bug in sprintf 2024a, can't use other variable and just repeating the first variable

this is the code that spoiled my food, I've been following this tutorial and I got stuck at this one for TWO HOURS! I can't use the second variable!

Nume = app.NumeratorEditField.Value;
    Domi = app.DenominatorEditField.Value;
    [r,p,k] = residue(Nume,Domi); % this is for partial fraction
    r1 = real(r) %I am trying to seperate the real # from the imaginary number (complex)
    r2 = imag(r)   %I am trying to seperate the real # from the imaginary number (complex)
    app.rTextArea.Value = sprintf('%g + %gi\n',r1, r2)    %this is where I am stuck

%%% Instead of giving me the result:

0.23123 + -1.123123i

0.344 + 1.7766i

0.234 + 0.00251234i

%%% It is giving me this:

0.23123 + 0.23123i

0.344 + 0.344i

0.234 + 0.234i

-1.123123 + -1.123123i

1.7766 + 1.7766i

0.00251234 + 0.00251234i

Take note about the bold&italic numbers this might be clue ;(

PLEEASE HELP ME PASSED THIS COURSE ToT

Answer 1

[Star Strider]

The residue funciton produces column vectors in its output. To use sprintf with them, concatenate them in a matrix and then transpose the matrix:

sprintf('%g + %gi\n',[r1, r2].')

Try this —

Nume = poly(randi(1,3))*10

Nume = 1×4

10.0000 -30.0000 -0.0000 -0.0000

Domi = poly(randi(1,4))*10

Domi = 1×5

10.0000 -40.0000 -0.0000 -0.0000 -0.0000

[r,p,k] = residue(Nume,Domi); % this is for partial fraction
r1 = real(r) %I am trying to seperate the real # from the imaginary number (complex)

r1 = 4×1

0.2500 0.7500 0.0000 -0.0000

r2 = imag(r)   %I am trying to seperate the real # from the imaginary number (complex)

r2 = 4×1

0 0 0 0

sprintf('%g + %gi\n',[r1, r2].')    %this is where I am stuck

ans = 
    '0.25 + 0i
     0.75 + 0i
     2.70696e-16 + 0i
     -3.58866e-32 + 0i
     '

.

[Daniel]

Oh my god I understand it and this actually work!!! Thank you so much Matlab master!!!

[Star Strider]

As always, my pleasure!

[Walter Roberson]

Alternative:

disp(compose("%g + %gi", r1, r2))