Question

Let $z$ be a complex number such that both real and imaginary parts of $\frac{z}{20}$ and $\frac{20}{\overline{z}}$ lies between $[0,1]$. Then the area of the region in the complex plane that consists of all points $z$ is
Assume $\pi =3.14$

Answer 1

Let $z=a+bi$
$\Rightarrow \frac{z}{20}=\frac{a}{20}+\frac{b}{20}i$

$\therefore 0\le \frac{a}{20},\frac{b}{20}\le 1$
$\Rightarrow 0\le a,b\le 20...\left(1\right)$
which is a square of side length $20$.

Also,
$\frac{20}{\overline{z}}=\frac{20}{a-bi}=\frac{20a}{a^2+b^2}+\frac{20b}{a^2+b^2}i$

$\therefore 0\le \frac{20a}{a^2+b^2},\frac{20b}{a^2+b^2}\le 1$
$\Rightarrow a,b\ge 0$
$\Rightarrow (a-10{)}^2+b^2\ge {10}^2...(2)$
$\Rightarrow a^2+(b-10{)}^2\ge {10}^2...(3)$


Area of shaded region
$={20}^2-[{10}^2+2\times \frac{1}{4}\times \pi \times 100]$
$=300-50\pi (\pi =3.14)$
$=143\text{sq. units}$