Question

Number of integral solutions satisfying the inequality $|Rez|+|Imz|<51$ is:

• A. $5050$
• B. $4901$
• C. $5101$
• D. $5100$

Answer 1

The correct option is C $5101$

Let $z=x+iy$

Hence, $Re\left(z\right)=x$ and $Im\left(z\right)=y$

$x+y<51$
$x\ge 0,y\ge 0$
or $x+y\le 50$
give one each to x and y
$x+y\le 48$
$x+y+z=48\Rightarrow$ number of solutions = ${}^{50}{C}_2$ = $\frac{50\times 49}{2}$
Number of solutions in all the four quadrants = $100.49=4900$
number of solutions except $(0,0)$ on x and y axis from $(-51,0)$
to $(51,0)$ and $(0,51)$ to $(0,-51)$ are $200$
Total solutions $=4900$ $+$ $200$ $+$ $1$ $=5101$