Complex double Array converts to double array after assignment

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Vids deo
Vids deo 2016-9-27
评论： Walter Roberson 2016-9-27
Hi Experts,
a = [1 2; 3 4];
b = [1 2; 3 4];
c = complex(a,b);
a1 = complex(a,0);
a1(1,1) = complex(abs(c(2,2)),0);
Above assignment somehow converts 'a1' to a DOUBLE array instead of COMPLEX double. I'm using the above method for assignment to a 3D array (S-parameter processing) which is facing the same issue in a code. Why would this be happening ??

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Walter Roberson
Walter Roberson 2016-9-27
complex arrays whose real parts are all completely 0 are fragile; as soon as you do anything to them, the all-zero imaginary part will fall off.
Consider that if you have
a = [1+2i, 3+0i]
a(imag(a)>0) = 0
then this would be like a(1) = 0, giving you a matrix [0, 3+0i]. People now expect that to degrade to [0, 3] rather than retaining its complex character just because it was formerly complex. Therefore, the complex parts are checked for all-zero after every assignment.
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Walter Roberson
Walter Roberson 2016-9-27
This is the way it is. It is considered a design feature.
If you want, you can use
if isreal(a1); a1 = complex(a1); end
But remember that if you then do a1+0 or a1*1 or anything with a1 other than assign all of it to another variable or pass it alone to a routine, then the complex part will promptly fall off again.

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Massimo Zanetti
Massimo Zanetti 2016-9-27
编辑：Massimo Zanetti 2016-9-27
Sure it does, because a complex number with a NULL imaginary part is indeed real. Notice that in the call COMPLEX(a,0), you in fact sets the imaginary part to 0. If you want to create a complex number without imaginary part do this:
a1=complex(a);
This will give a1 as COMPLEX.
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Vids deo
Vids deo 2016-9-27
Hey Massimo,
The statement that you said only works with using complex(a) on complete matrix. It doesn't work the same way when used "element-wise" That is -
a1(1,1)=complex(a(2,1));
My operations are element-wise. And its screwing up the whole matrix to a real double (instead of complex double).
--Vidya

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