# Complex Poles

A pair of complex poles is simple if it is not repeated; it is a double or multiple pole if repeated. Simple complex poles may be handled the same as simple real poles, but because complex algebra is involved the result is always cumbersome. An easier approach is a method known as completing the square. The idea is to express each complex pole pair (or quadratic term) in $D(s)$ as a complete square such as $(s+\alpha)^{2}+\beta^{2}$ and then use Table $2$ to find the inverse of the term.
Since $N(s)$ and $D(s)$ always have real coefficients and we know that the complex roots of polynomials with real coefficients must occur in conjugate pairs, $F(s)$ may have the general form
$$F(s)=\frac{A_{1} s+A_{2}}{s^{2}+a s+b}+F_{1}(s) \tag{1}$$
where $F_{1}(s)$ is the remaining part of $F(s)$ that does not have this pair of complex poles. If we complete the square by letting
$$s^{2}+a s+b=s^{2}+2 \alpha s+\alpha^{2}+\beta^{2}=(s+\alpha)^{2}+\beta^{2}$$
and we also let
$$A_{1} s+A_{2}=A_{1}(s+\alpha)+B_{1} \beta$$
then Eq. (1) becomes
$$F(s)=\frac{A_{1}(s+\alpha)}{(s+\alpha)^{2}+\beta^{2}}+\frac{B_{1} \beta}{(s+\alpha)^{2}+\beta^{2}}+F_{1}(s)$$
From Table $2$, the inverse transform is
$$f(t)=A_{1} e^{-\alpha t} \cos \beta t+B_{1} e^{-\alpha t} \sin \beta t+f_{1}(t)$$
The sine and cosine terms can be combined using Eq. (2).
$$\bbox[10px,border:1px solid grey]{C = \sqrt{A^2 + B^2}}, \,\bbox[10px,border:1px solid grey]{θ = \tan^{−1} {B \over A}} \tag{2}$$
Whether the pole is simple, repeated, or complex, a general approach that can always be used in finding the expansion coefficients is the method of algebra, illustrated in Examples $1$ to $3$. To apply the method, we first set $F(s)=N(s) / D(s)$ equal to an expansion containing unknown constants. We multiply the result through by a common denominator. Then we determine the unknown constants by equating coefficients (i.e., by algebraically solving a set of simultaneous equations for these coefficients at like powers of $s$ ). Another general approach is to substitute specific, convenient values of $s$ to obtain as many simultaneous equations as the number of unknown coefficients, and then solve for the unknown coefficients. We must make sure that each selected value of $s$ is not one of the poles of $F(s)$. Example $3$ illustrates this idea.