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Magazine Chapter 10: Problem 116
A function \(g\) is given by $$ g(z)=\frac{z^{4}-z^{2}}{z-1} $$ Find \(g(2-3 i)\)
Short Answer
Expert verified
g(2-3i) = -51 - 21i
Step by step solution
01
- Substitute the given value into the function
Substitute the value \(z = 2 - 3i\) into the function \(g(z)\). So, we need to calculate \[g(2-3i) = \frac{(2-3i)^4 - (2-3i)^2}{2-3i - 1}\]
02
- Calculate \( (2-3i)^2 \)
First, calculate \[(2-3i)^2 = (2-3i)(2-3i) = 4 - 6i - 6i + 9i^2 = 4 - 12i + 9(-1) = 4 - 12i - 9 = -5 - 12i\]
03
- Calculate \( (2-3i)^4 \)
Square the result from step 2: \[((2-3i)^2)^2 = (-5 - 12i)^2 = (-5 - 12i)(-5 - 12i) = 25 + 60i + 60i + 144i^2 = 25 + 120i + 144(-1) = 25 + 120i - 144 = -119 + 120i\]
04
- Simplify the numerator
Use the results from steps 2 and 3 to find the numerator of \( g(2-3i) \): \[ (-119 + 120i) - (-5 - 12i) = -119 + 120i + 5 + 12i = -114 + 132i\]
05
- Simplify the denominator
Simplify the denominator: \[2-3i - 1 = 1 - 3i\]
06
- Combine the numerator and denominator
So the function becomes: \[ g(2-3i) = \frac{-114 + 132i}{1 - 3i} \]
07
- Multiply by the conjugate
To simplify, multiply both the numerator and the denominator by the conjugate of the denominator \(1 + 3i\):\[g(2-3i) = \frac{(-114 + 132i)(1 + 3i)}{(1 - 3i)(1 + 3i)} \]
08
- Simplify the denominator
The denominator simplifies using the difference of squares: \[ (1 - 3i)(1 + 3i) = 1 - (3i)^2 = 1 + 9 = 10\]
09
- Simplify the numerator
Expand the numerator using the distributive property: \[(-114 + 132i)(1 + 3i) = -114 - 342i + 132i + 396i^2 = -114 - 210i + 396(-1) = -114 - 210i - 396 = -510 - 210i\]
10
- Divide
Now divide both parts of the numerator by the denominator (10): \[ g(2-3i) = \frac{-510 - 210i}{10} = -51 - 21i\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Complex Numbers
Complex numbers are numbers that have both a real part and an imaginary part. The general form of a complex number is given by \(a + bi\) where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit with the property that \(i^2 = -1\). In this exercise, we work with the complex number \(2 - 3i\). To handle operations with complex numbers:
- Add or subtract their real and imaginary parts separately.
- To multiply, use the distributive property and replace \(i^2\) with \(-1\).
Polynomial Functions
Polynomial functions involve expressions of the form \(a_nz^n + a_{n-1}z^{n-1} + ... + a_1z + a_0\), where \(a_n, a_{n-1}, ..., a_1, a_0\) are constants. Here, our function, \(g(z) = \frac{z^4 - z^2}{z-1}\), involves polynomials in the numerator and denominator. When working with polynomial functions:
- To simplify, factorize the polynomial as much as possible.
- Substitute complex numbers and simplify step-by-step.
Rational Functions
Rational functions are ratios of polynomial functions, such as \(g(z) = \frac{z^4 - z^2}{z-1}\). Simplifying rational functions often involves:
- Performing polynomial long division or factoring if needed.
- Checking for undefined points where the denominator equals zero.
- Using complex number properties for operations, such as multiplying by the conjugate when necessary.
Simplification
Simplification helps to make expressions more manageable. It often involves:
- Combining like terms.
- Applying algebraic identities.
- Performing arithmetic operations carefully, especially with complex numbers.
Conjugates
A conjugate of a complex number \(a + bi\) is \(a - bi\). Conjugates are useful in simplifying division of complex numbers. When we multiply a complex number by its conjugate:
\[ (a + bi)(a - bi) = a^2 - (bi)^2 = a^2 + b^2 \]
We used this idea in our exercise to simplify the denominator. By multiplying both the numerator and the denominator by the conjugate \((1 + 3i)\), we were able to remove the imaginary part from the denominator, making it easier to divide and finalize the result.
\[ (a + bi)(a - bi) = a^2 - (bi)^2 = a^2 + b^2 \]
We used this idea in our exercise to simplify the denominator. By multiplying both the numerator and the denominator by the conjugate \((1 + 3i)\), we were able to remove the imaginary part from the denominator, making it easier to divide and finalize the result.
