Complex Numbers via Matlab

Complex Numbers via MATLAB

Here we show how to

Finding roots of polynomials

MATLAB can find the roots of polynomials via the roots command. To find the roots of \(z^2+6z+25\) you enter the coefficients of \(z\)


>>eqn = [1 6 25]
eqn =
     1     6    25

and ask for the roots:


>>roots(eqn)
ans =
  -3.0000 + 4.0000i
  -3.0000 - 4.0000i

Notice that this is the same answer as given in the book.

MATLAB outputs \(i\) as the square root of \(-1\), but you can use \(j\) too.

Manipulating complex numbers

There are many commands for manipulating complex numbers. You can guess what they do by their names.


>>z = 3 + j*4
z =
   3.0000 + 4.0000i
>>real(z)
ans =
     3
>>imag(z)
ans =
     4
>>conj(z)
ans =
   3.0000 - 4.0000i
>>abs(z)
ans =
     5
>>angle(z)
ans =
    0.9273

Note the angle is in radians, you can convert to degrees by:


>>angle(z)*180/pi
ans =
   53.1301

If you want to input as polar use:


>>z2=10*exp(j*pi/4)
z2 =
   7.0711 + 7.0711i


>>10*exp(j*45/180*pi)
ans =
   7.0711 + 7.0711i

A handy way to see a complex number in several forms is zprint. zprint is not part of MATLAB, but comes with this CD-ROM


>>zprint([z,z2])
 Z =     X    +     jY     Magnitude    Phase    Ph/pi   Ph(deg)
           3           4           5    0.927    0.295    53.13
       7.071       7.071          10    0.785    0.250    45.00

Type help zprint for more details

Plotting vectors

You can plot the two vectors (z and z2) via zvect and zcat.
zvect plots each vector with it's tail at the origin.
zcat plots each vector head to tail.


zvect([z, z2])


zcat([z, z2])

You might want to use the axis command to put 0,0 in the center


zvect([z, z2])
axis([-8 8 -8 8])

The example from section 2.4.1 via MATLAB is:


>>z1 = 7*exp(j*4*pi/7)
z1 =
  -1.5576 + 6.8245i
>>z2 = 5*exp(-j*5*pi/11)
z2 =
   0.7116 - 4.9491i

Notice MATLAB automatically converts them to rectangular form. Now just add them.


>>z3 = z1 + z2
z3 =
  -0.8461 + 1.8754i

We can also show this graphically with:


>>zcat([z1, z2]), hold on
>>zvect(z3)

Finding roots of Unity

We know that the square root of \(1\) gives two answers, \(+1\) and \(-1\). Likewise, for the \(n\)-th root of \(1\), we get \(n\) answers. Here we find the third roots of unity, by solving \(z^3=1\)


>>p = [1 0 0 -1]
p =
     1     0     0    -1
>>z = roots(p)
z =
  -0.5000 + 0.8660i
  -0.5000 - 0.8660i
   1.0000

You can check to see if you have the right polynomial with the poly2sym command.


>>poly2sym(p,'z')
ans =
z^3-1

Good, it checks.

We might see a pattern if we convert to polar.


>>zprint(z)
 Z =     X    +     jY     Magnitude    Phase    Ph/pi   Ph(deg)
        -0.5       0.866           1    2.094    0.667   120.00
        -0.5      -0.866           1   -2.094   -0.667  -120.00
           1           0           1    0.000    0.000     0.00

Hmmm, do you see the pattern? Let's plot them.


zvect(z)

Do you see a relationship between the power to which z is raised and the number of vectors?
What if we solve \(z^{10}=1\)?


>>z10 = roots([1 0 0 0 0 0 0 0 0 0 -1])
zvect(z10)
z10 =
  -1.0000          
  -0.8090 + 0.5878i
  -0.8090 - 0.5878i
  -0.3090 + 0.9511i
  -0.3090 - 0.9511i
   0.3090 + 0.9511i
   0.3090 - 0.9511i
   1.0000          
   0.8090 + 0.5878i
   0.8090 - 0.5878i
>>zprint(z10)
 Z =     X    +     jY     Magnitude    Phase    Ph/pi   Ph(deg)
          -1           0           1    3.142    1.000   180.00
      -0.809      0.5878           1    2.513    0.800   144.00
      -0.809     -0.5878           1   -2.513   -0.800  -144.00
      -0.309      0.9511           1    1.885    0.600   108.00
      -0.309     -0.9511           1   -1.885   -0.600  -108.00
       0.309      0.9511           1    1.257    0.400    72.00
       0.309     -0.9511           1   -1.257   -0.400   -72.00
           1           0           1    0.000    0.000     0.00
       0.809      0.5878           1    0.628    0.200    36.00
       0.809     -0.5878           1   -0.628   -0.200   -36.00
>>zvect(z10)

Did you guess right?