3.43.Using the definition, find the derivative of each function at the indicated points,
 f(z)= 3z^-2 ; z=1+i

To find the derivative of the function \( f(z) = 3z^{-2} \) at the point \( z = 1 + i \), we will use the definition of the derivative in complex analysis, which is analogous to the definition in real calculus.

### Definition of the Derivative:
The derivative of a complex function \( f(z) \) at a point \( z_0 \) is given by:

\[
f'(z_0) = \lim_{\Delta z \to 0} \frac{f(z_0 + \Delta z) - f(z_0)}{\Delta z}
\]

Where \( \Delta z \) is a small complex number.

### Step-by-Step Solution:

1. **Function Definition:**
   \[
   f(z) = 3z^{-2} = \frac{3}{z^2}
   \]

2. **Apply the definition at \( z = 1 + i \):**
   We want to compute:

   \[
   f'(1 + i) = \lim_{\Delta z \to 0} \frac{f((1 + i) + \Delta z) - f(1 + i)}{\Delta z}
   \]

3. **Find \( f(1 + i) \):**
   First, calculate \( f(1 + i) \). Since \( f(z) = \frac{3}{z^2} \), we need to compute \( (1 + i)^2 \):

   \[
   (1 + i)^2 = 1^2 + 2(1)(i) + i^2 = 1 + 2i - 1 = 2i
   \]
   
   So,
   \[
   f(1 + i) = \frac{3}{(1 + i)^2} = \frac{3}{2i}
   \]
   
   Now simplify \( \frac{3}{2i} \) by multiplying both numerator and denominator by \( -i \) (to eliminate the imaginary number from the denominator):

   \[
   \frac{3}{2i} = \frac{3(-i)}{2i(-i)} = \frac{-3i}{-2} = \frac{3i}{2}
   \]

   Thus,
   \[
   f(1 + i) = \frac{3i}{2}
   \]

4. **Find the derivative:**
   Using the power rule for derivatives in complex analysis, we can differentiate \( f(z) = 3z^{-2} \) directly:

   \[
   f'(z) = -6z^{-3}
   \]

5. **Evaluate at \( z = 1 + i \):**
   Now we need to compute \( f'(1 + i) = -6(1 + i)^{-3} \). First, we need to find \( (1 + i)^3 \).

   \[
   (1 + i)^3 = (1 + i)(1 + i)^2 = (1 + i)(2i)
   \]

   Multiply \( (1 + i) \) by \( 2i \):

   \[
   (1 + i)(2i) = 1 \cdot 2i + i \cdot 2i = 2i - 2 = -2 + 2i
   \]

   Thus,
   \[
   (1 + i)^3 = -2 + 2i
   \]

6. **Substitute into the derivative formula:**
   Now, substitute \( (1 + i)^3 = -2 + 2i \) into the expression for the derivative:

   \[
   f'(1 + i) = -6 \cdot \frac{1}{-2 + 2i}
   \]

7. **Simplify:**
   To simplify \( \frac{1}{-2 + 2i} \), multiply both numerator and denominator by \( -2 - 2i \):

   \[
   \frac{1}{-2 + 2i} = \frac{-2 - 2i}{(-2 + 2i)(-2 - 2i)} = \frac{-2 - 2i}{4 + 4} = \frac{-2 - 2i}{8} = -\frac{1}{4} - \frac{i}{4}
   \]

   Therefore,
   \[
   f'(1 + i) = -6 \cdot \left( -\frac{1}{4} - \frac{i}{4} \right) = \frac{6}{4} + \frac{6i}{4} = \frac{3}{2} + \frac{3i}{2}
   \]

### Final Answer:
\[
f'(1 + i) = \frac{3}{2} + \frac{3i}{2}
\]

Would you like further clarification or details?

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Here are 5 related questions to further your understanding:

1. How do we calculate the derivative of a general power function \( f(z) = z^n \)?
2. Can we use the Cauchy-Riemann equations to verify differentiability of complex functions like \( f(z) = 3z^{-2} \)?
3. What are the key differences between real and complex differentiation?
4. How do we find the modulus of a complex number like \( z = 1 + i \)?
5. What is the geometric interpretation of complex derivatives?

**Tip:** In complex analysis, powers and reciprocals of complex numbers are often simplified by converting them into polar form!

3.43. Using the definition, find the derivative of each function at the indicated points, f(z)= 3z^-2 ; z=1+i

Find the derivative of the complex function f(z) = 3z^-2 at the point z = 1 + i using the definition of the derivative in complex analysis. This solution involves key concepts in complex numbers, power rule, and complex function derivatives, ideal for undergraduate students.

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