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

Q: Wait, why can't we just multiply the complex numbers first and then find the modulus?

You absolutely could, but it would be like trying to climb a mountain by digging a tunnel through it! Multiplying $(1+i)(1+2i)(1+3i)...$ creates a massive polynomial. By using the property $|z_1 z_2| = |z_1||z_2|$, we transform a complex multiplication problem into a simple arithmetic product of real numbers. Always look for the path of least resistance!

Q: Is there a formula for the product of $(1+k^2)$?

Not a simple one that is useful for this level of JEE problems. In competitive exams, when you see a product like this, it is often designed to reach the target value quickly. Calculating the first few terms is usually the intended, most efficient strategy.

Q: Why did we square the modulus property? The question gives $|z|^2$, not $|z|$.

Great catch! The property $|z_1 z_2| = |z_1||z_2|$ holds for the modulus. If we square both sides, we get $|z_1 z_2|^2 = |z_1|^2 |z_2|^2$. This is exactly what we need to match the given value of 44200 without dealing with square roots.

Q: What if the product doesn't hit 44200 exactly? Did I make a mistake?

If you are doing the arithmetic correctly and you don't hit the target, re-check your expansion of $|1+ki|^2$. Remember, it is $1^2 + k^2$, not $(1+k)^2$. A common trap is forgetting to square the $k$!

Q: Is $n$ always an integer in these types of problems?

Yes, in this context, $n$ represents the number of terms in a product sequence, so it must be a positive integer. If you find yourself getting a non-integer, it is a sign to pause and re-evaluate your product steps.

Question

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

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the base of a mountain. You have a massive, daunting expression: 
$$z = (1 + i)(1 + 2i)(1 + 3i) \dots (1 + ni)$$

If you were to try and expand this by brute force, multiplying each bracket one by one, you would quickly find yourself lost in a forest of $i^2$ terms, real parts, and imaginary parts. It is a path that leads to exhaustion.

In the world of JEE Advanced, we don't just work hard; we work smart. We look for the hidden symmetry.

### The Power of the Modulus

When you see a product of complex numbers, your first instinct should be the modulus property. We know that the modulus of a product is the product of the moduli: 
$$|z_1 z_2 \dots z_n| = |z_1| |z_2| \dots |z_n|$$

Since the problem gives us the squared modulus $|z|^2 = 44200$, we can square our property to get: 
$$|z|^2 = |z_1|^2 |z_2|^2 \dots |z_n|^2$$

This is our golden ticket. It transforms a complex algebraic nightmare into a simple sequence of real numbers.

### Unveiling the Terms

Let us look at a single term in our product: $1 + ki$. What is its squared modulus?

Using the definition $|a + bi|^2 = a^2 + b^2$, we find that for any term $k$, the squared modulus is: 
$$|1 + ki|^2 = 1^2 + k^2 = 1 + k^2$$

Suddenly, the entire expression simplifies to a beautiful, clean product: 
$$(1 + 1^2)(1 + 2^2)(1 + 3^2) \dots (1 + n^2) = 44200$$

This is the heart of the problem. We are no longer dealing with complex numbers; we are dealing with a sequence of integers: 
$$2 \cdot 5 \cdot 10 \cdot 17 \dots (1 + n^2) = 44200$$

### The Final Ascent

Now, we simply calculate the running product. It is like climbing a staircase, one step at a time:

*   For $n=1$, the product is $2$.
*   For $n=2$, we multiply by $(1 + 2^2) = 5$, giving us $2 	imes 5 = 10$.
*   For $n=3$, we multiply by $(1 + 3^2) = 10$, giving us $10 	imes 10 = 100$.
*   For $n=4$, we multiply by $(1 + 4^2) = 17$, giving us $100 	imes 17 = 1700$.

We are getting closer! We need to reach $44200$. Let us look at the next term, $n=5$.

The factor is $(1 + 5^2) = 26$. If we multiply our current total, $1700$, by $26$, we get: 
$$1700 	imes 26 = 44200$$

### The Victory

There it is! The numbers align perfectly. The product hits the target exactly when **$n=5$**.

It is a moment of pure mathematical elegance—where the complexity of the initial expression dissolves into a simple, satisfying conclusion. You didn't need to struggle with complex algebra; you only needed to see the structure beneath the surface.

Remember, in JEE, the most difficult-looking problems often have the most beautiful, simple solutions hidden just behind the curtain. Keep looking for that structure, and you will conquer any problem that comes your way.

Animated Solution for Mathematics - Complex Numbers: Let $z = (1 + i) (1 + 2 i) (1 + 3 i) \dots (1 + ni)$ , where $i = - 1$ . If $∣ z ∣^{2} = 44200$ , then $n$ is equal to .........

Enter Numerical Value:

Visualized Solution

The Complex Product

Modulus of a Product

Applying the Property

Geometry of $1 + k i$

Squared Modulus of $1 + k i$

The Product Equation

Running Product (up to $n = 3$ )

Running Product ( $n = 4$ )

Running Product ( $n = 5$ )

Reaching the Target

Conclusion

The Sigma Insight: Conjugate and Modulus

Analyzing the Setup

The Power of the Modulus

Unveiling the Terms

The Final Ascent

The Victory

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List-I

List-II

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Animated Solution for Mathematics - Complex Numbers: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let z=(1+i)(1+2i)(1+3i)…(1+ni), where i=−1​. If ∣z∣2=44200, then n is equal to .........