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: Take modulus of the given product

Given,

z = ∏_r=1ⁿ (1 + ri)

Using the property of modulus of a product:

|z| = ∏_r=1ⁿ |1 + ri|

Squaring both sides,

|z|² = ∏_r=1ⁿ |1 + ri|²

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Step 2: Compute the modulus squared of each term

|1 + ri|² = 1² + r²

= 1 + r²

Therefore,

|z|² = ∏_r=1ⁿ (1 + r²)

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Step 3: Use the given numerical value

It is given that:

∏_r=1ⁿ (1 + r²) = 44200

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Step 4: Evaluate the product successively

Compute the product term by term:

(1 + 1²)(1 + 2²)(1 + 3²)(1 + 4²)(1 + 5²)

= 2 × 5 × 10 × 17 × 26

= 44200

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Step 5: Determine the value of n

Since the product matches exactly for r = 1 to 5,

the number of terms is

n = 5

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Final Answer:

The value of n is
5