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Q: 10     If  \small z_1=2-i, z_2=1+i, find    \small \left |\frac{z_1+z_2+1}{z_1-z_2+1} \right |.

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It is given that
\small z_1=2-i, z_2=1+i
Then,
\small \left |\frac{z_1+z_2+1}{z_1-z_2+1} \right | =\left | \frac{2-i+1+i+1}{2-i-1-i+1} \right | = \left | \frac{4}{2(1-i)} \right |= \left | \frac{2}{(1-i)} \right |
Now, multiply the numerator and denominator  by  1+i 
\Rightarrow \left | \frac{2}{(1-i)} \times \frac{1+i}{1+i} \right |=\left |\frac{2(1+i)}{1^2-i^2} \right |=\left | \frac{2(1+i)}{1+1} \right |= \left| 1+i \right |
Now,
|1+i| = \sqrt{1^2+1^2}=\sqrt{1+1}=\sqrt{2}
Therefore, the value of

  \small \left |\frac{z_1+z_2+1}{z_1-z_2+1} \right |   is  \sqrt{2}

Posted by

Gautam harsolia

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