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Michael Hardy
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For $m=1$, this is the residue: $$\mathrm{Res} \left( z^{kn-1} \left( az+1 \right)^{-k}; -1/a \right)=\frac{(-1)^{k n-k} }{a^{kn}(k-1)!}\prod _{p=1}^{k-1} (k n-p).$$$$\operatorname{Res} \left( z^{kn-1} \left( az+1 \right)^{-k}; -1/a \right)=\frac{(-1)^{k n-k} }{a^{kn}(k-1)!}\prod _{p=1}^{k-1} (k n-p).$$

For $m=1$, this is the residue: $$\mathrm{Res} \left( z^{kn-1} \left( az+1 \right)^{-k}; -1/a \right)=\frac{(-1)^{k n-k} }{a^{kn}(k-1)!}\prod _{p=1}^{k-1} (k n-p).$$

For $m=1$, this is the residue: $$\operatorname{Res} \left( z^{kn-1} \left( az+1 \right)^{-k}; -1/a \right)=\frac{(-1)^{k n-k} }{a^{kn}(k-1)!}\prod _{p=1}^{k-1} (k n-p).$$

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Carlo Beenakker
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For $m=1$, this is the residue: $$\mathrm{Res} \left( z^{kn-1} \left( az+1 \right)^{-k}; -1/a \right)=\frac{(-1)^{k} }{a^{kn}(k-1)!}\prod _{p=1}^{k-1} (k n-p).$$$$\mathrm{Res} \left( z^{kn-1} \left( az+1 \right)^{-k}; -1/a \right)=\frac{(-1)^{k n-k} }{a^{kn}(k-1)!}\prod _{p=1}^{k-1} (k n-p).$$

For $m=1$, this is the residue: $$\mathrm{Res} \left( z^{kn-1} \left( az+1 \right)^{-k}; -1/a \right)=\frac{(-1)^{k} }{a^{kn}(k-1)!}\prod _{p=1}^{k-1} (k n-p).$$

For $m=1$, this is the residue: $$\mathrm{Res} \left( z^{kn-1} \left( az+1 \right)^{-k}; -1/a \right)=\frac{(-1)^{k n-k} }{a^{kn}(k-1)!}\prod _{p=1}^{k-1} (k n-p).$$

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Carlo Beenakker
  • 188.2k
  • 18
  • 448
  • 651

For $m=1$, this is the residue: $$\mathrm{Res} \left( z^{kn-1} \left( az+1 \right)^{-k}; -1/a \right)=\frac{(-1)^{k} }{a^{kn}(k-1)!}\prod _{p=1}^{k-1} (k n-p).$$