Let z = (1+i)(1+2i)(1+3i)dots(1+ni), where i | vedclass

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(B) We are given $z = \prod_{r=1}^n (1+ri)$.Taking the modulus squared on both sides, we get $|z|^2 = \prod_{r=1}^n |1+ri|^2$.Since $|1+ri|^2 = 1^2 +

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$5$

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(B) We are given $z = \prod_{r=1}^n (1+ri)$.Taking the modulus squared on both sides, we get $|z|^2 = \prod_{r=1}^n |1+ri|^2$.Since $|1+ri|^2 = 1^2 + r^2 = 1+r^2$, we have $|z|^2 = \prod_{r=1}^n (1+r^2) = 44200$.For $n=1$: $1+1^2 = 2$.For $n=2$: $2 	imes (1+2^2) = 2 	imes 5 = 10$.For $n=3$: $10 	imes (1+3^2) = 10 	imes 10 = 100$.For $n=4$: $100 	imes (1+4^2) = 100 	imes 17 = 1700$.For $n=5$: $1700 	imes (1+5^2) = 1700 	imes 26 = 44200$.Thus, $n=5$.

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

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