Question

For all complex numbers z, let f(z) = z^2 if z is not real, and f(z) = -z^2 if z is real. Find f(f(f(f(1+i)))).

Answer 1

To solve for \( f(f(f(f(1+i)))) \), we will work step by step through the function \( f \).

1. **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:
\[
f(1 + i) = 2i.
\]

2. **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:
\[
f(2i) = -4.
\]

3. **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:
\[
f(-4) = -16.
\]

4. **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:
\[
\boxed{-256}.
\]