Question

Let $z = (1+i)(1+2i)(1+3i)\cdots(1+ni)$ and $|z|^2 = 44200$. Find the value of $n$.

Hint:

For products involving complex numbers, convert to modulus squared to simplify multiplication.

Answer 1

Correct Answer: 5

Step 1: Use modulus property.
Given \[ z = \prod_{r=1}^{n} (1+ri) \] Taking modulus squared, \[ |z|^2 = \prod_{r=1}^{n} |1+ri|^2 \] Step 2: Evaluate $|1+ri|^2$.
\[ |1+ri|^2 = 1^2 + r^2 = 1+r^2 \] Hence, \[ |z|^2 = \prod_{r=1}^{n} (1+r^2) \] Step 3: Use the given value.
\[ \prod_{r=1}^{n} (1+r^2) = 44200 \] Prime factorizing, \[ 44200 = 2^3 \times 5^2 \times 13 \times 17 \] Step 4: Evaluate product term by term.
\[ (1+1^2)(1+2^2)(1+3^2)(1+4^2)(1+5^2) \] \[ = 2 \times 5 \times 10 \times 17 \times 26 \] \[ = 44200 \] Step 5: Final conclusion.
Hence, the value of $n$ is 5.