<p>Let \(z\) satisfy \(\text{Im}(z^2) = \text{Im}(z\bar{z})\). Which hold?</p>
Step-by-Step Solution
Key Concept: Im(z^2) = Im(z \cdot z̄) = Im(|z|^2) = 0. So Im(z^2) = 0. If z=x+iy: Im(z^2)=2xy=0 \Rightarrow x=0 or y=0 \Rightarrow z is real or purely imaginary.
<p>$z\bar{z}=|z|^2\in\mathbb{R}$, so $\text{Im}(z\bar{z})=0$. The condition becomes $\text{Im}(z^2)=0$. With $z=x+iy$: $z^2=x^2-y^2+2xyi\Rightarrow 2xy=0\Rightarrow x=0$ or $y=0$. So C=correct. D is also true in the sense that there are other cases. Key=CD.</p>
Correct Answer: CD