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JEE Main 202624 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
24 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Complex Number

JEE Main 2026 · 24 January · Evening Shift · Mathematics · Complex Number

Let z=(1+i)(1+2 i)(1+3 i) (1+n i) , where i= -1 . If |z|^ 2 =44200...

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

Correct answer

0

Step-by-step solution

z=(1+i)(1+2i)(1+3i)⋯(1+ni)z = (1 + i)(1 + 2i)(1 + 3i)\cdots(1 + ni) where i=−1i = \sqrt{-1}
∣z∣2=∏k=1n∣1+ki∣2=∏k=1n(1+k2)|z|^2 = \prod_{k=1}^{n} |1 + ki|^2 = \prod_{k=1}^{n} (1 + k^2)
Computing: (1+1)(1+4)(1+9)(1+16)(1+25)=2×5×10×17×26(1+1)(1+4)(1+9)(1+16)(1+25) = 2 \times 5 \times 10 \times 17 \times 26
=10×10×17×26=100×442=44200= 10 \times 10 \times 17 \times 26 = 100 \times 442 = 44200
Therefore n=5n = 5
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