Problem 63 - Precalculus

Question

Explain the mistake that is made. Find \((\sqrt{2}+i \sqrt{2})^{6}\) Solution: Raise each term to the sixth power. \((\sqrt{2})^{6}+i^{6}(\sqrt{2})^{6}\) Simplify. \(8+8 i^{6}\) Let \(i^{6}=i^{4} \cdot i^{2}=-1\) \(8-8=0\) This is incorrect. What mistake was made?

Step-by-Step Solution

Answer

The mistake was incorrectly applying the exponent to individual terms instead of the binomial as a whole using the binomial theorem or complex number properties.

Step 1: Identify the Error in Formulation

The mistake is in the step where each term of the binomial \( (\sqrt{2}+i \sqrt{2})^6 \) is incorrectly separated and raised to the sixth power individually, rather than expanding the binomial as a whole using the binomial theorem or complex exponentiation formulas.

Step 2: Correct Use of the Binomial Theorem

Recognize that to solve \((a + b)^n\), you must use the binomial theorem, not independently apply the exponent to each term. The binomial theorem \((x+y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^{k}\) can be used to correctly distribute the exponent over each term in the binomial.

Step 3: Simplify Using Properties of Complex Numbers

Instead of separating terms, note that \(i^2 = -1\), and calculations involving higher powers of \(i\) should consider this cycle: \(i, -1, -i, 1, i, -1, \ldots\). Therefore, when exponentiating complex expressions, simplify using this cycle to avoid incorrect simplifications seen in the solution attempt.

Step 4: Re-evaluate the Expression Correctly

The expression \((\sqrt{2}+i\sqrt{2})^6\) should first be recognized as a complex number \(z = r(cos \theta + i\sin \theta)\), where \(r = |z| = 2\) and \(\theta = \frac{\pi}{4}\). Using De Moivre's Theorem, \(z^n = r^n (cos(n\theta) + i\sin(n\theta))\). Therefore, \([(\sqrt{2}+i \sqrt{2})]^6 = (2)^6(cos(\frac{6\pi}{4}) + i\sin(\frac{6\pi}{4})) = 64(-1) = -64\).

Key Concepts

Binomial Theorem

The Binomial Theorem is a powerful tool used to expand expressions raised to a power. It provides a formula for expressing \((x + y)^n\) in a series format. The theorem is expressed as follows:

\[(x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^{k}\]

Here,

• \(n\) is the power to which the binomial is raised.
• \(k\) is an index of summation.
• \(\binom{n}{k}\) is a combination, indicating the number of ways to choose \(k\) elements from \(n\).
• \(x\) and \(y\) are the terms in the binomial.

When you expand a binomial using this theorem, you systematically apply the coefficient, combine the terms and powers to ensure all parts of the binomial are raised appropriately.

Using the binomial theorem ensures better accuracy when dealing with expressions involving higher exponentiation, especially in cases involving complex numbers.

Complex Exponentiation

Complex exponentiation involves raising a complex number to a certain power. Unlike real numbers, complex numbers include an imaginary part, denoted by \(i\). When exponentiating complex numbers, it is crucial to consider their polar form.

In rectangular form, a complex number is represented as \(a + bi\), where \(i\) is the imaginary unit. However, in polar form, a complex number \(z\) can be expressed as \(r(cos\theta + i\sin\theta)\), where

• \(r\) is the modulus \(|z|\) of the complex number and \(\theta\) is the argument.

Raising a complex number in polar form to a power is simplified using this formula:

\[z^n = r^n (\cos(n\theta) + i\sin(n\theta))\]

This approach makes it easier to handle the trigonometric components of complex numbers, reducing errors during calculations and allowing the expression to be accurately evaluated.

De Moivre's Theorem

De Moivre's Theorem provides a straightforward way to find powers and roots of complex numbers. This theorem states:

If a complex number is expressed in polar form as \(z = r(cos\theta + i\sin\theta)\), then the power of this complex number is given by:

\[z^n = r^n (\cos(n\theta) + i\sin(n\theta))\]

This theorem exploits the simplicity of polar representation to make exponentiation of complex numbers less cumbersome.

For instance, by applying De Moivre's Theorem, you can correctly compute higher powers like \((\sqrt{2}+i\sqrt{2})^6\). This approach greatly aids in managing both the magnitude and direction of the complex number through its polar form.

Powers of i

The imaginary unit \(i\) has special properties when raised to powers. Since \(i\) is defined as the square root of \(-1\), \(i^2 = -1\). Understanding how this functions with powers is fundamental:

• \(i^1 = i\)
• \(i^2 = -1\)
• \(i^3 = i \cdot i^2 = -i\)
• \(i^4 = i^2 \cdot i^2 = 1\)

The powers of \(i\) repeat every four cycles. Recognizing this cyclical nature assists in performing complex arithmetic since exponents can be easily simplified through division by four.

This simplification is particularly useful when dealing with more significant calculations in expressions where the cycle of powers impacts the result, ensuring accurate simplification and reduction of errors.