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KEAM 2013 Math Paper

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Question : 8 of 120

Marks: +1, -0

Let

z_1

z_2

be complex numbers such that

z_1+i(\overline{z_2})=0

\arg(\overline{z_1}z_2)=\frac{\pi}{3}.

\arg(\overline{z}_1)

is

$\frac{\pi}{3}$
$\pi$
$\frac{\pi}{2}$
$\frac{5\pi}{12}$
$\frac{5\pi}{6}$

Solution:

Given

z_1+i\;\overline{z}_2=0

\Rightarrow\;\;z_1=(-i)\overline{z}_2

Taking argument on both sides, we get

\arg(z_1)=\arg\{(-i)\cdot\overline{z}_2\}

\Rightarrow\;\; \arg(z_1)=\arg(-i)+\arg(\overline{z}_2)

(by property)

\Rightarrow \arg(z_1)-\arg(\overline{z}_2)=\tan^{-1}\left(\frac{-1}{0}\right)=-\frac{\pi}{2}

\Rightarrow \arg(z_1)+\arg(z_2)=-\frac{\pi}{2}

....(i)

[ \because \arg(\overline{z})=-\arg(z) ]

and

\arg(\overline{z}_1 z_2)=\frac{\pi}{3}

\Rightarrow\;\; \arg(\overline{z}_1)+\arg(z_2)=\frac{\pi}{3}

\Rightarrow\;\; -\arg(z_1)+\arg(z_2)=\frac{\pi}{3}

.......(ii)

On adding Eqs. (i) and (ii), we get

2\arg(z_2)=-\frac{\pi}{6}

\Rightarrow

\arg(z_2)=-\frac{\pi}{12}

\therefore

From Eq. (i),

\arg(z_1)=-\frac{\pi}{2}+\frac{\pi}{12}=-\frac{5\pi}{12}

\Rightarrow\;\; -\arg(z_1)=\frac{5\pi}{12}

\Rightarrow

\arg(z_1)=\frac{5\pi}{12}

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