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Mathematics · 2026

JEE Main · 24 January 2026, Shift 2 · Q21

Let z=(1+i)(1+2 i)(1+3 i) ….(1+n i), where i=√ -1. If |z|^2=44200, then n is equal to ____

Let $\displaystyle z=(1+i)(1+2 i)(1+3 i) \ldots .(1+n i)$, where $\displaystyle i=\sqrt{-1}$. If $\displaystyle |z|^2=44200$, then $\displaystyle n$ is equal to $\displaystyle \_\_\_\_$
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JEE Main 2026 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.