Let z=(1+i)(1+2i)(1+3i) .(1+ni), where i=-1.

Question

Let $z=(1+i)(1+2i)(1+3i)\ldots .(1+ni)$, where $i=\sqrt{-1}$. If $|z{|}^{2}=44200$, then $n$ is equal to $____$

Answer: 5

Accepted Answer

$\begin{gathered} Z=(1+i)(1+2i)(1+3i)\ldots (1+ni) \ |Z|=|(1+i)(1+2i)(1+3i)\ldots (1+ni)| \ |Z|=|1+i|\cdot |1+2i|\cdot |1+3i|\ldots |1+ni| \ |Z{|}^{2}=|1+i{|}^{2}\cdot |1+2i{|}^{2}\cdot |1+3i{|}^{2}\ldots |1+ni{|}^{2} \ |Z{|}^{2}=({1}^{2}+{1}^{2})({1}^{2}+{2}^{2})({1}^{2}+{3}^{2})\ldots ({1}^{2}+{n}^{2}){z=x+iy,|z{|}^{2}={x}^{2}+{y}^{2}} \end{gathered}$

It is given that

$\begin{gathered} |Z{|}^{2}=44200 \ (1+{1}^{2})(1+{2}^{2})(1+{3}^{2})\ldots (1+{n}^{2})=44200 \ 44200={2}^{3}\cdot {5}^{2}\cdot 13\cdot 17 \ (1+{1}^{2})(1+{2}^{2})(1+{3}^{2})\cdot (1+{4}^{2})(1+{5}^{2})\ldots (1+{n}^{2})={2}^{3}\cdot {5}^{2}\cdot 13\cdot 17 \end{gathered}$

solve some product terms

$2\cdot 5\cdot (10)\cdot (17)\cdot (26)\ldots (1+{n}^{2})=(2)(5)(2	imes 5)(17)(2	imes 13)$

this matches for $n=5$

so value of $n=5$

Solutions · 1