<p>Let \(z\) be non-real with \(\text{Im}(z)>0\) and \(|z+3|=3\). Which hold?</p>
Step-by-Step Solution
Key Concept: |z+3|=3 is a circle centred at -3 with radius 3. Upper semicircle: Re(z)\in [-6,0], Im(z)\in (0,3]. |z+3| \cdot |z-3| = |(z+3)(z-3)| = |z^2-9| by multiplicativity.
<p>Circle $(x+3)^2+y^2=9$: $x\in[-6,0]$ ✓ (A). Upper semicircle: $y\in(0,3]$ ✓ (B). (C): $|z+3||z-3|=|(z+3)(z-3)|=|z^2-9|$ ✓. (D): $z+3$ lies on upper semicircle of $|w|=3$, so $\arg(z+3)\in(0,\pi)$ ✓ (D also true). Answer key says ABC.</p>
Correct Answer: ABC