Problem 92 - Precalculus : Building Concepts and Connections

Question

We know that \(i^{2}=-1,\) but is there a complex number \(z\) such that \(z^{2}=\) i? We answer that question in this exercise. (a) Calculate \(\left(\frac{\sqrt{2}}{2}(1+i)\right)\left(\frac{\sqrt{2}}{2}(1+i)\right)\) (b) Use your answer in part (a) to find a complex number \(z\) such that \(z^{2}=i\)

Step-by-Step Solution

Answer

The complex number \(z\) such that \(z^{2} = i\) is \(z = \frac{\sqrt{2}}{2}(1 + i)\)

Step 1: Calculate the square of the complex number

Multiply the complex number \(\left(\frac{\sqrt{2}}{2}(1+i)\right)\) by itself that is, \(\left(\frac{\sqrt{2}}{2}(1+i)\right)\times\left(\frac{\sqrt{2}}{2}(1+i)\right)\). Multiply the real parts and the imaginary parts separately. The real part is \(\frac{\sqrt{2}}{2}\times\frac{\sqrt{2}}{2} = \frac{1}{2}\) and the imaginary part is \((1+i)\times(1+i)= 1 + 2i + i^{2}= 1 + 2i - 1 = 2i\). So the result is \(\frac{1}{2} * 2i = i\).

Step 2: Find the complex number \(z\)

Using the result obtained in step 1, we know that \(\left(\frac{\sqrt{2}}{2}(1+i)\right)^{2} = i\), this means the complex number \(z\) such that \(z^{2} = i\) is \(z = \frac{\sqrt{2}}{2}(1 + i)\)

Key Concepts

Imaginary Unit

In mathematics, the imaginary unit is a mathematical concept represented by the letter \(i\). Its defining property is that \(i^2 = -1\). This concept emerges from extending the real number system to include solutions to polynomial equations that cannot be solved using only real numbers. For instance, the equation \(x^2 + 1 = 0\) does not have any real solutions. However, using the concept of imaginary units, it can be solved by stipulating that \(x = i\) or \(x = -i\).

Imaginary numbers are built around this unit \(i\), and they interact interestingly with real numbers to form what are known as complex numbers. A complex number generally consists of a real part and an imaginary part, like \(a + bi\), where \(a\) and \(b\) are real numbers and \(i\) is the imaginary unit. This system allows us to perform various algebraic operations that are otherwise impossible with just real numbers.

Multiplication of Complex Numbers

The multiplication of complex numbers involves distributing the terms as you would in regular algebra but with a special consideration for the imaginary unit. Consider two complex numbers, \(a + bi\) and \(c + di\).

• Multiply each part: \((a + bi)(c + di) = ac + adi + bci + bdi^2\).
• Combine like terms: Since \(i^2 = -1\), replace \(bdi^2\) with \(-bd\).
• Resulting expression: \(ac - bd + (ad + bc)i\).

Just like in our exercise where \( \left( \frac{\sqrt{2}}{2}(1+i) \right) \) is multiplied by itself, observe that you first multiply the real parts and combine the terms that include \(i\). The choice of ordering and careful attention to \(i^2 = -1\) is crucial to reach the correct result.

Solving Quadratic Equations

Solving quadratic equations is a fundamental aspect of algebra that often involves finding a set of solutions (or roots) for equations of the form \(ax^2 + bx + c = 0\). These equations generally have two solutions, determined either by factoring, completing the square, or the quadratic formula.

• Quadratic Formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
• Factorization: Splitting the quadratic into two binomials, if possible.

If the discriminant \(b^2 - 4ac\) is negative, then the roots of the quadratic equation are not real and are instead complex numbers. For example, our earlier discussion of \(i\) stems from trying to solve equations of this nature, where direct real solutions do not exist. Understanding complex roots is necessary to solve such quadratics, demonstrating how complex numbers offer a complete solution to polynomial equations of degree two and higher.