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On 09/11/2012 08:53 PM, Daniel J Sebald
wrote:
On
09/11/2012 07:15 PM, Jordi Gutiérrez Hermoso wrote:
On 11 September
2012 20:00, Daniel J Sebald<address@hidden>
wrote:
octave:59>
[1,-1,-0]'*i
ans =
0 + 1i
-0 - 1i
-0 - 0i
Is this proper behavior? I would think
There is no actual "i" in Octave. Instead, "i" is a function
that
returns the equivalent of std::complex<double>(0.0, 1.0).
You can see
this behaviour in other ways, e.g. inf*i giving NaNs.
Oops, I take it back. Was
thinking of integers.
Forget what I said.
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