Question

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

2) Find the modulus and argument of the complex numbers:
ii) $\frac{1}{1+i}$

Answer 1

1) Step 1 Given data (let) $z=\frac{1+i}{1-i}$
Step 2 Using a rationalization method.
$z=\frac{1+i}{1-i}\times \frac{1+i}{1+i}$
$Z=\frac{(1+i{)}^2}{(1{)}^2-(i{)}^2}\left\{\text{using}\right(a-b\left)\right(a+b)=a^2-b^2\}$
$z=\frac{1+i^2+\mathrm{2.1.}i}{1+1}[\because (i{)}^2=-1]$
$z=\frac{1-1+2i}{2}$
$z=\frac{2i}{2}=i$
$\Rightarrow z=0+i=x+iy$ (let)
Step 3 Modulus of $z$:
$|z|=\sqrt{(0{)}^2+(1{)}^2}=1$
Step 4 Argument of $z$:
As we can see $z$ lies on the positive y-axis
So, $\text{arg}\left(z\right)=\frac{\pi }{2}$

2) Step 1 Given data: (let ) $z=\frac{1}{1+i}$
Step 2 Using rationalization method
$z=\frac{1}{1+i}\times \frac{1-i}{1-i}$
$z=\frac{1-i}{(1{)}^2-(i{)}^2}\left\{\text{using}\right(a-b\left)\right(a+b)=a^2-b^2\}$
$z=\frac{1-i}{1+1}$
$z=\frac{1-i}{2}$
$z=\frac{1}{2}-i\frac{1}{2}=x+iy$ (let)
Step 3 Modulus of $z$:
$|z|=\sqrt{{\left(\frac{1}{2}\right)}^2+{(-\frac{1}{2})}^2}=\sqrt{\frac{1}{4}+\frac{1}{4}}=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}$
Step 4 Argument of $z$:
As we can see $z$ is in the fourth quadrant
So, $\text{arg}\left(z\right)=-{\tan }^{-1}\left|\frac{y}{x}\right|$
$=-{\tan }^{-1}\left|\frac{-\frac{1}{2}}{\frac{1}{2}}\right|$
$=-{\tan }^{-1}\left(1\right)$
$=-\frac{\pi }{4}$