Purely Imaginary 77C212

1. **State the problem:** We need to prove that if $z$ is a complex number with magnitude $|z| = 1$ and $z \neq -1$, then the expression $$\frac{z - 1}{z + 1}$$ is purely imaginary. 2. **Recall the formula and important rules:** A complex number $w = a + bi$ is purely imaginary if its real part $a = 0$. 3. **Express $z$ in terms of its conjugate:** Since $|z| = 1$, we have $$z \overline{z} = 1,$$ where $\overline{z}$ is the complex conjugate of $z$. 4. **Calculate the conjugate of the expression:** Let $$w = \frac{z - 1}{z + 1}.$$ Then its conjugate is $$\overline{w} = \frac{\overline{z} - 1}{\overline{z} + 1}.$$ Using $\overline{z} = \frac{1}{z}$ (since $|z|=1$), we get $$\overline{w} = \frac{\frac{1}{z} - 1}{\frac{1}{z} + 1} = \frac{1 - z}{1 + z} \cdot \frac{1}{z} \cdot \frac{z}{z} = \frac{1 - z}{1 + z}.$$ Simplifying, $$\overline{w} = \frac{1 - z}{1 + z}.$$ 5. **Compare $w$ and $-\overline{w}$:** Note that $$-\overline{w} = -\frac{1 - z}{1 + z} = \frac{z - 1}{z + 1} = w.$$ 6. **Interpretation:** Since $w = -\overline{w}$, this means $$w + \overline{w} = 0,$$ so the real part of $w$ is zero. 7. **Conclusion:** Therefore, $$\frac{z - 1}{z + 1}$$ is purely imaginary. **Final answer:** The expression $$\frac{z - 1}{z + 1}$$ is purely imaginary when $|z|=1$ and $z \neq -1$.