Let

Given equation $|z|^3+2z^2+4\bar z-8=0$ Substitute $z=x+iy$.

First compute: $|z|^3=(x^2+y^2)^{3/2}$ $z^2=(x+iy)^2=x^2-y^2+2ixy$ $\bar z=x-iy$

Hence, $|z|^3+2(x^2-y^2+2ixy)+4(x-iy)-8=0$

Separate real and imaginary parts:

• Real part: $|z|^3+2(x^2-y^2)+4x-8=0$
• Imaginary part: $4xy-4y=0$ $4y(x-1)=0$

Since $y\neq 0$, we must have $x=1$

Use $x=1$ in the real part Let $r=|z|=\sqrt{x^2+y^2}=\sqrt{1+y^2}$ Then the real part becomes $r^3+2(1-y^2)+4-8=0$ $r^3-2y^2-2=0$

Since $r^2=1+y^2$, we have $y^2=r^2-1$ So, $r^3-2(r^2-1)-2=0$ $r^3-2r^2=0$ $r^2(r-2)=0$

As $r=|z|>0$, we get $r=2$ Therefore, $|z|^2=r^2=4$

Find $y$ $x^2+y^2=4$ $1+y^2=4$ $y^2=3$ Since $y\neq 0$, this is valid.

Now evaluate each quantity

(P) $|z|^2$

$|z|^2=4$ So, $P\to (2)$

(Q) $|z-\bar z|^2$

$z-\bar z=(x+iy)-(x-iy)=2iy$ $|z-\bar z|^2=|2iy|^2=4y^2=4\cdot 3=12$ So, $Q\to (1)$

(R) $|z|^2+|z+\bar z|^2$

$z+\bar z=2x=2$ $|z+\bar z|^2=|2|^2=4$ Hence, $|z|^2+|z+\bar z|^2=4+4=8$ So, $R\to (3)$

(S) $|z+1|^2$

$z+1=(x+1)+iy=2+iy$ $|z+1|^2=2^2+y^2=4+3=7$ So, $S\to (5)$

Final matching $P\to (2),\quad Q\to (1),\quad R\to (3),\quad S\to (5)$

This corresponds to Option B.