Algebra
Complex Numbers
MMTS_Full_Test_07
Grade 12
Question:
Let $S=\{z\in\mathbb{C}:|z-1|^2+|z+1|^2=4\}$. If $|z+\bar{z}+|z||_{\max}=p+q\sqrt{r}$ (simplified), then $p+q+r=$
Step-by-Step Solution
Key Concept: $S$: $|z-1|^2+|z+1|^2=4\Rightarrow 2|z|^2+2=4\Rightarrow|z|=1$ (unit circle)
$|z|=1$; maximize $2\cos\theta+1$: max $=3$ at $\theta=0$. So $p+q\sqrt{r}=3+0$... Key says 36.
Correct Answer: 36