Complex Number Equation and Modulus Expressions Matching Set

To solve the problem, we start with the given complex equation: $|z|^3 + 2z^2 + 4\bar{z} - 8 = 0$

Taking the complex conjugate of both sides of the equation (noting that $|z|^3$ and $8$ are real): $|z|^3 + 2\bar{z}^2 + 4z - 8 = 0$

Subtracting the conjugated equation from the original equation: $(2z^2 + 4\bar{z}) - (2\bar{z}^2 + 4z) = 0$ $2(z^2 - \bar{z}^2) - 4(z - \bar{z}) = 0$ $2(z - \bar{z})(z + \bar{z}) - 4(z - \bar{z}) = 0$ $2(z - \bar{z})(z + \bar{z} - 2) = 0$

Since the imaginary part of $z$ is non-zero ($\text{Im}(z) \neq 0$), we have $z \neq \bar{z}$, which implies $z - \bar{z} \neq 0$. Therefore: $z + \bar{z} - 2 = 0 \implies z + \bar{z} = 2$

Let $z = x + iy$, where $x, y \in \mathbb{R}$ and $y \neq 0$. Since $z + \bar{z} = 2x = 2$, we get: $x = 1 \implies z = 1 + iy$

Now, let $r = |z| = \sqrt{1^2 + y^2} = \sqrt{1 + y^2}$, which gives $y^2 = r^2 - 1$.

Substitute $z = 1 + iy$ and $\bar{z} = 1 - iy$ back into the original equation: $r^3 + 2(1 + iy)^2 + 4(1 - iy) - 8 = 0$ $r^3 + 2(1 - y^2 + 2iy) + 4 - 4iy - 8 = 0$ $r^3 + 2 - 2y^2 + 4iy + 4 - 4iy - 8 = 0$ $r^3 - 2y^2 - 2 = 0$

Substitute $y^2 = r^2 - 1$: $r^3 - 2(r^2 - 1) - 2 = 0$ $r^3 - 2r^2 = 0$ $r^2(r - 2) = 0$

Since $r = |z| > 0$, we have $r = 2$. Hence, $|z| = 2$ and $y^2 = 2^2 - 1 = 3 \implies y = \pm\sqrt{3}$. So, $z = 1 \pm i\sqrt{3}$.

Now we evaluate each expression in List-I:

1. (P) $|z|^2$: $|z|^2 = 2^2 = 4$ So, $(\text{P}) \rightarrow (2)$.

2. (Q) $|z - \bar{z}|^2$: $z - \bar{z} = (1 \pm i\sqrt{3}) - (1 \mp i\sqrt{3}) = \pm 2i\sqrt{3}$ $|z - \bar{z}|^2 = |\pm 2i\sqrt{3}|^2 = (2\sqrt{3})^2 = 12$ So, $(\text{Q}) \rightarrow (1)$.

3. (R) $|z|^2 + |z + \bar{z}|^2$: $|z|^2 + |z + \bar{z}|^2 = 4 + 2^2 = 4 + 4 = 8$ So, $(\text{R}) \rightarrow (3)$.

4. (S) $|z + 1|^2$: $z + 1 = 2 \pm i\sqrt{3}$ $|z + 1|^2 = 2^2 + (\sqrt{3})^2 = 4 + 3 = 7$ So, $(\text{S}) \rightarrow (5)$.

Thus, the correct matching is: $(\text{P}) \rightarrow (2) \quad (\text{Q}) \rightarrow (1) \quad (\text{R}) \rightarrow (3) \quad (\text{S}) \rightarrow (5)$

This corresponds to Option (B).