Editing Residue (complex analysis) (section)

==== Example 3 ====

The next example shows that, computing a residue by series expansion, a major role is played by the [[Formal series#The Lagrange inversion formula|Lagrange inversion theorem]]. Let<math display="block"> u(z) := \sum_{k\geq 1}u_k z^k</math>be an [[entire function]], and let<math display="block">v(z) := \sum_{k\geq 1}v_k z^k</math>with positive radius of convergence, and with <math display="inline"> v_1 \neq 0</math>. So <math display="inline"> v(z)</math> has a local inverse <math display="inline"> V(z)</math> at 0, and <math display="inline"> u(1/V(z))</math> is [[meromorphic]] at 0. Then we have:<math display="block">\operatorname{Res}_0 \big(u(1/V(z))\big) = \sum_{k=0}^\infty ku_k v_k. </math>Indeed,<math display="block">\operatorname{Res}_0\big(u(1/V(z))\big) = \operatorname{Res}_0 \left(\sum_{k\geq 1} u_k V(z)^{-k}\right) = \sum_{k\geq 1} u_k \operatorname{Res}_0 \big(V(z)^{-k}\big)</math>because the first series converges uniformly on any small circle around 0. Using the Lagrange  inversion theorem<math display="block">\operatorname{Res}_0 \big(V(z)^{-k}\big) = kv_k,</math>and we get the above expression. For example, if <math>u(z) = z + z^2</math> and also <math>v(z) = z + z^2</math>, then<math display="block">V(z) = \frac{2z}{1 + \sqrt{1 + 4z}}</math>and<math display="block">u(1/V(z)) = \frac{1 + \sqrt{1 + 4z}}{2z} + \frac{1 + 2z + \sqrt{1 + 4z}}{2z^2}.</math>The first term contributes 1 to the residue, and the second term contributes 2 since it is asymptotic to <math>1/z^2 + 2/z</math>.


Note that, with the corresponding stronger symmetric assumptions on <math display="inline"> u(z)</math> and <math display="inline"> v(z)</math>, it also follows<math display="block">\operatorname{Res}_0 \left(u(1/V)\right) = \operatorname{Res}_0\left(v(1/U)\right),</math>where <math display="inline"> U(z)</math> is a local inverse of <math display="inline"> u(z)</math> at&nbsp;0.