Let $z = (1+i)(1+2i)(1+3i)\cdots(1+ni)$ and $|z|^2 = 44200$. Find the value of $n$.

Question

If $50 \left( \frac{2x}{1 + 3i} + \frac{y}{1 - 2i} \right) = 31 + 17i$ where $x, y \in R$ & $i = \sqrt{-1}$ then value of $10(x + 3y)$ is ________

Answer 1

Let's solve the problem step by step by analyzing the given expression for the complex number $ z $ and determine $ n $ such that $ |z|^2 = 44200 $.

The given expression for $ z $ is:

$z = (1+i)(1+2i)(1+3i)\cdots(1+ni)$

The magnitude or modulus of a product of complex numbers is the product of the magnitudes of each complex number. Therefore,

$|z| = |1+i| \cdot |1+2i| \cdot |1+3i| \cdots |1+ni|$

Thus,

$|z|^2 = |1+i|^2 \cdot |1+2i|^2 \cdot |1+3i|^2 \cdots |1+ni|^2 = 44200$

The modulus of a complex number $ a+bi $ is given by:

$|a+bi| = \sqrt{a^2 + b^2}$

Therefore, for each $ k $ in $ 1+ki $,

Thus, we have:

$|z|^2 = (1+1^2)(1+2^2)(1+3^2)\cdots(1+n^2)$

Now, let's calculate the product step by step until it equals 44200, assuming we start from $ n = 1 $ and increase:

$(1+1^2) = 2$
$(1+2^2) = 5$
$(1+3^2) = 10$
$(1+4^2) = 17$
$(1+5^2) = 26$

Now compute the product:

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

Thus, the value of $ n $ that satisfies the equation is 5.

Hence, the correct answer is 5.

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