Problem 11
Question
Show that the given sequence \(\left\\{z_{n}\right\\}\) converges to a complex number \(L\) by computing \(\lim _{n \rightarrow \infty} \operatorname{Re}\left(z_{n}\right)\) and \(\lim _{n \rightarrow \infty} \operatorname{Im}\left(z_{n}\right)\). \(\left\\{\frac{4 n+3 n i}{2 n+i}\right\\}\)
Step-by-Step Solution
Verified Answer
The sequence converges to the complex number \(2 + \frac{3}{2}i\).
1Step 1: Separate Real and Imaginary Parts
Begin by dividing both the numerator and denominator of the sequence \(z_n = \frac{4n + 3ni}{2n + i}\) by \(n\) to simplify the expression, by separating real and imaginary parts. This gives:\[ z_n = \frac{4 + 3i}{2 + \frac{i}{n}}. \]
2Step 2: Compute the Real Part Limit
To find \(\lim_{n \to \infty} \operatorname{Re}(z_n)\), we first expand the expression:\[ z_n = \frac{(4 + 3i)(2 - \frac{i}{n})}{(2 + \frac{i}{n})(2 - \frac{i}{n})} = \frac{(4 + 3i)(2 - \frac{i}{n})}{4 + \frac{1}{n^2}}. \]Simplify the numerator:\[ (4 + 3i)(2 - \frac{i}{n}) = 8 + \frac{4i}{n} + 6i - \frac{3}{n} = 8 + (\frac{4}{n} + 6)i - \frac{3}{n}. \]Thus, the real part is \(\operatorname{Re}(z_n) = \frac{8 - \frac{3}{n}}{4 + \frac{1}{n^2}}\).Taking the limit as \(n \to \infty\), we get:\[ \lim_{n \to \infty} \operatorname{Re}(z_n) = \frac{8}{4} = 2. \]
3Step 3: Compute the Imaginary Part Limit
Now find \(\lim_{n \to \infty} \operatorname{Im}(z_n)\). Use the imaginary simplification found earlier:\[ \operatorname{Im}(z_n) = \frac{\frac{4}{n} + 6}{4 + \frac{1}{n^2}}. \]Taking the limit as \(n \to \infty\), the terms involving \(\frac{1}{n}\) disappear:\[ \lim_{n \to \infty} \operatorname{Im}(z_n) = \frac{6}{4} = \frac{3}{2}. \]
4Step 4: Conclude the Limit of the Sequence
Since we now have both limits, the sequence \(z_n\) converges to the complex number \(L = 2 + \frac{3}{2}i\). This conclusion comes from the real and imaginary limits we computed.
Key Concepts
Complex NumbersReal and Imaginary PartsLimits of SequencesSequence Convergence
Complex Numbers
A complex number is a fundamental concept in mathematics, particularly in complex analysis. It is expressed in the form \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit. The imaginary unit \(i\) is defined by the property \(i^2 = -1\).
Complex numbers provide a way to solve equations that have no real number solutions. They are essential in various fields such as engineering, physics, and applied mathematics. Complex numbers also enable us to explore more advanced mathematical concepts like complex analysis, which involves studying functions that operate in the complex plane.
Complex numbers provide a way to solve equations that have no real number solutions. They are essential in various fields such as engineering, physics, and applied mathematics. Complex numbers also enable us to explore more advanced mathematical concepts like complex analysis, which involves studying functions that operate in the complex plane.
- The real part of a complex number is \(a\).
- The imaginary part is \(b\).
- Complex numbers can be represented graphically in the complex plane, where the x-axis represents the real part and the y-axis represents the imaginary part.
Real and Imaginary Parts
In the given sequence \(\left\{z_{n}\right\}\), each term is a complex number comprised of real and imaginary components. By expressing a complex number as \(z = x + yi\), the real and imaginary parts can be distinguished:
- \(\operatorname{Re}(z_n)\) denotes the real part of the sequence terms.
- \(\operatorname{Im}(z_n)\) stands for the imaginary part of the terms.
Limits of Sequences
Limits are a critical concept in calculus and analysis, representing the value that a sequence or function approaches as the index or input grows indefinitely. Computing limits of sequences involving complex numbers requires careful handling of both real and imaginary components.
For the sequence \(\{z_n\}\), the goal is to determine the limits of both \(\operatorname{Re}(z_n)\) and \(\operatorname{Im}(z_n)\) separately. This requires simplifying the sequence, often starting by dividing both the real and imaginary components by \(n\), which helps in finding behavior as \(n\) becomes large.
The sequence convergence is confirmed if both the real and imaginary parts approach definite numbers.
For the sequence \(\{z_n\}\), the goal is to determine the limits of both \(\operatorname{Re}(z_n)\) and \(\operatorname{Im}(z_n)\) separately. This requires simplifying the sequence, often starting by dividing both the real and imaginary components by \(n\), which helps in finding behavior as \(n\) becomes large.
The sequence convergence is confirmed if both the real and imaginary parts approach definite numbers.
Sequence Convergence
Sequence convergence, in the context of complex numbers, implies that the values of the sequence approach a specific complex number as the index \(n\) increases. For sequence \(\{z_n\}\), it converges to \(L = 2 + \frac{3}{2}i\) because both its real part, \(\operatorname{Re}(z_n)\), and imaginary part, \(\operatorname{Im}(z_n)\), reach respective limits:
- \(\lim_{n \to \infty} \operatorname{Re}(z_n) = 2\)
- \(\lim_{n \to \infty} \operatorname{Im}(z_n) = \frac{3}{2}\)
Other exercises in this chapter
Problem 10
In Problems 5-10, determine whether the given sequence converges or diverges. $$ \left\\{e^{1 / n}+2\left(\tan ^{-1} n\right) i\right\\} $$
View solution Problem 11
Expand the given function in a Maclaurin series. Give the radius of convergence of each series. \(f(z)=\sin z^{2}\)
View solution Problem 11
Evaluate the Cauchy principal value of the given improper integral. \(\int_{-\infty}^{\infty} \frac{1}{x^{2}-2 x+2} d x\)
View solution Problem 11
The indicated number is a zero of the given function. Use a Maclaurin or Taylor series to determine the order of the zero. \(f(z)=1-e^{z-1} ; z=1\)
View solution