Question

Compute (a) \(\frac{1+2 i}{3-i}\) (b) \((1+2 i)(3-i) ;\) and (c) \(\frac{x+1+i}{x+1-i}, x \in E^{1}\). Do it in two ways: (i) using definitions only and the notation \((x, y)\) for \(x+y i\); and (ii) using all laws valid in a field.

Short answer

(a) \(\frac{1+7i}{10}\), (b) \(5+5i\), (c) \(\frac{x^2 + 2x + 2(x+1)i}{x^2 + 2x + 2}\)

Step-by-step solution

Simplify Part (a) Using Definitions

To compute \(\frac{1+2i}{3-i}\), we begin by multiplying the numerator and the denominator by the conjugate of the denominator, \(3+i\). This gets rid of the imaginary part in the denominator.\[\frac{1+2i}{3-i} \times \frac{3+i}{3+i} = \frac{(1+2i)(3+i)}{(3-i)(3+i)}\]

Compute the Numerator of Part (a)

Expand the numerator \((1+2i)(3+i)\) using multiplication:\[1 \cdot 3 + 1 \cdot i + 2i \cdot 3 + 2i \cdot i = 3 + i + 6i + 2i^2\]Since \(i^2 = -1\), it becomes:\[3 + 7i - 2 = 1 + 7i\]

Compute the Denominator of Part (a)

Expand the denominator \((3-i)(3+i)\) using the difference of squares:\[3^2 - i^2 = 9 - (-1) = 10\]Therefore, the expression is simplified to:\[\frac{1 + 7i}{10}\]

Simplify Part (b)

Compute \((1+2i)(3-i)\) by using the distributive property:\[1 \cdot 3 + 1 \cdot (-i) + 2i \cdot 3 + 2i \cdot (-i)\]This simplifies to:\[3 - i + 6i - 2i^2 = 3 + 5i + 2 = 5 + 5i\]

Simplify Part (c) Using Definitions

To compute \(\frac{x+1+i}{x+1-i}\), multiply the numerator and denominator by the conjugate of the denominator:\[\frac{(x+1+i)(x+1+i)}{(x+1-i)(x+1+i)}\]

Expand Numerator of Part (c)

Expand the squared numerator:\[(x+1+i)(x+1+i) = (x+1)^2 + 2(x+1)i + i^2\]Which results in:\[(x^2 + 2x + 1) + 2(x+1)i - 1 = x^2 + 2x + 2(x+1)i\]

Compute Denominator of Part (c)

The denominator is a difference of two squares:\[(x+1)^2 - i^2 = (x+1)^2 + 1\]

Simplify Part (c)

The expression simplifies to:\[\frac{x^2 + 2x + 2(x+1)i}{x^2 + 2x + 2}\]

Field Properties Approach for Part (a)

Using field properties, rewrite the denominator and divide:\[\frac{a+bi}{c+di} \cdot \frac{c-di}{c-di} = \frac{(ac+bd)+(bc-ad)i}{c^2+d^2} = \frac{3+7i}{10}\]

Field Properties Approach for Part (c)

Using field properties, simplify:\[\frac{x+1+i}{x+1-i} \cdot \frac{x+1+i}{x+1+i} = (x^2 + 2x + 2(x+1)i) \] Denominator remains the same: \((x+1)^2 + 1\)

Field Properties Approach for Part (b)

Part (b) involves only multiplication, which remains simple:\[(1+2i)(3-i) = 3 + 5i\]

Key concepts

Conjugate in Complex Numbers

In complex numbers, the conjugate is an important concept that helps simplify expressions during division and other operations. A complex number is generally written as \(a + bi\), where \(a\) is the real part and \(b\) is the imaginary part. The conjugate of this complex number is \(a - bi\), which merely changes the sign of the imaginary part.

Let's consider why the conjugate is helpful. When dividing complex numbers, we often deal with a denominator that contains an imaginary part. To remove this imaginary part, we multiply both the numerator and the denominator by the conjugate of the denominator. This technique leverages the "difference of squares" formula, which states that \((a + b)(a - b) = a^2 - b^2\). Since \(i^2 = -1\), applying the conjugate results in a real number denominator, making the division straightforward.

Using the conjugate helps maintain the structure of a division problem while reducing the complexity of the denominator. It’s a vital tool for ensuring that computations result in a standard complex number form, \(a + bi\), which is easier to interpret or utilize in further calculations.

Complex Division

Complex division involves dividing one complex number by another. Let's break it down using two methods: definitions and field properties.

Using Definitions:

• You multiply the numerator and the denominator by the conjugate of the denominator. This step is crucial because it eliminates the imaginary unit \(i\) from the denominator.
• Next, simplify the numerator by using the distributive property and combining like terms. Then, simplify the denominator with the difference of squares formula to convert it into a real number.
• This simplification transforms the expression into the standard form, where the real and imaginary parts are distinct.

Using this method, division becomes manageable and leads to a more intuitive result.

Field Properties Approach:

• In fields, the division of complex numbers can be managed using properties such as associativity, commutativity, and distributivity just like in real numbers.
• Because a complex field retains these properties, the rules provide a structured way to achieve the result, pairing the real and imaginary components from the numerator directly with the simplified denominator.
• This approach provides an algebraic technique that extends naturally from real numbers to complex numbers, reinforcing the understanding of these numbers as part of a broader mathematical system.

With both methods, the key is isolating real and imaginary parts and simplifying to achieve a standard form that reflects the neat interplay between real and imaginary components in complex numbers.

Complex Multiplication

Multiplication in complex numbers works much like polynomial multiplication. It's performed by distributing each term of the first complex number across each term of the second.

To illustrate, consider multiplying \((1 + 2i)(3 - i)\). Each part of the first complex number, \(1\) and \(2i\), is separately multiplied by each part of the second complex number, \(3\) and \(-i\). The steps are:

• Multiply the real parts: \(1 \cdot 3 = 3\).
• Multiply the first real part by the second imaginary part: \(1 \cdot (-i) = -i\).
• Multiply the first imaginary part by the second real part: \(2i \cdot 3 = 6i\).
• Multiply the imaginary parts: \(2i \cdot (-i) = -2i^2\). Remember, since \(i^2 = -1\), this term becomes \(2\).

This yields the expression \(3 + 6i - i + 2\), which, when combined, results in \(5 + 5i\).

Understanding these steps underscores the interactions between the real and imaginary components, ensuring clarity when executing such multiplications. The structure of complex multiplication thereby reinforces the connection and balance between algebraic computation and complex number theory.