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  • Where does z lie, if |(x+yi-5i)/(x+iy+5i)|=1.

Where does z lie, if |\frac{z-5i}{z+5i}|=1.

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\begin{array}{c} \text { Let } z=x+i y \quad\left|\frac{x+y i-5 i}{x+i y+5 i}\right|=1 \\ |x+(y-5) i|=|x+(y+5) i| \\ x^{2}+(y-5)^{2}=x^{2}+(y+5)^{2} \\ (y-5)^{2}=(y+5)^{2} \\ \qquad \begin{array}{r} y^{2}+25-10 y=y^{2}+25+10 y \\ 20 y=0 \quad y=0 \end{array} \end{array}

 Hence, z lies on x-axis i.e., real axis.

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infoexpert21

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