The complex numbers z = x + iy which satisfy | vedclass

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(A) Given the equation $\left| \frac{z - 5i}{z + 5i} \right| = 1$. Substituting $z = x + iy$,we get $\left| \frac{x + i(y - 5)}{x + i(y + 5)} \right|

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Real axis

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(A) Given the equation $\left| \frac{z - 5i}{z + 5i} \right| = 1$. Substituting $z = x + iy$,we get $\left| \frac{x + i(y - 5)}{x + i(y + 5)} \right| = 1$. Using the property $\left| \frac{z_1}{z_2} \right| = \frac{|z_1|}{|z_2|}$,we have $|x + i(y - 5)| = |x + i(y + 5)|$. Squaring both sides,we get $x^2 + (y - 5)^2 = x^2 + (y + 5)^2$. Expanding the squares: $x^2 + y^2 - 10y + 25 = x^2 + y^2 + 10y + 25$. Simplifying the equation: $-10y = 10y$,which implies $20y = 0$,so $y = 0$. The equation $y = 0$ represents the real axis.

The complex numbers $z = x + iy$ which satisfy the equation $\left| \frac{z - 5i}{z + 5i} \right| = 1$ lie on

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