> > Is that not the same as > np.abs(z)**2 ? > It is, but since that involves taking sqrt, it is *much* slower. Even now, ``` In [32]: r = np.arange(10000)*(1+1j)
In [33]: %timeit np.abs(r)**2 1000 loops, best of 3: 213 µs per loop In [34]: %timeit r.real**2 + r.imag**2 10000 loops, best of 3: 47.5 µs per loop ``` -- Marten
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