Question

Divide and simplify. $$(-2+3 i) \div(1-i)$$

Short answer

-1+i

Step-by-step solution

Multiply numerator and denominator by the complex conjugate of the denominator

The complex conjugate of the denominator is obtained by changing the sign of the imaginary part. So, the conjugate of 1-i is 1+i. We multiply both the numerator and denominator by this conjugate to eliminate the imaginary number in the denominator.

Apply the distributive property (FOIL) to both the numerator and denominator

Using FOIL (First, Outer, Inner, Last) to expand the numerator and denominator: (-2+3i)(1+i) for the numerator and (1-i)(1+i) for the denominator.

Simplify the expressions

After expanding, combine like terms in both the numerator and the denominator. The denominator simplifies to a real number as the imaginary parts cancel out.

Write the final simplified form

After simplification, express your result in the form a + bi, where a and b are real numbers.

Key concepts

Complex Conjugate

The concept of a complex conjugate is crucial when working with complex numbers, especially in operations like division. A complex conjugate is formed by changing the sign of the imaginary part of a complex number. So, for the number \(1-i\), its complex conjugate is \(1+i\).

Why do we use it? When you multiply a complex number by its conjugate, the result is always a real number. This is because the product of \(i\) and \(–i\) is \(–i^2\), which simplifies to \(+1\), since \(i^2 = -1\). This property is very valuable in simplifying expressions that involve complex numbers, as it helps to remove the imaginary part from the denominator, making division possible.

FOIL Method

The FOIL method stands for First, Outer, Inner, Last and refers to a technique for multiplying two binomials. Let’s take the binomials from the original exercise as an example. Using FOIL, we multiply the terms in the following order:

• First: The first terms of each binomial (\( -2 \times 1 \))
• Outer: The first term of the first binomial and the second term of the second binomial (\( -2 \times i \))
• Inner: The second term of the first binomial and the first term of the second binomial (\( 3i \times 1 \))
• Last: The second terms of each binomial (\( 3i \times i \))

By following this method, we ensure all parts of the binomials are multiplied correctly to produce a simplified form in a systematic way.

Imaginary Numbers

Imaginary numbers are fundamental to complex arithmetic. They are defined by the existence of \(i\), which is the square root of \(–1\). In the context of our division problem, \(i\) allows us to represent and manipulate square roots of negative numbers.

Without imaginary numbers, we couldn't have complex numbers, which are expressed in the form \(a + bi\), where \(a\) and \(b\) are real numbers. Understanding the behavior of \(i\), such as \(i^2 = -1\), is key to performing operations like multiplication and division with complex numbers.

Simplifying Expressions

Simplifying complex number expressions involves reducing them to their simplest form. After using the FOIL method and multiplying by the complex conjugate, we often have terms that can be combined. For example, similar real terms or those involving \(i\) are grouped and simplified.

In our exercise, \(i^2\) appears during multiplication, which should be immediately recognized as \(–1\), helping to simplify further. The goal is to have the final answer in \(a + bi\) form, with no parentheses and as few terms as possible, making the result clear and easy to work with in subsequent calculations.

Distributive Property

The distributive property is a cornerstone of algebra, and it’s used when multiplying a single term by each term in a binomial, trinomial, or polynomial. It’s important when applying the FOIL method because it justifies the step-by-step process of multiplication. This property is expressed as \(a(b + c) = ab + ac\).

In complex number division, once the complex conjugate is in place, we distribute (or 'multiply through') with both the numerator and the denominator. This helps in eliminating the imaginary units in the denominator and allows us to simplify the expression into a more manageable form without affecting any of the included values.