Step by step solution to "Evaluating a complex integral∫|z|=1z1112z12−4z9+2z6−4z3+1dz"

jisu61hbke

Answered question

2022-04-05

Evaluating a complex integral

\int_{| z | = 1} \frac{z^{11}}{12 z^{12} - 4 z^{9} + 2 z^{6} - 4 z^{3} + 1} d z

anghoelv1lw

Beginner2022-04-06Added 19 answers

Your integral is off by a factor of -1, since the variable change from z to

\frac{1}{z}

reverses the orientation of the circle. So what you need is

\int_{| w | = 1} \frac{1}{w (w^{12} - 4 w^{9} + 2 w^{6} - 4 w^{3} + 12)} d w

You could use Rouche but you don't have to: If

| w | \leq 1

then

| w^{12} - 4 w^{9} + 2 w^{6} - 4 w^{3} | \leq {| w |}^{12} + 4 {| w |}^{9} + 2 {| w |}^{6} + 4 {| w |}^{3} \leq 1 + 4 + 2 + 4 = 11

< 12

So the factor

w^{12} - 4 w^{9} + 2 w^{6} - 4 w^{3} + 12

is never zero. As the result the integrand just has the one pole at

2 = 0

and you can use the residue theorem for just that pole to get the value of the integral. The residue is

\frac{1}{w^{12} - 4 w^{9} + 2 w^{6} - 4 w^{3} + 12}

evaluated at

w = 0

, or

\frac{1}{12}

, and the overall integral is

2 π i \cdot \frac{1}{12} = \frac{π i}{6}

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