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Product (1+i)(1+2i)…(1+ni) Modulus = α²+β² | JEE

JEE Maths question with a full step-by-step solution.

Question

(1+i)(1+2i)(1+3i)\cdots(1+ni) = \alpha + i\beta

, then

2 \cdot 5 \cdot 10 \cdots (1+n^2)

(where

\alpha, \beta, n \in \mathbb{R}

) is equal to

\alpha - i\beta

\alpha^2 - \beta^2

\alpha^2 + \beta^2

correct

Dnone of these

Solution

Step 1: Take the modulus of both sides. The modulus is multiplicative,

|w_1 w_2 \cdots| = |w_1||w_2|\cdots

|(1+i)(1+2i)\cdots(1+ni)| = |1+i|\,|1+2i|\,\cdots\,|1+ni| = |\alpha + i\beta|

Step 2: Square both sides so the moduli become the convenient products:

|1+i|^2\,|1+2i|^2\,\cdots\,|1+ni|^2 = |\alpha + i\beta|^2

Step 3: Evaluate each factor with

|p + qi|^2 = p^2 + q^2

|1 + ki|^2 = 1^2 + k^2 = 1 + k^2 \quad\text{for } k = 1, 2, \ldots, n

So the left side is:

(1 + 1^2)(1 + 2^2)(1 + 3^2)\cdots(1 + n^2) = 2 \cdot 5 \cdot 10 \cdots (1 + n^2)

Step 4: Evaluate the right side:

|\alpha + i\beta|^2 = \alpha^2 + \beta^2

Step 5: Equate:

2 \cdot 5 \cdot 10 \cdots (1 + n^2) = \alpha^2 + \beta^2

Correct answer: (3)

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