Let's solve the problem **step-by-step** using the **derivative form of the Cauchy integral formula**. --- ## **Step 1: Recall the Derivative Form of Cauchy's Integral Formula** For a function \( f(z) \) analytic inside and on a simple closed curve \( C \), and \( a \) inside \( C \): \[ f^{(n)}(a) = \frac{n!}{2\pi i} \int_C \frac{f(z)}{(z-a)^{n+1}}\,dz \] --- ## **Step 2: Match the Given Integral** Given: \[ \frac{1}{2\pi i} \int_{|z|=3} \frac{e^z}{(z-1)^2} dz \] Compare with the formula: - \( f(z) = e^z \) - \( a = 1 \) - \( n = 1 \) (because \( (z-a)^{n+1} = (z-1)^2 \)) --- ## **Step 3: Apply the Formula** So: \[ \frac{1}{2\pi i} \int_{|z|=3} \frac{e^z}{(z-1)^2} dz = f'(1) \] where \( f(z) = e^z \). --- ## **Step 4: Compute the Derivative \( f'(z) \) and Evaluate at \( z = 1 \)** \[ f'(z) = \frac{d}{dz}e^z = e^z \] \[ f'(1) = e^1 = e \] --- ## **Step 5: Final Answer** \[ \boxed{e} \] --- ### **Summary Table** | Step | Action | Result | |------|--------|----------------| | 1 | Identify formula | Cauchy integral derivative form | | 2 | Match terms | \( f(z) = e^z,\, a = 1,\, n = 1 \) | | 3 | Apply formula | Equals \( f'(1) \) | | 4 | Differentiate and substitute | \( e \) | | 5 | Final answer | \( \boxed{e} \) | --- **So, the value of the integral is \( \boxed{e} \).**