z=(1+i)(1+2i)(1+3i)⋯(1+ni) where i=−1
∣z∣2=k=1∏n∣1+ki∣2=k=1∏n(1+k2)
Computing: (1+1)(1+4)(1+9)(1+16)(1+25)=2×5×10×17×26
=10×10×17×26=100×442=44200
Therefore n=5
Let z=(1+i)(1+2i)(1+3i)…(1+ni), where i=−1. If ∣z∣2=44200, then n is equal to ____
Held on 24 Jan 2026 · Verified 6 Jul 2026.
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