For pole and zero vectors from the origin to the pole or zer0 there is no calculation. the real part is the x-component and the imaginary part is the y-component. The y-components is also known as the jω-component since it is along the jω-axis in the complex plane.
Many filter tables use the letters a and b for the coefficients of a second order polynomial which shows up in the denominator of a transfer function. They relate to that polynomial in a standard form. For a a transfer function in the form:
\[ 1+as+bs^2 \]
The a coefficient is associated with the linear term and the b coefficient is associated with the squared term.
If the transfer function is expressed as:
\[ s^2+\frac{\omega_0}{Q}s\;+\;\omega_0^2 \]
you can transform it to the other form by dividing each term by \( \omega_0^2 \) and setting the corresponding coefficients equal to each other.
With either formulation you can relate the pole locations to a and b or to \( \omega_0 \;\text and\; Q \).
You can also use the quadratic formula on either form to get the pole locations.
Hello there just to elaborate if I may be so bold.
Quadratics are degree-two polynomials.
A "quadratic" is a polynomial that looks like "ax2 + bx + c", where "a", "b", and "c" are just numbers.
To end up with a quadratic that had a leading coefficient of 1 (and no fractions), each of the original binomials also had to have had a leading coefficient of 1