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This may be a parser limitation...
>>> z = -0.-0.j
>>> z
(-0+0j)
>>> z.imag
0.0
>>> z = 0.-0.j
>>> z.imag
0.0

This is not a bug, but a known and difficult-to-avoid issue with signed zeros and creating complex numbers from real and imaginary parts. The safe way to do it is to use the `complex(real_part, imag_part)` construction.
In the first example, you're subtracting a complex number (0.j) from a float (-0.0). The float gets promoted to a complex (by filling in an imaginary part of +0.0), and the imaginary literal is similarly treated as a complex number (by filling in a 0.0 for the real part). So the subtraction is doing:
 complex(-0.0, 0.0) - complex(0.0, -0.0)
which as expected gives a real part of -0.0, and an imaginary part of +0.0.

Sorry; that should be:
 complex(-0.0, 0.0) - complex(0.0, 0.0)
The imaginary part is then worked out as (0.0 - 0.0), which (in all rounding modes but round-towards-negative) is 0.0.

P.S. The C standardisation folks tried to introduce the Imaginary type (optional for a conforming implementation) to get around exactly this type of problem, but it doesn't seem to have caught on in most mainstream compilers.

FTR, a similar issue with regular floats:
>>> -0.0-0.0
-0.0
>>> (-0.0)-0.0
-0.0
>>> z = -0.0
>>> z - z
0.0

Nope, no float issue here. :-) Which part of the output do you think is wrong?
Your last (z - z) is doing ((-0.0) - (-0.0)), not (-0.0 - 0.0). This is all following the IEEE 754 rules.

Yes, the last part is ok. It's the preceding parts that are a bit surprising.

Oops, I get it now. Sorry.

> Oops, I get it now.
Okay. :-)
Just for the record, for anyone encountering this issue in the future, here's the relevant extract from section 6.3 of IEEE 754-2008:
"""
When the sum of two operands with opposite signs (or the difference of two operands with like signs) is exactly zero, the sign of that sum (or difference) shall be +0 in all rounding-direction attributes except roundTowardNegative; under that attribute, the sign of an exact zero sum (or difference) shall be −0. However, x + x = x − (−x) retains the same sign as x even when x is zero.
"""
It doesn't cover the "sum with like signs" or "difference with opposite signs" cases, I suppose because it should be obvious what happens in those cases.

> It doesn't cover the "sum with like signs"
Ah, sorry, yes it does; it's in the previous paragraph.
"""
[...] the sign of a sum, or of a difference x−y regarded as a sum x+(−y), differs from at most one of the addends’ signs
"""