Solve for x: (1+x)^n=(1-x)^n, where n=+ive integer, plz use knowledge of complex number.

Let's delve into the equation (1 + x)ⁿ = (1 - x)ⁿ, where n is a positive integer. Our goal here is to find the values of x that satisfy this equation. To start solving, we can expand both sides using the Binomial Theorem, which provides a clear path to understanding how the terms interact.

Applying the Binomial Theorem

The Binomial Theorem states that for any integer n:

• (a + b)ⁿ = Σ (n choose k) * a^k * b^n-k for k = 0 to n

In our case, we can expand both sides:

• For (1 + x)ⁿ: Σ (n choose k) * 1^k * x^n-k = Σ (n choose k) * x^n-k
• For (1 - x)ⁿ: Σ (n choose k) * 1^k * (-x)^n-k = Σ (n choose k) * (-1)^n-k * x^n-k

Setting the Two Expansions Equal

By equating the two expansions, we have:

Σ (n choose k) * x^n-k = Σ (n choose k) * (-1)^n-k * x^n-k

This means that for each term in the series where k varies from 0 to n, the coefficients of similar powers of x must be equal. This gives us a series of equations for the coefficients.

Analyzing the Coefficients

Let’s consider the coefficients of a specific case: when k = 0 and k = n. For k = 0, we have:

• For k = 0: 1 = (-1)ⁿ
• If n is even, this holds true (1 = 1); if n is odd, it does not (1 = -1 is false).

Now looking at k = n, we also find:

• For k = n: (n choose n) * x⁰ = (n choose n) * (-1)⁰ holds for all n.

Considering Roots of Unity

Next, we can analyze the situation by setting x to certain values. Since we're dealing with powers and symmetry, we can look at specific values of x:

• If x = 0, both sides equal 1.
• If x = 1, we get (1 + 1)ⁿ = (1 - 1)ⁿ, which simplifies to 2ⁿ = 0 (not valid).
• If x = -1, we find (1 - 1)ⁿ = (1 + 1)ⁿ, leading to 0 = 2ⁿ (also not valid).

Finding the Solutions

To summarize, the only real solution to the equation (1 + x)ⁿ = (1 - x)ⁿ that holds for all positive integers n is x = 0. For odd n, we could also have complex solutions, as both sides can become complex numbers but are symmetric around the real axis.

This symmetry implies that x could also take on complex values, specifically involving roots of unity. The solutions in complex form could be expressed using the imaginary unit i, where x could equal values along the unit circle in the complex plane.

Final Thoughts

Thus, the primary solutions are:

• x = 0 (the real solution)
• Complex solutions may exist, particularly related to roots of unity.

In summary, this exploration of the equation leads us to a deeper understanding of how both real and complex numbers can be applied in evaluating polynomial identities. If you have any further questions or would like to explore complex solutions more deeply, feel free to ask!