Problem 87 - A Graphical Approach to College Algebra

Question

Show that \(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2} i\) is a square root of \(i\).

Step-by-Step Solution

Answer

\(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i\) squared equals \(i\), confirming it is a root.

Step 1: Recognize the Square Root Property

To show that \( \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i \) is a square root of \( i \), we need to square it and check if the result is equal to \( i \). So, we must find \((\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i)^2\).

Step 2: Apply the Binomial Theorem

Apply the formula \((a+b)^2 = a^2 + 2ab + b^2\), where \(a = \frac{\sqrt{2}}{2}\) and \(b = \frac{\sqrt{2}}{2} i\). Then, calculate each term: \(a^2\), \(2ab\), and \(b^2\).

Step 3: Calculate Each Term Separately

- For \(a^2: \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{2}{4} = \frac{1}{2}\).- For \(2ab: 2 \times \frac{\sqrt{2}}{2} \times \frac{\sqrt{2}}{2} i = 2 \times \frac{1}{2} i = i\).- For \(b^2: \left(\frac{\sqrt{2}}{2} i\right)^2 = \frac{2}{4} i^2 = \frac{1}{2} (-1) = -\frac{1}{2}\).

Step 4: Add the Terms Together

Combine the results of the squared terms: \(a^2 + 2ab + b^2 = \frac{1}{2} + i - \frac{1}{2}\). This simplifies to \(i\), which verifies that \( \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i\) is indeed a square root of \(i\).

Key Concepts

Square Root of Complex Numbers

When it comes to complex numbers, finding the square root can be a bit tricky but exciting. Complex numbers can be expressed in the form of \(a + bi\), where \(a\) and \(b\) are real numbers. The imaginary unit \(i\) signifies \( \sqrt{-1} \). Therefore, the challenge is to find two numbers \(x\) and \(y\) such that \((x + yi)^2 = a + bi\).

To solve this, you expand \((x + yi)^2 = x^2 + 2xyi + (yi)^2\). When he square of imaginary component \((yi)^2\) is computed, it results in \(-y^2\) since \(i^2 = -1\). So, \((x + yi)^2\) can be re-expressed as \(x^2 - y^2 + 2xyi\).

If you equate this to a complex number \(a + bi\), you can deduce two separate equations: \(x^2 - y^2 = a\) and \(2xy = b\). Solving these equations simultaneously gets you the potential values of \(x\) and \(y\), which are the real and imaginary parts of the square root respectively.

• Squared real component \(a^2\)
• Term resulting from the product of real and imaginary component \(2ab\)
• Squared imaginary component \(b^2\)

Binomial Theorem

The Binomial Theorem is a powerful tool used to expand expressions that are raised to a power. It is especially handy when dealing with powers of a binomial, which is an algebraic expression of the sum or difference of two terms, like \((a + b)^n\). Simply stated, the theorem provides a formula for these expansions without having to multiply everything out manually.

The Binomial Theorem tells us that \((a + b)^2 = a^2 + 2ab + b^2\). This is particularly useful for calculating the powers of sums, especially when they contain complex numbers. In the exercise provided, the binomial terms are \(a = \frac{\sqrt{2}}{2}\) and \(b = \frac{\sqrt{2}}{2} i\).

By squaring this binomial, we can apply the theorem directly to work with each part individually:

• The first term is \(a^2 = \left( \frac{\sqrt{2}}{2} \right)^2 = \frac{1}{2}\).
• The second term, being the interaction of both terms, results in \(2ab = 2 \times \frac{\sqrt{2}}{2} \times \frac{\sqrt{2}}{2} i = i\).
• The last term involves squaring the imaginary component \(b^2 = \left( \frac{\sqrt{2}}{2} i\right)^2 = -\frac{1}{2}\).

This precise breakdown of the squares and cross-terms shows how the binomial theorem can greatly simplify complex calculations.

Algebraic Proofs

Algebraic proofs are methodical ways to verify mathematical statements using established algebraic properties. When working with complex numbers, these proofs help us to establish validity confidently and with logical rigor. In the provided exercise, algebraic proofs explain why \( \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i \) is indeed the square root of \(i\).

The proof process begins with identifying the relationships and properties involved, such as recognizing that squaring \( \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i \) should translate back to \(i\).

By squaring the complex number, and carefully applying the algebraic rule of sums: \((a + b)^2 = a^2 + 2ab + b^2\), each component is calculated sequentially.

• The first component \(a^2\) provides \(\frac{1}{2}\).
• The mixed term \(2ab\) produces \(i\), an essential part of the imaginary outcome.
• The last component brought by \(b^2\) negates some of the previous real component with \(-\frac{1}{2}\).

By successfully proving all parts add up to \(i\), it confirms the initial assumption, showcasing the power of algebra in validating complex number identities.