Question

A region $S$ in complex plane is defined by $S=\{x+iy:-1\le x,y\le 1\}$. A complex number $z=x+iy$ is chosen uniformly at random from $S.$ If $P$ be the probability that the complex number $\frac{3}{4}(1+i)z$ is also in $S,$ then the value of $27P$ is

Answer 1

Clearly, $S$ denotes a region bounded by $x=\pm 1$ and $y=\pm 1$ i.e. square with centre at origin of side length $2$ on the complex plane.
Hence, $n\left(S\right)=4...\left(1\right)$


Now, let $w=\frac{3}{4}(1+i)(x+iy)$
$\Rightarrow w=\frac{3}{4}[(x-y)+i(x+y)]$
In order that $w$ lies on $S,$ we have
$-1\le Re\left(w\right)\le 1\text{and}-1\le Im\left(w\right)\le 1$
$\Rightarrow -1\le \frac{3}{4}(x-y)\le 1\text{and}-1\le \frac{3}{4}(x+y)\le 1$
$\Rightarrow -\frac{4}{3}\le x-y\le \frac{4}{3}\text{and}-\frac{4}{3}\le x+y\le \frac{4}{3}$


Consider, $x-y=\frac{4}{3};x-y=-\frac{4}{3}$
and $x+y=\frac{4}{3};x+y=-\frac{4}{3}$
Area $EFGH=4(\frac{1}{2}\times \frac{4}{3}\times \frac{4}{3})=\frac{32}{9}$

Area of small triangles which is outside the region $S$ is
$4(\frac{1}{2}\times \frac{2}{3}\times \frac{1}{3})=\frac{4}{9}$
$\therefore n\left(A\right)=\frac{32}{9}-\frac{4}{9}=\frac{28}{9}$

So, the required probability is,
$P=\frac{n\left(A\right)}{n\left(S\right)}=\frac{\frac{28}{9}}{4}=\frac{7}{9}$
$\therefore 27P=21$