The region of the complex plane for which lef | vedclass

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(B) Given $\left| \frac{z - a}{z + \overline{a}} \right| = 1$. This implies $|z - a| = |z + \overline{a}|$. Squaring both sides: $|z - a|^2 = |z + \ov

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$y$-axis

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(B) Given $\left| \frac{z - a}{z + \overline{a}} ight| = 1$. This implies $|z - a| = |z + \overline{a}|$. Squaring both sides: $|z - a|^2 = |z + \overline{a}|^2$. Using the property $|w|^2 = w \cdot \overline{w}$,we get: $(z - a)(\overline{z} - \overline{a}) = (z + \overline{a})(\overline{z} + a)$. Expanding both sides: $z\overline{z} - z\overline{a} - a\overline{z} + a\overline{a} = z\overline{z} + za + \overline{a}\overline{z} + \overline{a}a$. Canceling $z\overline{z}$ and $a\overline{a}$ from both sides: $-z\overline{a} - a\overline{z} = za + \overline{a}\overline{z}$. Rearranging terms: $za + z\overline{a} + a\overline{z} + \overline{a}\overline{z} = 0$. $(z + \overline{z})(a + \overline{a}) = 0$. Since $a + \overline{a} = 2	ext{Re}(a) 
eq 0$,we must have $z + \overline{z} = 0$. Let $z = x + iy$,then $z + \overline{z} = (x + iy) + (x - iy) = 2x = 0$. Thus,$x = 0$,which is the equation of the $y$-axis.

The region of the complex plane for which $\left| \frac{z - a}{z + \overline{a}} \right| = 1$ where $\text{Re}(a) \neq 0$ is

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