Question

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

Hint:

For products of complex numbers, always square the modulus to simplify multiplication into real factors.

Answer 1

Correct Answer: 20

Step 1: Use modulus property.
For a complex number $a+bi$, \[ |a+bi|^2=a^2+b^2 \] Thus, \[ |1+ki|^2=1+k^2 \] Step 2: Express $|z|^2$.
\[ |z|^2=\prod_{k=1}^{n}(1+k^2) \] Step 3: Factorize the given value.
\[ 44200=2^3\times5^2\times13\times17 \] Step 4: Match with the product.
\[ (1+1^2)(1+2^2)(1+3^2)\cdots(1+n^2) =2\times5\times10\times17\times\cdots \] Matching factors gives \[ n=20 \] Step 5: Final conclusion.
\[ \boxed{20} \]