Find the modulus of each of the following:

Solution:

Given: $\frac{(2-i)(1+i)}{(1+i)}$

Firstly, we calculate $\frac{(2-i)(1+i)}{(1+i)}$ and then find its modulus

$\frac{(2-i)(1+i)}{(1+i)}=\frac{2(1)+2(i)+(-i)(1)+(-i)(i)}{(1+i)}$

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

$=\frac{2+i-(-1)}{1+i}\left[\because i^{2}=-1\right]$

$=\frac{3+i}{1+i}$

Now, we rationalize the above by multiplying and divide by the conjugate of 1 + i

$=\frac{3+i}{1+i} \times \frac{1-i}{1-i}$

$=\frac{(3+i)(1-i)}{(1+i)(1-i)} \ldots(\mathrm{i})$

Now, we know that,

$(a+b)(a-b)=\left(a^{2}-b^{2}\right)$

So, eq. (i) become

$=\frac{3(1-i)+i(1-i)}{(1)^{2}-(i)^{2}}$

$=\frac{3(1)+3(-i)+i(1)+i(-i)}{1-i^{2}}$

$=\frac{3-3 i+i-i^{2}}{1-(-1)}\left[\because \mathrm{i}^{2}=-1\right]$

$=\frac{3-2 i-(-1)}{1+1}\left[\because \mathrm{i}^{2}=-1\right]$

$=\frac{3-2 i+1}{2}$

$=\frac{4-2 i}{2}$

$=2-i$

Now, we have to find the modulus of $(2-\mathrm{i})$

So, $|z|=|2-i|=|2+(-1) i|=\sqrt{(2)^{2}+(-1)^{2}}=\sqrt{4+1}=\sqrt{5}$