JEE Main 2026 (Online) 24th January Evening ShiftQuestion 1
Let $z=(1+i)(1+2i)(1+3i)\ldots(1+ni)$, where $i=\sqrt{-1}$. If $|z|^2=44200$, then $n$ is equal to $\_\_\_\_\_\_$ .
Complex Numbers
Answer$5$
Solution
$|z|^2=\prod_{k=1}^n(1+k^2)=2\cdot5\cdot10\cdot17\cdot26=44200$, so $n=5$.