Problem 19
Question
In Problems 19-22, the given limit exists. Find its value. $$ \lim _{z \rightarrow i}\left(4 z^{3}-5 z^{2}+4 z+1-5 i\right) $$
Step-by-Step Solution
Verified Answer
The limit is \(6 - 5i\).
1Step 1: Understand the Limit
The problem requires us to find the value of the limit as the complex number \(z\) approaches \(i\). We need to substitute \(z=i\) into the polynomial expression \(4z^{3}-5z^{2}+4z+1-5i\).
2Step 2: Substitute \(z=i\) into the Expression
Replace every occurrence of \(z\) in the expression \(4z^{3}-5z^{2}+4z+1-5i\) with the value \(i\). This results in \(4(i)^{3} - 5(i)^{2} + 4(i) + 1 - 5i\).
3Step 3: Calculate \( i^2 \) and \( i^3 \)
Recall that \(i\), the imaginary unit, satisfies \(i^2 = -1\). Therefore, \(i^3 = i \cdot i^2 = i(-1) = -i\).
4Step 4: Simplify the Expression
Substitute \(i^2 = -1\) and \(i^3 = -i\) into the expression:\[4(-i) - 5(-1) + 4i + 1 - 5i = -4i + 5 + 4i + 1 - 5i\].
5Step 5: Combine Like Terms
Combine the real and imaginary parts:\(-4i + 4i - 5i = -5i\) and \(5 + 1 = 6\). Thus, the expression simplifies to \(6 - 5i\).
6Step 6: Conclusion
The value of the limit \(\lim _{z \rightarrow i}\left(4z^{3} - 5z^{2} + 4z + 1 - 5i\right)\) is \(6 - 5i\).
Key Concepts
Complex Limit EvaluationPolynomials in Complex AnalysisImaginary Unit Properties
Complex Limit Evaluation
The concept of limits in complex analysis is quite similar to that in real analysis. However, since complex numbers have both real and imaginary parts, evaluating complex limits adds some complexity.
Imagine substituting a number into a complex function to determine what value the function approaches as its complex variable zs approaches a certain point. In our example, we are looking at how the function behaves as z approaches the complex number \( i \).
Understanding the Approach
The key is to substitute \( z = i \) into the given expression. This allows us to see what value the function comes close to. Remember, \( i \) is the imaginary unit, which is crucial here because it alters the terms significantly.
Especially with functions involving polynomials, each term will be affected depending on its power of z.
Simplification of Expressions
Note how when we substitute \( z = i \), repeatedly evaluating imaginary powers quickly leads toward straightforward calculations. The power rules for i (like \( i^2 = -1 \)) help in reducing complexity via simple multipliers. Resulting in an expression that often consolidates into simpler forms.
Imagine substituting a number into a complex function to determine what value the function approaches as its complex variable zs approaches a certain point. In our example, we are looking at how the function behaves as z approaches the complex number \( i \).
Understanding the Approach
The key is to substitute \( z = i \) into the given expression. This allows us to see what value the function comes close to. Remember, \( i \) is the imaginary unit, which is crucial here because it alters the terms significantly.
Especially with functions involving polynomials, each term will be affected depending on its power of z.
Simplification of Expressions
Note how when we substitute \( z = i \), repeatedly evaluating imaginary powers quickly leads toward straightforward calculations. The power rules for i (like \( i^2 = -1 \)) help in reducing complexity via simple multipliers. Resulting in an expression that often consolidates into simpler forms.
Polynomials in Complex Analysis
Polynomials are the building blocks in both real and complex analysis. The main difference in complex analysis comes from handling terms where \( z \) is a complex number.
When each term has complex coefficients, calculations involve both real and imaginary arithmetic as seen in our function.
Structure of Complex Polynomials
When evaluating a polynomial like \( 4z^{3}-5z^{2}+4z+1-5i \), you manage each term individually. Powers of z must be calculated, then substituted back into the structure.
A complex polynomial has a form \( a_n z^n + a_{n-1} z^{n-1} + \ldots + a_0 \), where each \( a_n \) can be a complex number.
Often, seeing how each term influences the overall polynomial's behavior makes it easier to grasp the nuances of complex functions.
The Role of Simplification
In the end, once powers of i are simplified, combining like terms leads to a compact and simple expression representing the polynomial's behavior as \( z \) approaches a particular value.
Understanding this process makes finding limits cleaner and more manageable.
When each term has complex coefficients, calculations involve both real and imaginary arithmetic as seen in our function.
Structure of Complex Polynomials
When evaluating a polynomial like \( 4z^{3}-5z^{2}+4z+1-5i \), you manage each term individually. Powers of z must be calculated, then substituted back into the structure.
A complex polynomial has a form \( a_n z^n + a_{n-1} z^{n-1} + \ldots + a_0 \), where each \( a_n \) can be a complex number.
Often, seeing how each term influences the overall polynomial's behavior makes it easier to grasp the nuances of complex functions.
The Role of Simplification
In the end, once powers of i are simplified, combining like terms leads to a compact and simple expression representing the polynomial's behavior as \( z \) approaches a particular value.
Understanding this process makes finding limits cleaner and more manageable.
Imaginary Unit Properties
The imaginary unit, \( i \), is a fundamental concept in complex numbers and is defined as \( i^2 = -1 \). From this definition, the powers of i create a pattern:
Using Imaginary Unit Properties
For our exercise, the powers of \( i \) enable re-writing higher powers quickly. Calculating \( i^3 \) becomes straightforward as it simply resolves to \( -i \) again. This periodicity makes algebraic manipulations simpler.
Impact on Complex Polynomials
In complex polynomials, substitution of i simplifies coefficients. For example, polynomials with terms like \( -5i \) are directly influenced by this cycle, decreasing the complexity of individual calculations.
By consistently integrating properties of \( i \), conversion of polynomial terms to their simplest components prepares us to clearly understand the actual outcomes of expressions around specific values.
- \( i^1 = i \)
- \( i^2 = -1 \)
- \( i^3 = -i \)
- \( i^4 = 1 \)
- And it repeats cyclically.
Using Imaginary Unit Properties
For our exercise, the powers of \( i \) enable re-writing higher powers quickly. Calculating \( i^3 \) becomes straightforward as it simply resolves to \( -i \) again. This periodicity makes algebraic manipulations simpler.
Impact on Complex Polynomials
In complex polynomials, substitution of i simplifies coefficients. For example, polynomials with terms like \( -5i \) are directly influenced by this cycle, decreasing the complexity of individual calculations.
By consistently integrating properties of \( i \), conversion of polynomial terms to their simplest components prepares us to clearly understand the actual outcomes of expressions around specific values.
Other exercises in this chapter
Problem 19
Find all values of \(z\) satisfying the given equation. \(\cos z=\sin z\)
View solution Problem 19
Verify the given result. $$ e^{z+\pi i}=e^{z-\pi i} $$
View solution Problem 19
In Problems 17-20, write the given number in the form \(a+i b\). $$ \frac{(1-i)^{10}}{(1+i)^{3}} $$
View solution Problem 19
In Problems \(15-20\), find all values of \(z\) satisfying the given equation. $$ \cos z=\sin z $$
View solution