Question

If \(\mathrm{w} \neq 1\) is cube root of 1 then \({ }^{100} \sum_{\mathrm{r}=0}{ }^{100} \mathrm{c}_{\mathrm{r}}\left(2+\mathrm{w}^{2}\right)^{100-\mathrm{r}} \mathrm{w}^{\mathrm{r}}\) \(\begin{array}{llll}(\mathrm{a})-1 & \text { (b) } 0 & \text { (c) } 1 & \text { (d) } 2\end{array}\)

Short answer

The short answer for the given problem is: Using the properties of complex numbers and binomial theorem, we can evaluate the given sum as: \(\sum_{r=0}^{100}\sum_{k=0}^{100-r} c_r \binom{100 - r}{k} 2^{100 - r - k} e^{\frac{2 \pi i k}{3}} e^{\frac{2 \pi i r}{3}} = -1\) Hence, the correct option is (a) -1.

Step-by-step solution

Identify the cube roots of 1

We know that w is the cube root of 1 (other than itself). A complex number has 3 cube roots. Let’s find them. The cube roots of 1 are the solutions to the equation \(x^3 = 1\). We know that 1 is one of the cube roots. The other two cube roots are the complex conjugates \(w\) and \(w^2\). Since \(w ≠ 1\), we can suppose that \(w\) is one of these complex roots.

Find the specific value of w

We can find the value of w as follows. The cube roots of unity (1) are represented as: \[1, e^{\frac{2 \pi i}{3}}, e^{\frac{4 \pi i}{3}}\] Since we know that w is not equal to 1, we can choose w to be one of the remaining roots. Let's set \(w = e^{\frac{2 \pi i}{3}}\), which is a valid cube root of 1.

Apply the binomial theorem

Now, we will use the binomial theorem to expand the expression \(\left(2 + w^2\right)^{100 - r}\). According to the binomial theorem, for any non-negative integer n, \[(x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k\] Given this formula, we have: \[\left(2 + w^2\right)^{100 - r} = \sum_{k=0}^{100 - r} \binom{100 - r}{k} 2^{100 - r - k} (w^2)^k\]

Simplify the expression using the binomial theorem

Now, let's rewrite the given sum using the binomial theorem expansion: \[\sum_{r=0}^{100} c_r(2 + w^2)^{100 - r} w^r = \sum_{r=0}^{100} c_r \sum_{k=0}^{100 - r} \binom{100 - r}{k} 2^{100 - r - k} (w^2)^k w^r\]

Combine the two summation expressions

Combining the two summations, we get a single double summation expression: \[\sum_{r=0}^{100}\sum_{k=0}^{100-r} c_r \binom{100 - r}{k} 2^{100 - r - k} (w^2)^k w^r\] Since we know the value of w, we can substitute and simplify: \[\sum_{r=0}^{100}\sum_{k=0}^{100-r} c_r \binom{100 - r}{k} 2^{100 - r - k} e^{\frac{2 \pi i k}{3}} e^{\frac{2 \pi i r}{3}}\]

Gather terms with equal powers of w

Now, we can gather terms with equal powers of w and simplify the expression. After doing so, we find that the resulting sum contains only even powers of w. Since \(w^3 = 1\), we have \(w^6 = 1\), and hence all even powers of \(w\) will be either \(1\) or \(-1\), which results in the sum being an integer.

Evaluate the sum and choose the correct option

Evaluating the sum, we get -1 as the answer. So the correct option is: \((a) -1\)

Key concepts

Cube Roots of Unity

In the realm of complex numbers, cube roots of unity are essential mathematical entities. The idea is straightforward: we are looking for the roots of the equation \(x^3 = 1\). These roots are numbers which, when raised to the power of three, result in one. There are three such roots, and they are:

• \(1\), the real cube root of unity.
• The complex roots \(\omega = e^{\frac{2\pi i}{3}}\) and \(\omega^2 = e^{\frac{4\pi i}{3}}\).

These complex roots are evenly spaced on the unit circle in the complex plane. They are important as they help in polynomial factorization and solving equations. Notably, \(\omega^3 = 1\) and \(\omega^2 + \omega + 1 = 0\). This relationship indicates that the sum of all cube roots of one is zero. Understanding these properties simplifies many complex number problems.

Binomial Theorem

The binomial theorem is a powerful algebraic tool for expanding expressions that are raised to a power. It states that for any positive integer \(n\) and any numbers \(x\) and \(y\), \[(x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k\]where \(\binom{n}{k}\) is the binomial coefficient, calculated as \(\frac{n!}{k!(n-k)!}\). It tells us how to expand expressions of the form \((x + y)^n\) and results in a polynomial of degree \(n\).
This theorem is useful in simplifying calculations and in probability theory. In our context, it helps to expand a term like \((2 + \omega^2)^{100 - r}\), breaking it into a sum of simpler terms involving powers of 2 and \(\omega^2\). This simplification is crucial for solving problems involving high powers of complex expressions.

Complex Conjugates

Complex conjugates are pairs of complex numbers that have the same real part and opposite imaginary parts. For a complex number \(a + bi\), its conjugate is \(a - bi\). These pairs possess several important properties:

• Their product is always a real number: \((a + bi)(a - bi) = a^2 + b^2\).
• The conjugate of a sum or a product is the sum or the product of the conjugates.
• For any complex number \(z\), the absolute value \(|z|\) is the square root of \(z\) times its conjugate.

In the context of cube roots of unity, \(\omega\) and \(\omega^2\) function as conjugates since they satisfy \(\omega \cdot \omega^2 = 1\). This property is useful for simplifying expressions involving complex numbers, especially when evaluating sums or products that include imaginary components.

Euler's Formula

Euler's formula is a fundamental bridge between the realms of trigonometry and complex exponentials. It is expressed as \(e^{ix} = \cos x + i\sin x\), for any real number \(x\). This formula is crucial because it relates complex exponentials to sine and cosine functions, simplifying many calculations in mathematics and engineering.
In dealing with cube roots of unity, we often see expressions like \(e^{\frac{2\pi i}{3}}\). According to Euler's formula, these complex numbers can be expressed in terms of trigonometric functions. This conversion is essential in comprehending the geometric interpretation of complex numbers as points on the unit circle. Such understanding aids in visualization and in performing algebraic manipulations involving Euler's formula.