Problem 98 - Algebra and Trigonometry Real Mathematics, Real People

Question

Use the Binomial Theorem to expand the complex number. Simplify your result. (Remember that \(i=\sqrt{-1 .})\) \((5-\sqrt{3} i)^{4}\)

Step-by-Step Solution

Answer

The result of \( (5-\sqrt{3} i)^{4} \) is \(130 + (-132\sqrt{3})i\).

Step 1: Identify the Terms in the Binomial and the Exponent

In the expression \( (5-\sqrt{3} i)^{4} \), the first term (a) is 5, the second term (b) is \(-\sqrt{3}i\) and the exponent (n) is 4.

Step 2: Apply the Binomial Theorem and Expand

Recall that the Binomial Theorem for \((a-b)^n\) can be expressed as \(\sum_{k=0}^n (-1)^k {n \choose k} a^{n-k} \cdot b^k\). Expanding \( (5-\sqrt{3} i)^{4}\), we have \({4 \choose 0} (5^4) (-\sqrt{3}i)^0 - {4 \choose 1} (5^3) (-\sqrt{3}i)^1 + {4 \choose 2} (5^2) (-\sqrt{3}i)^2 - {4 \choose 3} (5^1) (-\sqrt{3}i)^3 + {4 \choose 4} (5^0) (-\sqrt{3}i)^4\). This is based on the alternating sign and the binomial coefficients, changing exponents extended from the Binomial Theorem.

Step 3: Simplify the individual terms

Now simplify each term. Remembering that \(i^2 = -1, i^3 = -i, i^4 = 1\), we have \(625 + 300\sqrt{3} i - 450 - 540\sqrt{3} i - 45 + 108\sqrt{3} i\).

Step 4: Combine Like Terms and Simplify the Result

Combining the real parts and imaginary parts together, we get \(130 + (-132\sqrt{3})i\), which is the simplified form of \( (5-\sqrt{3} i)^{4} \)

Key Concepts

Complex Numbers Simplification

Complex numbers are numbers that have both a real part and an imaginary part. They are typically written in the form of \(a + bi\), where \(a\) is the real part, \(b\) is the imaginary part, and \(i\) represents \(\sqrt{-1}\). Simplifying expressions involving complex numbers often involves combining like terms and using properties of \(i\), such as \(i^2 = -1\), to simplify powers of \(i\).

For example, if we expand a binomial expression like \((5 - \sqrt{3} i)^4\), we end up with a series of terms that include higher powers of \(i\). To simplify, we recognize that \(i^4 = 1\), \(i^3 = -i\), and so on. By substituting these values into our expression and combining like terms, we can simplify it to a standard form consisting of a real part and an imaginary part as illustrated in the step-by-step solution provided.

Binomial Coefficients

Binomial coefficients, symbolized by \({n \choose k}\) or \(C(n, k)\), are a key part of the Binomial Theorem and represent the number of ways to choose \(k\) elements out of a set of \(n\) elements without considering the order of selection. In the context of the Binomial Theorem expansion, they determine the coefficients in the expansion of a binomial expression raised to a power.

For instance, when expanding \((5 - \sqrt{3}i)^4\) using the Binomial Theorem, the binomial coefficients for each term will be \({4 \choose 0}\), \({4 \choose 1}\), \({4 \choose 2}\), \({4 \choose 3}\), and \({4 \choose 4}\). Calculating these coefficients gives us the multipliers for the corresponding terms of our expanded binomial expression. Remember, the coefficient \({n \choose 0}\) and \({n \choose n}\) is always 1, while the coefficient \({n \choose k}\) can be calculated by the formula \(\frac{n!}{k!(n-k)!}\).

Powers of i in Complex Numbers

In complex numbers, \(i\) is defined as \(i = \sqrt{-1}\), which is not a real number. The powers of \(i\) follow a predictable pattern that repeats every four exponents: \(i^1 = i\), \(i^2 = -1\), \(i^3 = -i\), and \(i^4 = 1\). This means that for powers higher than 4, we can simplify by dividing the exponent by 4 and using the remainder to find the equivalent power of \(i\) that lies between 1 and 4.

For example, to simplify \(i^{17}\), we would divide 17 by 4, which gives us a remainder of 1. So, \(i^{17}\) simplifies to just \(i\). Understanding this cyclical nature of the powers of \(i\) is crucial when simplifying complex numbers raised to higher powers, as we can reduce any power of \(i\) to one of these four fundamental results, making the simplification process much more manageable.