Chapter 4: Problem 80
If \(z=a+b i\) (with \(a, b\) real numbers, not both 0 ), express \(1 / z\) in standard form.
Short Answer
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Question: Given a complex number \(z = a + bi\), express the reciprocal of this complex number \(\frac{1}{z}\) in standard form.
Answer: The reciprocal of the given complex number in standard form is \(\frac{1}{z} = \frac{a^2}{a^2 + b^2} - \frac{b^2}{a^2 + b^2}i\).
Step by step solution
01
Write the given complex number as a fraction
Write the given complex number \(z = a + bi\) as a fraction. We can write this as a fraction by dividing by 1: \(z = \frac{a + bi}{1}\).
02
Find the complex conjugate of z
In order to be able to express the reciprocal of the complex number in standard form, we need to find the complex conjugate of \(z\), which is denoted as \(\bar{z}\).
The complex conjugate of a complex number \(a + bi\) is given by \(a - bi\). So, the complex conjugate of \(z = a+bi\) is: \(\bar{z} = a - bi\).
03
Multiply the numerator and the denominator of the fraction z by the complex conjugate
Multiply the numerator and the denominator of the fraction \(\frac{a + bi}{1}\) by the complex conjugate \(\bar{z} = a - bi\).
Doing this, we get: \(\frac{a + bi}{1} \cdot \frac{a - bi}{a - bi} = \frac{(a + bi)(a - bi)}{(a - bi)(a - bi)}\).
04
Expand the expressions in the numerator and the denominator
Expand the expressions in the numerator and the denominator using the distributive property:
Numerator: \((a + bi)(a - bi) = a(a) + a(-bi) + (bi)(a) + (bi)(-bi) = a^2 - abi + abi - b^2i^2 = a^2 - b^2i^2\).
Since \(i^2 = -1\), we can further simplify the numerator expression to: \(a^2 - b^2(-1) = a^2 + b^2\).
Denominator: Because \(\bar{z} = z\) for real numbers, the denominator becomes \((a-bi)(a-bi) = a^2 - 2abi + b^2i^2\). Since we know that \(2abi\) is imaginary, the result simplifies to \((a^2 - b^2(-1)) = a^2 + b^2\).
05
Simplify the fraction
Now, we can simplify the fraction we found in Step 4: \(\frac{a^2 + b^2}{a^2 + b^2}\). This is equal to 1 because any number divided by itself is 1.
So we have \(\frac{1}{z} = 1 = \frac{a^2 + b^2}{a^2 + b^2} = \frac{a^2}{a^2 + b^2} + \frac{b^2i^2}{a^2 + b^2}\). Since \(i^2 = -1\), the final expression is: \(\frac{1}{z} = \frac{a^2}{a^2 + b^2} - \frac{b^2}{a^2 + b^2}i\).
Hence, we have expressed the reciprocal of the given complex number in the standard form, as \(\frac{1}{z} = \frac{a^2}{a^2 + b^2} - \frac{b^2}{a^2 + b^2}i\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Complex Conjugate
When dealing with complex numbers, the concept of a **complex conjugate** is pivotal. If you have a complex number written as \( z = a + bi \), the complex conjugate of \( z \) is expressed as \( \bar{z} = a - bi \). Here, \( a \) and \( b \) are real numbers, and \( i \) is the imaginary unit. The conjugate basically flips the sign of the imaginary part.
The reason the complex conjugate is useful, especially in expressions like \( \frac{1}{z} \), is because multiplying a complex number by its conjugate results in a real number. Specifically, \( z \cdot \bar{z} = (a + bi)(a - bi) = a^2 + b^2 \). This product eliminates the imaginary part, allowing us to simplify complex fractions easily. This elimination happens because the middle terms cancel each other out: \( -abi + abi = 0 \).
So, the next time you see a complex number, you can think of its conjugate as a handy tool for simplification and conversion into standard form.
The reason the complex conjugate is useful, especially in expressions like \( \frac{1}{z} \), is because multiplying a complex number by its conjugate results in a real number. Specifically, \( z \cdot \bar{z} = (a + bi)(a - bi) = a^2 + b^2 \). This product eliminates the imaginary part, allowing us to simplify complex fractions easily. This elimination happens because the middle terms cancel each other out: \( -abi + abi = 0 \).
So, the next time you see a complex number, you can think of its conjugate as a handy tool for simplification and conversion into standard form.
Reciprocal of a Complex Number
Finding the **reciprocal of a complex number** is a common operation in mathematics, and understanding it fully helps in mastering complex numbers. The reciprocal of a complex number \( z = a + bi \) is defined as \( \frac{1}{z} \). To simplify this expression, the key is to multiply both the numerator and the denominator by the complex conjugate of \( z \). This step transforms the expression into a more manageable form.
For instance, if we have \( \frac{1}{a + bi} \), multiplying both the numerator and the denominator by \( a - bi \) (the complex conjugate) transforms it into:
Therefore, the reciprocal of \( z \) is transformed into standard form making calculations and interpretations easier. Always remember, in standardized form, a complex number has no imaginary unit in the denominator.
For instance, if we have \( \frac{1}{a + bi} \), multiplying both the numerator and the denominator by \( a - bi \) (the complex conjugate) transforms it into:
- Numerator: \((a - bi)\)
- Denominator: \(a^2 + b^2\)
Therefore, the reciprocal of \( z \) is transformed into standard form making calculations and interpretations easier. Always remember, in standardized form, a complex number has no imaginary unit in the denominator.
Imaginary Unit
Let's delve into the **imaginary unit**, \( i \). This is a fundamental element in complex numbers. The imaginary unit is defined as \( i = \sqrt{-1} \), which means when squared, \( i^2 = -1 \). It is this property that gives rise to imaginary numbers when dealing with negative square roots.
In complex numbers, \( i \) represents the vertical component, hence every complex number can be thought of as having both a real part and an imaginary part. When you encounter expressions involving \( i \), like \( i^2 \), remember that it simplifies to \(-1\). This knowledge is crucial when expanding complex expressions, as it helps in reducing terms effectively.
Recognizing \( i \) simply as a specific feature of complex numbers allows for ease in handling operations involving square roots of negative numbers. It eliminates confusion and equips you with a structured way to tackle complex equations. Remember, without \( i \), the world of complex numbers wouldn't be as fascinating or robust as it is.
In complex numbers, \( i \) represents the vertical component, hence every complex number can be thought of as having both a real part and an imaginary part. When you encounter expressions involving \( i \), like \( i^2 \), remember that it simplifies to \(-1\). This knowledge is crucial when expanding complex expressions, as it helps in reducing terms effectively.
Recognizing \( i \) simply as a specific feature of complex numbers allows for ease in handling operations involving square roots of negative numbers. It eliminates confusion and equips you with a structured way to tackle complex equations. Remember, without \( i \), the world of complex numbers wouldn't be as fascinating or robust as it is.