Area bounded by $-8z^2-8\bar{z}^2+20z\bar{z}=144$ (where $z$ is a complex number)
Step-by-Step Solution
Key Concept: $z\bar{z}=|z|^2$, $z^2+\bar{z}^2=2\text{Re}(z^2)=2(x^2-y^2)$
$-8\cdot 2(x^2-y^2)+20(x^2+y^2)=144\Rightarrow 4x^2+36y^2=144\Rightarrow\frac{x^2}{36}+\frac{y^2}{4}=1$. Area$=\pi\cdot 6\cdot 2=12\pi$.
Correct Answer: 3