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If z = x + iy lies in the third quadrant, the   \frac{\bar{z}}{z}  also lies in the third quadrant if
A. x > y > 0
B. x < y < 0
C. y < x < 0
D. y > x > 0

 

Answers (1)

The answer is the option (b).

If z lies in the third quadrant,So,x<0 \: \: and\: \: y<0\: \: , z=x+iy

\begin{aligned} \frac{\bar{z}}{z} &=\frac{x-i y}{x+i y}=\frac{x-i y}{x+i y} * \frac{x-i y}{x-i y} \\ &=\frac{x^{2}+i^{2} y^{2}-2 x y i}{x^{2}-i^{2} y^{2}} \\ &=\frac{x^{2}-y^{2}}{x^{2}+y^{2}}-\frac{2 x y}{x^{2}+y^{2}} i \end{aligned}

When z lies in third quadrant then \frac{\bar{z}}{z} will also be in third quadrant.

\begin{array}{c} \frac{x^{2}-y^{2}}{x^{2}+y^{2}}<0 \text { and } \frac{2 x y}{x^{2}+y^{2}}>0 \\\\ x^{2}-y^{2}<0 \text { and } 2 x y>0 \\\\ x^{2}<y^{2} \text { and } x y>0 \\\\ \text { So, } x<y<0 \end{array}

Hence, b is correct..

 

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infoexpert21

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