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Q: 12    Let   \small z_1=2-i,z_2=-2+i.  Find

               (i)  \small Re\left ( \frac{z_1z_2}{\bar{z_1}} \right )

Answers (1)

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It is given that
\small z_1=2-i \ and \ z_2=-2+i
Now,
z_1z_2= (2-i)(-2+i)= -4+2i+2i-i^2=-4+4i+1= -3+4i
And
\bar z_1 = 2+i
Now,
\frac{z_1z_2}{\bar z_1}= \frac{-3+4i}{2+i}= \frac{-3+4i}{2+i}\times \frac{2-i}{2-i}= \frac{-6+3i+8i-4i^2}{2^2-i^2}= \frac{-6+11i+4}{4+1}= \frac{-2+11i}{5}= -\frac{2}{5}+i\frac{11}{5}
Now,
Re\left ( \frac{z_1z_2}{z_1} \right )= -\frac{2}{5}
Therefore, the answer is

 -\frac{2}{5}

Posted by

Gautam harsolia

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