Problem 15 - Algebra 2 and Trigonometry

Question

In \(3-17,\) find each sum or difference of the complex numbers in \(a+b i\) form. $$ \left(-\frac{3}{2}+\frac{5}{3} i\right)-\left(\frac{9}{4}-\frac{1}{5} i\right) $$

Step-by-Step Solution

Answer

The difference is \\(-\frac{15}{4} + \frac{28}{15} i\\).

Step 1: Identify Real and Imaginary Parts

Let's first identify the real and imaginary parts of each complex number. For the first number, \(-\frac{3}{2} + \frac{5}{3} i\), the real part is \(-\frac{3}{2}\) and the imaginary part is \(+\frac{5}{3} i\). For the second number, \(\frac{9}{4} - \frac{1}{5} i\), the real part is \(\frac{9}{4}\) and the imaginary part is \(-\frac{1}{5} i\).

Step 2: Subtract the Real Parts

Subtract the real part of the second number from the real part of the first number: \(-\frac{3}{2} - \frac{9}{4}\). Convert both numbers to a common denominator before subtracting. The LCM of 2 and 4 is 4. Thus: \[-\frac{3}{2} = -\frac{6}{4}\]So, performing the operation: \[-\frac{6}{4} - \frac{9}{4} = -\frac{15}{4}\]

Step 3: Subtract the Imaginary Parts

Subtract the imaginary part of the second number from the imaginary part of the first number: \(\frac{5}{3} - (-\frac{1}{5})\), which simplifies to \(\frac{5}{3} + \frac{1}{5}\). Convert both fractions to a common denominator, which is 15:\[\frac{5}{3} = \frac{25}{15}, \quad \frac{1}{5} = \frac{3}{15}\]Adding these gives:\[\frac{25}{15} + \frac{3}{15} = \frac{28}{15}\]

Step 4: Combine the Results

Combine the results of the subtraction to express the difference of the complex numbers in \(a + bi\) form. The real part is \(-\frac{15}{4}\) and the imaginary part is \(\frac{28}{15} i\). Thus, the result is:\[-\frac{15}{4} + \frac{28}{15} i\]

Key Concepts

Real and Imaginary Parts of Complex Numbers

When dealing with complex numbers, it's important to break them down into their fundamental components: the real and imaginary parts. A complex number is typically expressed in the form \(a + bi\), where \(a\) represents the real part and \(bi\) represents the imaginary part. In this case, \(b\) is the coefficient of the imaginary unit \(i\), where \(i\) is the square root of \(-1\). This means that complex numbers have two dimensions, much like coordinates on a plane.
For example, in the complex number \(-\frac{3}{2} + \frac{5}{3} i\), \(-\frac{3}{2}\) is the real part, and \(\frac{5}{3} i\) is the imaginary part. Similarly, for \(\frac{9}{4} - \frac{1}{5} i\), the real part is \(\frac{9}{4}\), and the imaginary part is \(-\frac{1}{5} i\). By identifying these parts first, we can then handle real and imaginary components separately as we perform mathematical operations.

Subtracting Complex Numbers

Subtracting complex numbers involves treating the real and imaginary components independently. This process is similar to vector subtraction, where you subtract corresponding components.
To perform the subtraction, you subtract the real parts from each other and the imaginary parts from each other. For instance, consider two complex numbers: \(-(\frac{3}{2}) + \frac{5}{3}i\) and \(\frac{9}{4} - \frac{1}{5}i\). You first handle the real parts: \(-\frac{3}{2} - \frac{9}{4}\).
Handling each component separately allows you to simplify the expressions step by step. Once you've managed the real components, move on to the imaginary parts: \(\frac{5}{3} - (-\frac{1}{5})\), which simplifies to \(\frac{5}{3} + \frac{1}{5}\). Finally, once both subtractions are done, you can combine them back into a single complex number: \(-\frac{15}{4} + \frac{28}{15}i\). This methodically structured approach simplifies complex arithmetic.

Fractions with Common Denominators

Managing fractions, especially when dealing with different denominators, can often seem tricky. To make subtraction or addition of fractions smoother, you should first find a common denominator. This allows you to combine fractions more effortlessly by transforming them into comparable forms.
For example, when subtracting \(-\frac{3}{2}\) and \(\frac{9}{4}\), you need a common denominator. Since the denominators are 2 and 4, the least common multiple (LCM) is 4. By converting \(-\frac{3}{2}\) to \(-\frac{6}{4}\) and leaving \(\frac{9}{4}\) as is, both fractions now have a denominator of 4, allowing direct subtraction: \(-\frac{6}{4} - \frac{9}{4} = -\frac{15}{4}\).
Similarly, for \(\frac{5}{3} + \frac{1}{5}\), the LCM of 3 and 5 is 15. Transform \(\frac{5}{3}\) into \(\frac{25}{15}\), and \(\frac{1}{5}\) into \(\frac{3}{15}\), so you now have compatible terms: \(\frac{25}{15} + \frac{3}{15} = \frac{28}{15}\). This step ensures both operations and results are logical and simplified.