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KEAM 2011 Math Paper

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Question : 83 of 120

Marks: +1, -0

If

z_1

z_2

are two non-zero complex numbers such that

|z_1+z_2|=|z_1|+|z_2|,

\arg\left(\frac{z_1}{z_2}\right)

is equal to

0
$-\pi$
$-\frac{\pi}{2}$
$\frac{\pi}{2}$
$\pi$

Solution:

Given,

|z_{1}+z_{2}|=|z_{1}|+|z_{2}|

Let

\;\; z_{1}=x_{1}+i y_{1}

z_{2}=x_{2}+i y_{2}

|(x_{1}+x_{2})+i(y_{1}+y_{2})|=|x_{1}+i y_{1}|+|x_{2}+i y_{2}|

\sqrt{(x_{1}+x_{2})^{2}+(y_{1}+y_{2})^{2}}=\sqrt{x_{1}^{2}+y_{1}^{2}}+\sqrt{x_{2}^{2}+y_{2}^{2}}

Squaring on both sides

x_{1}^{2}+x_{2}^{2}+2 x_{1} x_{2}+y_{1}^{2}+y_{2}^{2}+2 y_{1} y_{2}

=x_{1}^{2}+y_{1}^{2}+x_{2}^{2}+y_{2}^{2}+2\sqrt{(x_{1}^{2}+y_{1}^{2})(x_{2}^{2}+y_{2}^{2})}

\Rightarrow\;\; 2(x_{1} x_{2}+y_{1} y_{2})=2\sqrt{(x_{1}^{2}+y_{1}^{2})(x_{2}^{2}+y_{2}^{2})}

\Rightarrow\;\; (x_{1} x_{2}+y_{1} y_{2})^{2}=(x_{1}^{2}+y_{1}^{2})(x_{2}^{2}+y_{2}^{2})

\Rightarrow\;\; x_{1}^{2} x_{2}^{2}+y_{1}^{2} y_{2}^{2}+2 x_{1} x_{2} y_{1} y_{2}

=x_{1}^{2} x_{2}^{2}+x_{2}^{2} y_{1}^{2}+x_{1}^{2} y_{2}^{2}+y_{1}^{2} y_{2}^{2}

\Rightarrow\;\; 2 x_{1} x_{2} y_{1} y_{2}=x_{2}^{2} y_{1}^{2}+x_{1}^{2} y_{2}^{2}

\Rightarrow\;\; (-x_{1} y_{2}+y_{1} x_{2})^{2}=0

\Rightarrow\;\; x_{2} y_{1}-y_{2} x_{1}=0

......(i)

\;\frac{z_{1}}{z_{2}}=\;\frac{x_{1}+i y_{1}}{x_{2}+i y_{2}}

=\;\frac{(x_{1}+i y_{1})(x_{2}-i y_{2})}{x_{2}^{2}-y_{2}^{2}}

=\;\frac{x_{1} x_{2}+y_{1} y_{2}}{x_{2}^{2}-y_{2}^{2}}+i\;\frac{y_{1} x_{2}-x_{1} y_{2}}{x_{2}^{2}-y_{2}^{2}}

\arg\left(\frac{z_{1}}{z_{2}}\right)=\tan^{-1}\left(\frac{\operatorname{Im}\left(\frac{z_{1}}{z_{2}}\right)}{\operatorname{Re}\left(\frac{z_{1}}{z_{2}}\right)}\right)

=\tan^{-1}\left(\frac{y_{1} x_{2}-x_{1} y_{2}}{x_{1} x_{2}+y_{1} y_{2}}\right)

=\tan^{-1}(0)=0\;\; [

from

\text{Eq}

(i)]

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