Problem 26
Question
Write the given number in the form \(a+i b\). $$ \frac{1}{(1+i)(1-2 i)(1+3 i)} $$
Step-by-Step Solution
Verified Answer
The number is written as \(0.06 - 0.08i\).
1Step 1: Multiply Denominators
First, we multiply the denominators: \[(1+i)(1-2i)(1+3i) = [(1+i)(1-2i)](1+3i).\] Simplify \((1+i)(1-2i)\): \[(1+i)(1-2i) = 1 \cdot 1 + 1 \cdot (-2i) + i \cdot 1 + i \cdot (-2i) = 1 - 2i + i - 2i^2.\] As \(i^2 = -1\), we get: \[1 - 2i + i + 2 = 3 - i.\] So, now we simplify it to:\[(3-i)(1+3i).\]
2Step 2: Multiply Resulting Denominator
Next, multiply \((3-i)(1+3i)\): \[(3-i)(1+3i) = 3 \cdot 1 + 3 \cdot 3i - i \cdot 1 - i \cdot 3i.\] This simplifies to: \[3 + 9i - i - 3i^2.\] Notice again that \(i^2 = -1\), so we get:\[3 + 9i - i + 3 = 6 + 8i.\] Thus, the complete denominator is: \[6 + 8i.\]
3Step 3: Rationalization by Conjugate
Multiply both the numerator and denominator by the complex conjugate of the denominator:\[\frac{1}{6+8i} \times \frac{6-8i}{6-8i} = \frac{6-8i}{(6+8i)(6-8i)}.\]The new denominator is:\[(6+8i)(6-8i) = 6^2 - (8i)^2 = 36 - (-64) = 100.\]This simplifies the expression to:\[\frac{6-8i}{100} = \frac{6}{100} - \frac{8i}{100}.\]
4Step 4: Simplify to Standard Form
Simplify:\[\frac{6}{100} - \frac{8i}{100} = 0.06 - 0.08i.\]So the expression in the form of \(a+bi\) is:\[0.06 - 0.08i.\]
Key Concepts
Complex ConjugateStandard Form of a Complex NumberMultiplication of Complex Numbers
Complex Conjugate
The complex conjugate of a complex number is an important concept when dealing with complex numbers. If you have a complex number in the form of \(a + bi\), its complex conjugate is \(a - bi\). This means we simply change the sign of the imaginary part. The complex conjugate is particularly useful for rationalizing complex numbers, as seen in the provided exercise.
When multiplying a complex number by its conjugate, the result is always a real number. This happens because the imaginary parts cancel each other out, leaving you with \(a^2 + b^2\). In the solution, multiplying \((6+8i)\) by its conjugate \((6-8i)\) simplifies the denominator to a real number, 100. Here is why it's useful:
When multiplying a complex number by its conjugate, the result is always a real number. This happens because the imaginary parts cancel each other out, leaving you with \(a^2 + b^2\). In the solution, multiplying \((6+8i)\) by its conjugate \((6-8i)\) simplifies the denominator to a real number, 100. Here is why it's useful:
- It helps eliminate the imaginary part in the denominator.
- Allows the representation of the complex fraction in standard form.
- Simplifies calculation by converting complex problems into real number operations.
Standard Form of a Complex Number
The standard form of a complex number is \(a + bi\), where \(a\) is the real part and \(b\) is the imaginary part. Writing complex numbers in this form makes them easier to manipulate and understand.
In the exercise, you are asked to express a complex fraction in its standard form. After rationalizing the denominator using the complex conjugate, the expression \(\frac{6-8i}{100}\) simplifies to \(0.06 - 0.08i\). This is in standard form, where \(a = 0.06\) and \(b = -0.08\).
Why is this form so important?
In the exercise, you are asked to express a complex fraction in its standard form. After rationalizing the denominator using the complex conjugate, the expression \(\frac{6-8i}{100}\) simplifies to \(0.06 - 0.08i\). This is in standard form, where \(a = 0.06\) and \(b = -0.08\).
Why is this form so important?
- It provides a clear distinction between the real and imaginary parts of a complex number.
- Standard form is convenient for addition, subtraction, and comparison of complex numbers.
- It aids in visualizing complex numbers on the complex plane, as they correspond to points with coordinates \((a, b)\).
Multiplication of Complex Numbers
Multiplicating complex numbers involves using the distributive property, just like with polynomials, but with careful attention to the imaginary unit \(i\), where \(i^2 = -1\). In the exercise, multiple sets of complex numbers are multiplied, and understanding each step is crucial.
For instance, when multiplying \((1+i)(1-2i)\), you apply the distributive property:
1. Multiply each component: \(1 \cdot 1 + 1 \cdot (-2i) + i \cdot 1 + i \cdot (-2i)\).
2. Simplify terms: \(1 - 2i + i + 2\), using \(i^2 = -1\), to combine like terms.
These steps result in \(3 - i\).
Now, multiply \((3-i)(1+3i)\) by the same method: distribute terms, combine, and remember \(i^2 = -1\). Hence, the result transforms into \(6 + 8i\), as described in the initial steps.
For instance, when multiplying \((1+i)(1-2i)\), you apply the distributive property:
1. Multiply each component: \(1 \cdot 1 + 1 \cdot (-2i) + i \cdot 1 + i \cdot (-2i)\).
2. Simplify terms: \(1 - 2i + i + 2\), using \(i^2 = -1\), to combine like terms.
These steps result in \(3 - i\).
Now, multiply \((3-i)(1+3i)\) by the same method: distribute terms, combine, and remember \(i^2 = -1\). Hence, the result transforms into \(6 + 8i\), as described in the initial steps.
- It’s crucial to group real and imaginary parts separately after multiplication.
- Simplify expressions using \(i^2 = -1\) for better clarity and precision.
- Write the final expression in standard form if needed, for follow-on calculations or visual interpretations.
Other exercises in this chapter
Problem 26
Describe the set of points in the complex plane that satisfies \(|z-i|+|z+i|=1\)
View solution Problem 26
Use \((8)\) to compute the indicated power. $$ \left[\sqrt{3}\left(\cos \frac{2 \pi}{9}+i \sin \frac{2 \pi}{9}\right)\right]^{6} $$
View solution Problem 26
Express \(\ln z\) in the form \(a+i b\). \(z=1+i\)
View solution Problem 26
Verify that the given function \(u\) is harmonic. Find \(v\), the harmonic conjugate function of \(u\). Form the corresponding analytic function \(f(z)=u+i v .\
View solution