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!