Problem 66 - Algebra 2

Question

Verify that \(-1+\sqrt{3} i\) is a cube root of 8 by expanding \((-1+\sqrt{3} i)^{3}\)

Step-by-Step Solution

Answer

The complex number \(-1+\sqrt{3} i\) is not a cube root of 8, as the expansion of \((-1+\sqrt{3} i)^3\) results in \(2 + 6\sqrt{3} i\) not \(8\).

Step 1: Understand complex numbers

The number \(-1+\sqrt{3} i\) is a complex number where -1 represents the real part and \(\sqrt{3} i\) represents the imaginary part. A complex number takes the form \(a + bi\) where a and b are real numbers and \(i\) is the imaginary unit with the property \(i^{2} = -1\). Thus, the power of \(i\) cycles every four, \(i, -1, -i, 1, i, -1 ..., \)

Step 2: Apply the binomial theorem

The cube of the complex number is \((-1 + \sqrt{3} i)^{3}\). The binomial theorem states that \((a + b )^{3} = a^{3} + 3a^{2}b + 3ab^{2} + b^{3}\). Applying the theorem where \(a = -1\) and \(b = \sqrt{3} i\), we get \((-1)^{3} + 3(-1)^{2}(\sqrt{3} i) + 3(-1)(\sqrt{3} i)^{2} + (\sqrt{3} i)^{3}\).

Step 3: Simplify the result

Expanding and simplifying the equation, we obtain \(-1 + 3\sqrt{3} i - 3i^{2} - (\sqrt{3})^{3} i^{3}\). Since \(i^{2} = -1\), the equation simplifies further to \(-1 + 3\sqrt{3} i + 3 + (-3\sqrt{3}i)\). This simplifies to \(2\).

Step 4: Reconsider the steps

Since the expansion results in \(2\) and not \(8\), there seems to be a mistake in the calculation. Going back to Step 2, the error can be spotted - we incorrectly calculated \((\sqrt{3} i)^{3}\). To correct, \((\sqrt{3} i)^{3} = \sqrt{3}^{3}. i^{3} = 3\sqrt{3}. (-i) = -3\sqrt{3}i\).

Step 5: Fix the error and finalize

The corrected expansion and simplification of \((-1 + \sqrt{3} i)^{3}\) is \(-1 + 3\sqrt{3} i + 3 + 3\sqrt{3}i = 2 + 6\sqrt{3} i\). This confirms \(-1 + \sqrt{3} i\) is not a cube root of \(8\) neither a root of \(2\).

Key Concepts

Cube Roots

Cube roots are numbers that when multiplied by themselves three times, yield a given number. For example, the cube root of 8 is 2, because when you multiply 2 by itself three times (2 x 2 x 2), you get 8. Similarly, when dealing with complex numbers, cube roots can also result in complex outputs. Finding cube roots for complex numbers involves more than multiplication. It often requires using algebraic identities and sometimes even numerical calculators for simplicity.

• Cube roots come in three forms: one real, and two with a complex conjugate pair.
• Complex cube roots are frequently used in advanced mathematics including solving polynomial equations.

Understanding cube roots in the context of complex numbers enriches problem-solving in algebra and calculus.

Binomial Theorem

The binomial theorem is a powerful tool in algebra that allows us to expand expressions raised to a power. It expresses \[(a + b)^n\]in terms of its coefficients and powers. Essential in expanding expressions like \((-1 + \sqrt{3}i)^3\), it dictates how each term in the expansion comes together.

• A basic formula is: \[(a + b)^n = C(n, 0)a^n + C(n, 1)a^{n-1}b^1 + \dots + C(n, n) b^n\]
• Here, \[C(n, k)\] is the binomial coefficient, calculated as \[\binom{n}{k} = \frac{n!}{k!(n-k)!}\].
• It's common to calculate individual terms separately and then combine them for clarity.

Applying the binomial theorem helps solve complex arithmetic quickly and accurately.

Complex Number Expansion

Complex number expansion takes into account both real and imaginary components, using a structured approach to turn powers of complex numbers into comprehensible terms. For example, expanding \((-1 + \sqrt{3}i)^3\) requires considering each imaginary and real part separately.

• Step by step calculation: Employ the binomial theorem on complex numbers, focusing first on individual components.
• Imaginary Parts: Recognize that terms involving \[(i)^2\]and higher powers can simplify using \[i^2 = -1\].
• Simplification: Always simplify the powers of \[i\], as they repeat every four due to \[i, -1, -i, 1\].

Through expansion, complex number calculations become simplified and more approachable.

Imaginary Unit

The imaginary unit, represented as \(i\), is the backbone of complex numbers. Defined by \(i^2 = -1\), it allows expressions outside the real number system.

• Core Property: \(i\) causes shifts between real and imaginary components when squared.
• Cyclic Nature: Powers of \(i\) cycle every four steps: \[i, -1, -i, 1\].This cyclic behavior is key when simplifying components.
• Applications: Beyond theoretical math, \(i\)is useful in physics and electrical engineering, especially in representing signals.

Understanding the imaginary unit's properties enables solving complex equations effectively and exploring broader mathematical and scientific applications.