$\cos \theta =\frac{1}{\sqrt{2}}$

Similarly, equating the imaginary parts we get,

$-\frac{1}{2}=r\sin \theta$

$⟹\sin \theta =-\frac{1}{\sqrt{2}}$

Here, $\sin \theta$ is $-ve$ and$\cos \theta$ is $+ve$. Hence, $\theta$ is in 4th quadrant.

So, $Argument=-{45}^{\circ }=-\frac{\pi }{4}radians$