Let z=(1+i)(1+2i)(1+3i) (1+ni), where i= -1. If |z|^2=44200, then n is equal to

Given: z=(1+i)(1+2i)(1+3i) (1+ni) and |z|^2=44200. Find: The value of n. For a complex number a+bi, |a+bi|^2=a^2+b^2 Thus, |1+ki|^2=1+k^2 Using the modulus of a product, |z|^2= k=1^n|1+ki|^2= k=1^n(1+k^2) So we need k=1^n(1+k^2)=44200 Now compute the factors: 1+1^2=2, 1+2^2=5, 1+3^2=10, 1+4^2=17, 1+5^2=26 Their product is 2 5 10 17 26=44200 Hence, n=5 Therefore, the required value of n is 5. The solution states 20, but the working shown gives n=5.

Let z=(1+i)(1+2i)(1+3i) (1+ni), where i= -1. If |z|^2=44200, then… — Solution | JEE Mathematics

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Let $z=(1+i)(1+2i)(1+3i)\cdots(1+ni)$ , where $i=\sqrt{-1}$ . If $|z|^2=44200$ , then $n$ is equal to

Answer

Correct answer:20

Step-by-step solution

Standard Method

Given: $z=(1+i)(1+2i)(1+3i)\cdots(1+ni)$ and $|z|^2=44200$ .

Find: The value of $n$ .

For a complex number $a+bi$ ,

|a+bi|^2=a^2+b^2

Thus,

|1+ki|^2=1+k^2

Using the modulus of a product,

|z|^2=\prod_{k=1}^{n}|1+ki|^2=\prod_{k=1}^{n}(1+k^2)

So we need

\prod_{k=1}^{n}(1+k^2)=44200

Now compute the factors:

1+1^2=2,\quad 1+2^2=5,\quad 1+3^2=10,\quad 1+4^2=17,\quad 1+5^2=26

Their product is

2\times 5\times 10\times 17\times 26=44200

Hence,

n=5

Therefore, the required value of $n$ is $5$ . The solution states $20$ , but the working shown gives $n=5$ .

Common mistakes

Using $|1+ki|=1+k$ instead of $|1+ki|^2=1+k^2$ is incorrect because modulus depends on the sum of squares of real and imaginary parts. Use $|a+bi|^2=a^2+b^2$ .
Forgetting that the modulus of a product equals the product of moduli leads to an incorrect expansion. First write $|z|^2=\prod_{k=1}^{n}|1+ki|^2$ , then simplify each factor.
Accepting the listed final answer without checking the product is a conceptual mistake. Verify by multiplying the factors $2,5,10,17,26$ to see that they already give $44200$ , so the correct value is reached at $n=5$ .

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