Problem 43 - Algebra 2 and Trigonometry

Question

In \(38-45,\) find the multiplicative inverse of each of the following in \(a+b i\) form. $$ 2-\frac{1}{2} i $$

Step-by-Step Solution

Answer

The multiplicative inverse is \( \frac{8}{17} + \frac{2}{17}i \).

Step 1: Identify the given complex number

The given complex number is \( 2 - \frac{1}{2}i \). We need to find its multiplicative inverse.

Step 2: Use the formula for multiplicative inverse

The multiplicative inverse of a complex number \( a + bi \) is given by \( \frac{a}{a^2 + b^2} + \frac{-b}{a^2 + b^2}i \). Thus for our problem, we have \( a = 2 \) and \( b = -\frac{1}{2} \).

Step 3: Calculate denominator for the inverse

First, calculate \( a^2 + b^2 \):\[(2)^2 + \left(-\frac{1}{2}\right)^2 = 4 + \frac{1}{4} = \frac{16}{4} + \frac{1}{4} = \frac{17}{4}\]

Step 4: Calculate the inverse

Substitute \( a \) and \( b \) into the inverse formula:\[\frac{2}{17/4} + \frac{\frac{1}{2}}{17/4}i = \frac{2 \times 4}{17} + \frac{\frac{1}{2} \times 4}{17}i = \frac{8}{17} + \frac{2}{17}i\]

Key Concepts

Multiplicative Inverse

The multiplicative inverse of a complex number plays an essential role just like the reciprocal for regular numbers. When you're dealing with complex numbers of the form \(a + bi\), the multiplicative inverse is the number which, when multiplied by \(a + bi\), gives you 1. That might seem tricky at first, but let's simplify it.

To find it, you use the formula:

• For \(a + bi\), the multiplicative inverse is \(\frac{a}{a^2 + b^2} + \frac{-b}{a^2 + b^2}i\)

The idea is to essentially divide the complex number by the sum of the squares of its real and imaginary parts. This results in the real and imaginary parts being scaled down, mirroring the process of finding reciprocals.

In the given exercise, for the complex number \(2 - \frac{1}{2}i\), we set \(a = 2\) and \(b = -\frac{1}{2}\). The result becomes \(\frac{8}{17} + \frac{2}{17}i\) after performing the calculations. This means, multiplying \(2 - \frac{1}{2}i\) by this result will indeed yield 1.

Complex Conjugate

The concept of a complex conjugate is straightforward but powerful. If you have a complex number \(a + bi\), its complex conjugate is simply \(a - bi\). This switch from \(+bi\) to \(-bi\) has interesting effects.

Why is it important? When multiplied by its conjugate, the imaginary parts cancel out, leaving a real number. This operation is useful for rationalizing denominators in the complex plane and is integral in various calculations involving complex numbers. For instance, calculating magnitudes and inverses often involves this technique.

• Example: For \(2 - \frac{1}{2}i\), the complex conjugate is \(2 + \frac{1}{2}i\).
• Usage: Find the multiplicative inverse by simplifying expressions like \(\frac{a - bi}{a^2 + b^2}\).

Working with complex conjugates helps simplify complex arithmetic and aids in deriving results that are more intuitive when presented in real number terms.

Complex Number Formula

Dealing with complex numbers isn't as daunting when you have the right formulas at your disposal. Each complex number \(a + bi\) can be harnessed in calculations with a few handy rules.

Here are crucial uses of the complex number formula:

• Adding: \((a + bi) + (c + di) = (a+c) + (b+d)i\)
• Subtracting: \((a + bi) - (c + di) = (a-c) + (b-d)i\)
• Multiplying: \((a + bi)(c + di) = (ac-bd) + (ad+bc)i\)
• Dividing needs conjugates: \(\frac{a + bi}{c + di} = \frac{(a + bi)(c - di)}{c^2 + d^2}\)

These formulas form the backbone of operations involving complex numbers. They ensure both the real and imaginary parts are accounted for evenly, keeping your calculations structured and intuitive.
In the current exercise, the multiplication formula is disguised in the operations forming the multiplicative inverse. Calculating such inverse relies heavily on these principles, ensuring accuracy in deriving results like \(\frac{8}{17} + \frac{2}{17}i\) from our initial \(2 - \frac{1}{2}i\).