Problem 18 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \((1+i)(1+2 i)(1+3 i) \ldots(1+n i)=\alpha+i \beta\) then \(2 \times 5\) \(\times 10 \ldots\left(1+n^{2}\right)=\) (A) \(\alpha-i \beta\) (B) \(\alpha^{2}-\beta^{2}\) (C) \(\alpha^{2}+\beta^{2}\) (D) None of these

Step-by-Step Solution

Answer

(C) \(\alpha^{2} + \beta^{2}\).

Step 1: Recognize the Product

You are given the product \((1+i)(1+2i)(1+3i)\ldots(1+ni)\) which is equal to \(\alpha + i\beta\). Each term in this product is of the form \((1+ki)\) where \(k\) is an integer from 1 to \(n\).

Step 2: Find the Conjugate

The complex conjugate of \(\alpha + i\beta\) is \(\alpha - i\beta\). In terms of the product, this is equivalent to the conjugate of \((1+i)(1+2i)(1+3i)\ldots(1+ni)\). This is \((1-i)(1-2i)(1-3i)\ldots(1-ni)\).

Step 3: Multiply Product with Conjugate

Compute \((1+i)(1+2i)(1+3i)\ldots(1+ni) \times (1-i)(1-2i)(1-3i)\ldots(1-ni)\). This gives \((1^{2}+1^{2})(1^{2}+2^{2})\ldots(1^{2}+n^{2}) \), which simplifies to \(2 \times 5 \times 10 \ldots(1+n^{2})\). This expression equals the product of \(\alpha + i\beta\) and its conjugate, which is \(\alpha^{2} + \beta^{2}\).

Step 4: Match with Options

Identify that \(\alpha^{2} + \beta^{2}\) is exactly the expression you derived in the previous step. This matches with option (C).

Key Concepts

Complex Conjugate

When dealing with complex numbers, one important concept to understand is the complex conjugate. The conjugate of a complex number is formed by changing the sign of the imaginary part. For instance, if you have a complex number expressed as \( a + bi \), its complex conjugate will be \( a - bi \).
This concept is crucial because multiplying a complex number by its conjugate results in a real number. This is because the imaginary parts cancel each other out, leaving only the sum of the squares of the real and imaginary components. In mathematical terms, \((a+bi)(a-bi) = a^2 + b^2\).
In our solution, we computed the conjugate of \((1+i)(1+2i)(1+3i)\ldots(1+ni)\) to form \((1-i)(1-2i)(1-3i)\ldots(1-ni)\). This step was essential to simplify further calculations and achieve a pure product expression.

Product of Complex Numbers

The product of complex numbers follows the same rules as multiplying any binomials, often using the distributive property. When you multiply two complex numbers, \((a+bi)\) and \((c+di)\), the result is:

• ac + adi + bci - bd\(i^{2}\)
• Because \(i^2 = -1\), this simplifies to \( (ac-bd) + (ad+bc)i \)

In the solution, we saw the product \((1+i)(1+2i)(1+3i)\ldots(1+ni)\), where each term \((1+ki)\) is a simple complex number with real part 1, and imaginary part \(ki\).
Thus, it involved multiplying several complex numbers together, creating a potentially large equation. The simplification of these products helps to express the resulting product in terms of real numbers by utilizing the expression of conjugates together.

Magnitude of Complex Numbers

The magnitude of a complex number is like its `length` on the complex plane, and it's crucial for understanding its geometric representation. To find the magnitude of a complex number \(a + bi\), you use the formula \( \sqrt{a^2 + b^2} \).
This is the distance from the origin to the point \((a, b)\) on the plane and is always a positive real number.
In regards to the exercise, when the product \( (1+i)(1+2i)(1+3i)\ldots(1+ni) \) was multiplied by its conjugate, it effectively calculated the square of the magnitude of this entire product. The result was a real number, \(\alpha^2 + \beta^2\), which was identified as option (C). Understanding the magnitude helps visualize how multiplication and conjugation of complex numbers affect their position and size in the complex plane.