JEE Advanced 2023 — Paper 1 — Question 16

$\text{Given\; } |z|^{3}+2z^{2}+4\bar z-8=0,\ \Im z\ne0.$

$\text{Take\; conjugate:\;}$

$$|z|^{3}+2\bar z^{2}+4z-8=0.$$

Subtract:

$$2(z^{2}-\bar z^{2})+4(\bar z-z)=0 \;\Rightarrow\; (z-\bar z)\bigl(2(z+\bar z)-4\bigr)=0.$$

Since $\Im z\ne0$, $z\ne\bar z$, hence

$$z+\bar z=2 \Rightarrow \Re z=1.$$

Let $z=1+iy,\ y\ne0.$

$$|z|^{2}=1+y^{2},\quad |z|^{3}=(1+y^{2})^{3/2}.$$

Substitute in the original equation:

$$(1+y^{2})^{3/2}+2(1+iy)^{2}+4(1-iy)-8=0.$$ $$(1+iy)^{2}=1-y^{2}+2iy \Rightarrow 2(1+iy)^{2}+4(1-iy)-8=-2y^{2}-2.$$

So

$$(1+y^{2})^{3/2}-2y^{2}-2=0 \Rightarrow (1+y^{2})^{3/2}=2(y^{2}+1).$$

Divide by $y^{2}+1>0$:

$$\sqrt{1+y^{2}}=2 \Rightarrow 1+y^{2}=4 \Rightarrow y^{2}=3.$$

Thus

$$z = 1\pm i\sqrt{3}.$$

Now the required quantities:

1. $|z|^{2}$:

$$|z|^{2}=1+y^{2}=1+3=4.$$

2. $|z-\bar z|^{2}$:

$$z-\bar z = 2iy \Rightarrow |z-\bar z|^{2}=|2iy|^{2}=4y^{2}=12.$$

3. $|z|^{2}+|z+\bar z|^{2}$:

$$z+\bar z=2 \Rightarrow |z+\bar z|^{2}=4,$$ $$|z|^{2}+|z+\bar z|^{2}=4+4=8.$$

4. $|z+1|^{2}$:

$$z+1=2\pm i\sqrt{3} \Rightarrow |z+1|^{2}=2^{2}+(\sqrt{3})^{2}=4+3=7.$$

Matchings (List–I $\to$ List–II): $P)\ |z|^{2} \to 4\ (2),\qquad$ $Q)\ |z-\bar z|^{2} \to 12\ (1),$ $R)\ |z|^{2}+|z+\bar z|^{2} \to 8\ (3),\qquad$ $S)\ |z+1|^{2} \to 7\ (5).$