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Let $\;z_{1}\;$ and $\;z_{2}\;$ be two complex numbers such that $\;\overline{z_{1} }+ i\overline{z_{2}}=0\;$ and $\;arg(z_{1} z_{2}) = \pi\;$ . Then find $\;arg(z_{1})$

$(a)\;\large\frac{3 \pi}{2} \qquad(b)\; \large\frac{3 \pi}{4}\qquad(c)\;\large\frac{ \pi}{2}\qquad(d)\;\large\frac{5 \pi}{2}$

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Answer : $\;\large\frac{3 \pi}{4}$
Explanation :
Given that , $\;\overline{z_{1} }+i \overline{z_{2}}=0\;$
$z_{1}= i z_{2} \;$ i.e $\; z_{2} = -iz_{1}$
Thus , $\;arg(z_{1} z_{2}) = arg(z_{1} ) + arg(-iz_{1}) = \pi$
$arg(-iz^{2}_{1} )= \pi$
$arg(-i) + arg(z^{2}_{1}) = \pi$
$arg(-i) + 2\;arg(z_{1}) = \pi$
$- \large\frac{\pi}{2} + 2\;arg(z_{1}) = \pi$
$\;arg(z_{2})= \large\frac{3 \pi}{4} \;.$
answered Apr 17, 2014 by yamini.v
 

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