JEE Main 2024 (Online) 8th April Morning ShiftQuestion 1
If the set $R=\{(a,b):a+5b=42,\ a,b\in\mathbb{N}\}$ has $m$ elements and $\sum_{n=1}^{m}\left(1-i^{n!}\right)=x+iy$, where $i=\sqrt{-1}$, then the value of $m+x+y$ is
Complex Numbers
Solution
$a+5b=42$, $b=1,\dots,8$: $m=8$. $i^{n!}$: $n=1\to i$; $n=2\to-1$; $n=3\ (6)\to-1$; $n\ge4$ ($4\mid n!$) $\to1$ (5 terms). Sum $=i-1-1+5=3+i$. $\sum(1-i^{n!})=8-(3+i)=5-i$, so $x=5,y=-1$. $m+x+y=12$. Option A.