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If $\;z_{1}\;$ and $\;z_{2}\;$ both satisfy $\;z+\overline{z} =2|z-1|\;arg(z_{1}-z_{2}) = \large\frac{\pi}{4}\;,$ then find $\;Im(z_{1}+z_{2})\;.$

$(a)\;2\qquad(b)\;\pm6\qquad(c)\;\pm3\qquad(d)\;4$

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Answer : 2
Explanation :
Let $\;z=x+iy\;,z_{1}=x_{1}+iy_{1}\;$ and $\;z_{2} = x_{2}+iy_{2}$
Then , $\;z+\overline{z} =2|z-1|$
$x+iy + x-iy = 2|x-1+iy|$
$2x=1+y^{2}$---(1)
Since $\;z_{1}\;$ and $\;z_{2}\;$ both satisfy (1) ,we have
$2x_{1} =1+y^{2}_{1}...\;$ and $\;2x_{2} = 1+y^{2}_{2}$
$2(x_{1}+x_{2}) = (y_{1}+y_{2})(y_{1}-y_{2})$
$2= (y_{1}+y_{2}) ( \large\frac{y_{1}-y_{2}}{x_{1}-x_{2}})$---(2)
$z_{1}-z_{2} = (x_{1}-x_{2}) + i(y_{1}-y_{2})$
Therefore , $\;tan \theta = \large\frac{y_{1} -y_{2}}{x_{1}-x_{2}}\;$ where $\; \theta = arg(z_{1}-z_{2})$
$tan \large\frac{\pi}{4} = \large\frac{y_{1} -y_{2}}{x_{1}-x_{2}} \quad $ [$\; \theta = \large\frac{\pi}{4}$]
$1 = \large\frac{y_{1} -y_{2}}{x_{1}-x_{2}} $
From (2) , we get $\;2= y_{1}+y_{2}$ i.e $\;Im(z_{1}+z_{2}) =2$
answered Apr 17, 2014 by yamini.v
 

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