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All complex number $z$ which satisfy the equation $\left|\frac{z-6 i}{z+6 i}\right|=1$ lie on the
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Verified Answer
The correct answer is:
real axis
We have,
$$
\begin{aligned}
&\quad\left|\frac{z-6 i}{z+6 i}\right|=1 \\
&\text { or } \quad|z-6 i|=|z+6 i| \\
&\text { Let } \quad \quad z=x+i y \\
&\text { or } \quad|x+i y-6 i|=|x+i y+6 i| \\
&\qquad|x+(y-6) i|=|x+(y+6) i| \\
&\Rightarrow \sqrt{x^{2}+(y-6)^{2}}=\sqrt{x^{2}+(y+6)^{2}} \\
&\Rightarrow \quad x^{2}+(y-6)^{2}=x^{2}+(y+6)^{2} \\
&\Rightarrow y^{2}+36-12 y=y^{2}+36+12 y \\
&\Rightarrow \quad \quad 24 y=0 \\
&\Rightarrow \quad y=0
\end{aligned}
$$
Hence, it is lie on the real axis.
$$
\begin{aligned}
&\quad\left|\frac{z-6 i}{z+6 i}\right|=1 \\
&\text { or } \quad|z-6 i|=|z+6 i| \\
&\text { Let } \quad \quad z=x+i y \\
&\text { or } \quad|x+i y-6 i|=|x+i y+6 i| \\
&\qquad|x+(y-6) i|=|x+(y+6) i| \\
&\Rightarrow \sqrt{x^{2}+(y-6)^{2}}=\sqrt{x^{2}+(y+6)^{2}} \\
&\Rightarrow \quad x^{2}+(y-6)^{2}=x^{2}+(y+6)^{2} \\
&\Rightarrow y^{2}+36-12 y=y^{2}+36+12 y \\
&\Rightarrow \quad \quad 24 y=0 \\
&\Rightarrow \quad y=0
\end{aligned}
$$
Hence, it is lie on the real axis.
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