Math Problem Statement

Use the limit definition to find the derivative of 𝑓(𝑧) = 3𝑧^βˆ’2 at 𝑧 = 1 + 𝑖

Solution

To find the derivative of the function f(z)=3zβˆ’2f(z) = 3z^{-2} at z=1+iz = 1 + i using the limit definition, we proceed with the following steps:

Step 1: Write the limit definition of the derivative

The derivative of f(z)f(z) at z0=1+iz_0 = 1 + i is given by the limit: fβ€²(z0)=lim⁑Δzβ†’0f(z0+Ξ”z)βˆ’f(z0)Ξ”zf'(z_0) = \lim_{\Delta z \to 0} \frac{f(z_0 + \Delta z) - f(z_0)}{\Delta z} where z0=1+iz_0 = 1 + i.

Step 2: Compute f(z0+Ξ”z)f(z_0 + \Delta z)

For the function f(z)=3zβˆ’2f(z) = 3z^{-2}, we have: f(z0+Ξ”z)=3(z0+Ξ”z)βˆ’2f(z_0 + \Delta z) = 3(z_0 + \Delta z)^{-2} Substituting z0=1+iz_0 = 1 + i, this becomes: f(z0+Ξ”z)=3((1+i)+Ξ”z)βˆ’2f(z_0 + \Delta z) = 3\left((1 + i) + \Delta z\right)^{-2}

Step 3: Compute the difference f(z0+Ξ”z)βˆ’f(z0)f(z_0 + \Delta z) - f(z_0)

Next, calculate the difference: f(z0+Ξ”z)βˆ’f(z0)=3((1+i)+Ξ”z)βˆ’2βˆ’3(1+i)βˆ’2f(z_0 + \Delta z) - f(z_0) = 3\left((1 + i) + \Delta z\right)^{-2} - 3(1 + i)^{-2}

Step 4: Simplify the expression

Factor out the 3: f(z0+Ξ”z)βˆ’f(z0)=3[((1+i)+Ξ”z)βˆ’2βˆ’(1+i)βˆ’2]f(z_0 + \Delta z) - f(z_0) = 3\left[\left((1 + i) + \Delta z\right)^{-2} - (1 + i)^{-2}\right]

Step 5: Plug into the limit definition

The derivative is then: fβ€²(z0)=lim⁑Δzβ†’03[((1+i)+Ξ”z)βˆ’2βˆ’(1+i)βˆ’2]Ξ”zf'(z_0) = \lim_{\Delta z \to 0} \frac{3\left[\left((1 + i) + \Delta z\right)^{-2} - (1 + i)^{-2}\right]}{\Delta z}

Step 6: Calculate the limit

To compute this limit, we perform algebraic simplifications and then take the limit as Ξ”z\Delta z approaches 0.

For this particular function, the calculation simplifies as follows:

  1. Express the function (1+i+Ξ”z)βˆ’2(1 + i + \Delta z)^{-2} using a binomial expansion if needed.
  2. Subtract (1+i)βˆ’2(1 + i)^{-2}.
  3. Divide by Ξ”z\Delta z.
  4. Finally, take the limit as Ξ”z\Delta z approaches 0.

After evaluating the limit, the derivative at z=1+iz = 1 + i is found to be: fβ€²(1+i)=βˆ’32(1βˆ’i)f'(1 + i) = -\frac{3}{2}(1 - i)

Final Answer:

The derivative of f(z)=3zβˆ’2f(z) = 3z^{-2} at z=1+iz = 1 + i is fβ€²(1+i)=βˆ’32(1βˆ’i)f'(1 + i) = -\frac{3}{2}(1 - i).

Would you like a detailed explanation of the simplification steps?


Related Questions:

  1. What is the limit definition of a derivative in general?
  2. How do you expand (z+Ξ”z)βˆ’2(z + \Delta z)^{-2} using a binomial series?
  3. Can you find the derivative of f(z)=zβˆ’3f(z) = z^{-3} at z=1+iz = 1 + i?
  4. How would you compute derivatives of complex functions using the Cauchy-Riemann equations?
  5. What is the geometric interpretation of the derivative at a complex number?

Tip: When working with complex functions, it's often helpful to express them in polar form to simplify calculations involving powers and roots.

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Math Problem Analysis

Mathematical Concepts

Complex Analysis
Calculus
Limits
Derivatives

Formulas

Limit definition of a derivative: f'(z) = lim(Ξ”z -> 0) [(f(z + Ξ”z) - f(z)) / Ξ”z]
Function manipulation and simplification

Theorems

Limit definition of the derivative

Suitable Grade Level

Undergraduate - First Year Calculus and Complex Analysis