# Basis Functions for a Transient Analysis of Linear Commensurate Fractional-Order Systems
## 3 Mathematical Prerequisites
### Some Fractional Goniometric and Other Functions  
In Section 4, we show that when computing some responses, it is convenient to compute the MLF as a complex function of the complex argument and, in a second step, to extract its real or imaginary part from the result. Consider the \(m\)-th derivative of a two-parameter MLF with complex argument \(z\) or complex conjugate \(z^{*}\),  
\[z=x+\mathrm{i}y=|z|e^{\mathrm{i}\varphi},z^{*}=x-\mathrm{i}y=|z|e^{-\mathrm{i} \varphi}. \tag{12}\]  
Then, the defining relation in Table 1 implies  
\[E_{\alpha,\beta}^{(m)}(z)=\mathrm{Re}\Big{(}E_{\alpha,\beta}^{(m)}(z)\Big{)}+ \mathrm{i}\operatorname{Im}\Big{(}E_{\alpha,\beta}^{(m)}(z)\Big{)}, \tag{13}\]  
\[E_{\alpha,\beta}^{(m)}(z^{*})=\mathrm{Re}\Big{(}E_{\alpha,\beta}^{(m)}(z) \Big{)}-\mathrm{i}\operatorname{Im}\Big{(}E_{\alpha,\beta}^{(m)}(z)\Big{)}=E _{\alpha,\beta}^{*(m)}(z), \tag{14}\]  
\[E_{\alpha,\beta}^{(m)}(z)+E_{\alpha,\beta}^{(m)}(z^{*})=2\mathrm{Re}\Big{(}E_ {\alpha,\beta}^{(m)}(z)\Big{)}, \tag{15}\]  
\[E_{\alpha,\beta}^{(m)}(z)-E_{\alpha,\beta}^{(m)}(z^{*})=2\mathrm{i}\operatorname {Im}\Big{(}E_{\alpha,\beta}^{(m)}(z)\Big{)}, \tag{16}\]  
where  
\[\mathrm{Re}\Big{(}E_{\alpha,\beta}^{(m)}(z)\Big{)}=\sum\limits_{k=0}^{\infty} \frac{(k+m)!}{k!}\frac{|z|^{k}\cos(k\varphi)}{\Gamma(\alpha k+\alpha m+\beta)}, \tag{17}\]  
\[\operatorname{Im}\Big{(}E_{\alpha,\beta}^{(m)}(z)\Big{)}=\sum\limits_{k=0}^{ \infty}\frac{(k+m)!}{k!}\frac{|z|^{k}\sin(k\varphi)}{\Gamma(\alpha k+\alpha m+ \beta)}. \tag{18}\]  
Although relations (15) and (16) are trivial, they are mentioned here since the real and imaginary parts of the corresponding MLFs and their derivatives described by these formulas play an important role in the semi-symbolic analysis of systems with complex poles (see Section 4). The case of MLF with the purely imaginary argument \(z=\mathrm{i}y\), which corresponds to modeling the time response of a classical IOS without a damping factor, also deserves attention. In this case,  
\[ \begin{array}{l}E_{\alpha,\beta}^{(m)}(\mathrm{i}y)=\sum\limits_{k=0}^{ \infty}\frac{(k+m)!}{k!}\frac{\mathrm{i}^{k}y^{k}}{\Gamma(\alpha k+\alpha m+ \beta)}=\sum\limits_{k=0}^{\infty}(-1)^{k}\frac{(2k+m)!}{(2k)!}\frac{y^{2k}}{ \Gamma((2k+m)\alpha+\beta)}\\ +\mathrm{i}\sum\limits_{k=0}^{\infty}(-1)^{k}\frac{(2k+m)!}{(2k+1)!}\frac{y^{ 2k+1}}{\Gamma((2k+m+1)\alpha+\beta)}.\end{array}  \tag{19}\]  
For the integer-order case, i.e., \(\alpha=\beta=1\), the \(m\)-th derivative of the function \(E_{\alpha,\beta}\) (\(\mathrm{i}y\)) on the left-hand side of (19) translates into the classical exponential function of the imaginary argument, and the sums on the right-hand side translate into the Maclaurin series of functions of the cosine and sine type:  
\[ \begin{array}{l}E_{1,1}^{(m)}(\mathrm{i}y)=\sum\limits_{k=0}^{\infty}(-1)^{ k}\frac{(2k+m)!}{(2k)!}\frac{y^{2k}}{\Gamma(2k+m+1)}+\mathrm{i}\sum\limits_{k=0}^{ \infty}(-1)^{k}\frac{(2k+1+m)!}{(2k+1)!}\frac{y^{2k+1}}{\Gamma(2k+m+2)}=\\ \sum\limits_{k=0}^{\infty}(-1)^{k}\frac{y^{2k}}{(2k)!}+\mathrm{i}\sum\limits_{ k=0}^{\infty}(-1)^{k}\frac{y^{2k+1}}{(2k+1)!}=\cos(y)+\mathrm{i}\sin(y)=e^{ \mathrm{i}y}.\end{array}