# Solve this

Question:

If $z_{1}=(2-i)$ and $z_{2}=(1+i)$, find $\left|\frac{z_{1}+z_{2}+1}{z_{1}-z_{2}+i}\right|$.

Solution:

Given: $z_{1}=(2-i)$ and $z_{2}=(1+i)$

To find: $\left|\frac{z_{1}+z_{2}+1}{z_{1}-z_{2}+i}\right|$

Consider,

$\left|\frac{z_{1}+z_{2}+1}{z_{1}-z_{2}+i}\right|$

Putting the value of $z_{1}$ and $z_{2}$, we get

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

$=\left|\frac{4}{2-i-1-i+i}\right|$

$=\left|\frac{4}{1-i}\right|$

Now, rationalizing by multiply and divide by the conjugate of $1-\mathrm{i}$

$=\left|\frac{4}{1-i} \times \frac{1+i}{1+i}\right|$

$=\left|\frac{4(1+i)}{(1-i)(1+i)}\right|$

$=\left|\frac{4(1+i)}{(1)^{2}-(i)^{2}}\right|_{\left[(a-b)(a+b)=a^{2}-b^{2}\right]}$

$=\left|\frac{4(1+i)}{1-i^{2}}\right|$

$=\left|\frac{4(1+i)}{1-(-1)}\right|_{\left[\text {Putting } i^{2}=-1\right]}$

$=\left|\frac{4(1+i)}{2}\right|$

$=|2(1+i)|$

$=|2+2 i|$

Now, we have to find the modulus of $(2+2 i)$

So, $|z|=|2+2 i|=\sqrt{(2)^{2}+(2)^{2}}=\sqrt{4+4}=\sqrt{8}=2 \sqrt{2}$

Hence, the value of $\left|\frac{z_{1}+z_{2}+1}{z_{1}-z_{2}+i}\right|=2 \sqrt{2}$