Math Problem Statement

Определить тип особой точки z = 0 для данной функции

Определить тип особой точки z = 0 для данной функции f(z) = (e^(z^3) - 1 - z^3) * cot(2z^3)

Solution

To determine the type of singularity at z=0z = 0 for the given function

f(z)=(ez31z3)cot(2z3),f(z) = \left(e^{z^3} - 1 - z^3 \right) \cdot \cot(2z^3),

we analyze the behavior of the function around z=0z = 0:

Step-by-step Analysis:

  1. Expand ez3e^{z^3} in a Taylor series around z=0z = 0: ez3=1+z3+(z3)22!+(z3)33!+e^{z^3} = 1 + z^3 + \frac{(z^3)^2}{2!} + \frac{(z^3)^3}{3!} + \cdots Therefore: ez31z3=(z3)22!+(z3)33!+=z62+z96+.e^{z^3} - 1 - z^3 = \frac{(z^3)^2}{2!} + \frac{(z^3)^3}{3!} + \cdots = \frac{z^6}{2} + \frac{z^9}{6} + \cdots.

  2. Simplify the numerator of f(z)f(z): The numerator of f(z)f(z) becomes: (ez31z3)=z62+z96+.\left(e^{z^3} - 1 - z^3\right) = \frac{z^6}{2} + \frac{z^9}{6} + \cdots.

  3. Expand cot(2z3)\cot(2z^3): Recall the series expansion of cot(x)\cot(x): cot(x)=1xx3x345+.\cot(x) = \frac{1}{x} - \frac{x}{3} - \frac{x^3}{45} + \cdots. Substituting x=2z3x = 2z^3: cot(2z3)=12z32z33(2z3)345+.\cot(2z^3) = \frac{1}{2z^3} - \frac{2z^3}{3} - \frac{(2z^3)^3}{45} + \cdots.

  4. Behavior of f(z)f(z) near z=0z = 0: Combining the two terms, the function becomes: f(z)=(z62+z96+)(12z32z33(2z3)345+).f(z) = \left(\frac{z^6}{2} + \frac{z^9}{6} + \cdots \right) \cdot \left(\frac{1}{2z^3} - \frac{2z^3}{3} - \frac{(2z^3)^3}{45} + \cdots \right). Multiply the terms:

    • The leading term in the product is: z6212z3=z34.\frac{z^6}{2} \cdot \frac{1}{2z^3} = \frac{z^3}{4}.
    • Higher-order terms are of the form z6,z9,z^6, z^9, etc.

    Hence, near z=0z = 0, f(z)z34f(z) \sim \frac{z^3}{4}, meaning z=0z = 0 is a removable singularity because the function is analytic there after division by z3z^3.

Conclusion:

The singularity at z=0z = 0 for the given function is a removable singularity.


Would you like further clarification or additional examples?

Related Questions:

  1. What are the different types of singularities in complex analysis?
  2. How can Taylor expansions help analyze singularities?
  3. How does the behavior of cot(z)\cot(z) near z=0z = 0 affect singularities?
  4. Could this function have other singularities besides z=0z = 0?
  5. How do Laurent series differ from Taylor series in classifying singularities?

Tip:

When analyzing singularities, always focus on the leading-order term of the Laurent series to determine the type of singularity.

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Math Problem Analysis

Mathematical Concepts

Complex Analysis
Singularities
Taylor Series
Laurent Series

Formulas

e^x = 1 + x + x^2/2! + x^3/3! + ...
cot(x) = 1/x - x/3 - x^3/45 + ...
Laurent series expansion to analyze singularities

Theorems

Definition of removable singularities
Taylor series expansion theorem
Properties of cotangent function near zero

Suitable Grade Level

Undergraduate Mathematics