To solve for \( f(f(f(f(1+i)))) \), we will work step by step through the function \( f \).
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First, calculate \( f(1 + i) \): Since \( 1 + i \) is not a real number, we use the first part of the function definition: \[ f(1 + i) = (1 + i)^2. \] Now, compute \( (1 + i)^2 \): \[ (1 + i)^2 = 1^2 + 2 \cdot 1 \cdot i + i^2 = 1 + 2i - 1 = 2i. \] So, we have: \[ f(1 + i) = 2i. \]
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Next, calculate \( f(f(1+i)) = f(2i) \): Since \( 2i \) is not a real number, again we use the first part of the function definition: \[ f(2i) = (2i)^2. \] Now calculate \( (2i)^2 \): \[ (2i)^2 = 4i^2 = 4(-1) = -4. \] Thus, we find: \[ f(2i) = -4. \]
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Next, calculate \( f(f(f(1+i))) = f(-4) \): Since \( -4 \) is a real number, we use the second part of the function definition: \[ f(-4) = -(-4)^2. \] Now calculate \( -(-4)^2 \): \[ -(-4)^2 = -16. \] Therefore: \[ f(-4) = -16. \]
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Finally, calculate \( f(f(f(f(1+i)))) = f(-16) \): Since \( -16 \) is also a real number, we apply the second part of the function definition once again: \[ f(-16) = -(-16)^2. \] Now compute \( -(-16)^2 \): \[ -(-16)^2 = -256. \] Thus, we conclude: \[ f(-16) = -256. \]
Putting it all together, we find: \[ f(f(f(f(1+i)))) = -256. \]
The final answer is: \[ \boxed{-256}. \]