Find the indices of the imaginary element of the matrix

Suppose I have a Matrix like this:

A=[1    2    3;
   4+i  5    6-i;
   7    8+i  9]

Now, I want to know if there is a way to find the indecies of the imaginary elements (by this I mean the complex numbers which have imaginary part). I am looking for an answer like that:

A(2,1)

A(2,3)

A(3,2)

or any other form the conveys the indecies information of the imaginary elements.

Thank you so much.

Answer 1

[David Hill]

Lidx=find(imag(A)~=0);

[David Hill]

If your application will allow, I always perfer linear indexing.

[Hassan Alkomy]

Thank you so much.

Answer 2

[Joseph Cheng]

you can do a comparison to the real(A) like

A=[1    2    3;
   4+i  5    6-i;
   7    8+i  9]

A = 
   1.0000 + 0.0000i   2.0000 + 0.0000i   3.0000 + 0.0000i
   4.0000 + 1.0000i   5.0000 + 0.0000i   6.0000 - 1.0000i
   7.0000 + 0.0000i   8.0000 + 1.0000i   9.0000 + 0.0000i

[row col]=find(A~=real(A))

row = 3×1

2 3 2

col = 3×1

1 2 3

then with find you can get the row and column index values of the as real(A)~=A

[Hassan Alkomy]

Thank you so much.

Answer 3

[Scott MacKenzie]

% test matrix
A=[1    2    3;
   4+i  5    6-i;
   7    8+i  9]

% identify imaginary elements in A (+1 or -1)
B = imag(A)

% generate column vector of indices of imaginary elements in A
find(B~=0)

Output:

A =

1 + 0i 2 + 0i 3 + 0i

4 + 1i 5 + 0i 6 - 1i

7 + 0i 8 + 1i 9 + 0i

B =

0 0 0

1 0 -1

0 1 0

ans =

2

6

8

[Hassan Alkomy]

Thank you so much.

Answer 4

[Hassan Alkomy]

Actually, the three solutions by David, Joseph and Scott give what i want. I chose David's answer because it is a one-line answer, but the rest of the solutions gives the same result.

Thank you all.