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Chapter 8: Problem 44

Combine the following complex numbers. $$[(4-5 i)+(2+i)]-(2+5 i)$$

Short Answer

Expert verified
The result is \(4 - 9i\).

Step by step solution

01

Simplify the expression inside the brackets

First, focus on simplifying the expression inside the outer brackets: \((4 - 5i) + (2 + i)\). Combine the real numbers and the imaginary numbers separately:- Real parts: \(4 + 2 = 6\)- Imaginary parts: \(-5i + i = -4i\) Thus, the simplified expression is \(6 - 4i\).
02

Subtract the remaining complex number

Now, take the result from Step 1 and subtract the complex number outside the brackets:\((6 - 4i) - (2 + 5i)\). Again, combine the real and imaginary parts separately:- Real parts: \(6 - 2 = 4\)- Imaginary parts: \(-4i - 5i = -9i\).Thus, the result of the subtraction is \(4 - 9i\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Addition of Complex Numbers
When adding complex numbers, the first step is to deal with each part separately. Complex numbers have two components: the real part and the imaginary part. To add two complex numbers, such as
  • \((a + bi) + (c + di)\),
you will add the real parts \((a + c)\) and the imaginary parts \((b + d)\) separately. This simplifies our example from the original exercise:
  • Real parts: \(4 + 2 = 6\)
  • Imaginary parts: \(-5i + i = -4i\).
Therefore, the addition of the complex numbers results in the new complex number \(6 - 4i\). Remember, always handle the numbers within their distinct categories (real vs. imaginary) to avoid any confusion. Be sure to keep your imaginary unit \(i\) clearly distinguished from the real number operations.
Subtraction of Complex Numbers
Subtracting complex numbers works much like addition, but you subtract corresponding parts instead. For two complex numbers:
  • \((a + bi) - (c + di)\),
you will subtract the real parts \((a - c)\) and the imaginary parts \((b - d)\). Let's apply this in the context of our example. After previously simplifying to \((6 - 4i)\), we subtract the complex number \((2 + 5i)\):
  • Real parts: \(6 - 2 = 4\)
  • Imaginary parts: \(-4i - 5i = -9i\).
Now, we are left with the complex result \(4 - 9i\). Be cautious with the signs during subtraction, especially with negatives, as they can often lead to errors if overlooked.
Imaginary Numbers
Imaginary numbers might seem a bit intimidating at first, but they are actually quite straightforward. An imaginary number is composed of the imaginary unit \(i\), which is defined as the square root of \(-1\). Essentially, it helps to "extend" the real number line into a more complex plane, allowing us to solve equations involving negative square roots.
  • For any real number \(b\), multiplying it by \(i\) gives you \(bi\), representing the imaginary part of a complex number.
  • In a complex number \(a + bi\), \(a\) is the real part, and \(bi\) is the imaginary part.
By understanding the concept of imaginary numbers, we can appreciate how they allow us to represent numbers that are not possible to express using only real numbers. This makes complex numbers a powerful tool in mathematics, particularly when tackling quadratic equations and electrical engineering problems.

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Most popular questions from this chapter

Graph each equation.\(r^{2}=4 \cos 2 \theta\)

Match each equation with its appropriate graph. Assume \(a\) and \(b\) are constants, and that \(n\) is a natural number. a. \(r=a\) i. Cardioid b. \(\theta=a\) ii. Circle c. \(r=a \cos 5 \theta\) iii. Lemniscate d. \(r=a \sin 6 \theta\) iv. Lima莽on e. \(r^{2}=a \cos 2 \theta\) v. Line f. \(r=a+a \cos \theta\) vi. Rose (even number of leaves) g. \(r=a+b \cos \theta\) vii. Rose (odd number of leaves)

Determine whether the graph of the given equation will be a line, circle, rose curve, lemniscate, cardioid, or lima莽on. Use Figures 15-21 to help you identify each equation, but do not actually sketch the graph.\(r=6\)

Let \(z_{1}=2+3 i, z_{2}=2-3 i\), and \(z_{3}=4+5 i\), and find $$z_{3}\left(z_{1}-z_{2}\right)$$

Convert each equation to polar coordinates and then sketch the graph.\(x^{2}+y^{2}=16\)
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