Search any question & find its solution
Question:
Answered & Verified by Expert
If $z_1=2-i, z_2=1+i$ find $\left|\frac{z_1+z_2+1}{z_1-z_2+i}\right|$
Solution:
1105 Upvotes
Verified Answer
$\left|\frac{z_1+z_2+1}{z_1-z_2+i}\right|=\left|\frac{(2-i)+(1+i)+1}{(2-i)-(1+i)+i}\right|$
$\Rightarrow\left|\frac{2-i+1+i+1}{2-i-1-i+i}\right|=\left|\frac{4}{1-i}\right|=\left|\frac{4}{1-i} \times \frac{1+i}{1+i}\right|=\left|\frac{4(1+i)}{1-i^2}\right|$
$\Rightarrow 4\left|\frac{(1+i)}{1+1}\right|=4\left|\frac{(1+i)}{2}\right|=2|(1+i)|$
$\therefore \quad\left|\frac{z_1+z_2+1}{z_1-z_2+i}\right|$
$\quad=2|(1+i)|=2 \cdot \sqrt{(1)^2+(1)^2}=2 \sqrt{2}$
$\Rightarrow\left|\frac{2-i+1+i+1}{2-i-1-i+i}\right|=\left|\frac{4}{1-i}\right|=\left|\frac{4}{1-i} \times \frac{1+i}{1+i}\right|=\left|\frac{4(1+i)}{1-i^2}\right|$
$\Rightarrow 4\left|\frac{(1+i)}{1+1}\right|=4\left|\frac{(1+i)}{2}\right|=2|(1+i)|$
$\therefore \quad\left|\frac{z_1+z_2+1}{z_1-z_2+i}\right|$
$\quad=2|(1+i)|=2 \cdot \sqrt{(1)^2+(1)^2}=2 \sqrt{2}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.