Całka po konturze
: 8 gru 2022, o 18:14
Proszę o pomoc, w którym miejscu popełniam błąd?
\(\displaystyle{ \int_{C}^{} \frac{dz}{(z ^{9} + 9) ^{2} } }\) po konturze \(\displaystyle{ \left| z-3i \right| =1}\)
\(\displaystyle{ Res(z=3i) = \lim_{z \to 3i} \frac{d}{dz} (z-3i) ^{2} \frac{1}{(z-3i) ^{2} (z+3i) ^{2} } = \lim_{z \to 3i} \frac{d}{dz} (z+3i) ^{-2} = \lim_{z \to 3i} - \frac{2}{(z+3i) ^{3} } = - \frac{2}{(6i) ^{3} } = - \frac{i}{108} }\)
\(\displaystyle{ \int_{C}^{} \frac{dz}{(z ^{9} + 9) ^{2} } = 2 \pi i \cdot\left( - \frac{i}{108}\right) = - \frac{ \pi }{54} }\)
\(\displaystyle{ \int_{C}^{} \frac{dz}{(z ^{9} + 9) ^{2} } }\) po konturze \(\displaystyle{ \left| z-3i \right| =1}\)
\(\displaystyle{ Res(z=3i) = \lim_{z \to 3i} \frac{d}{dz} (z-3i) ^{2} \frac{1}{(z-3i) ^{2} (z+3i) ^{2} } = \lim_{z \to 3i} \frac{d}{dz} (z+3i) ^{-2} = \lim_{z \to 3i} - \frac{2}{(z+3i) ^{3} } = - \frac{2}{(6i) ^{3} } = - \frac{i}{108} }\)
\(\displaystyle{ \int_{C}^{} \frac{dz}{(z ^{9} + 9) ^{2} } = 2 \pi i \cdot\left( - \frac{i}{108}\right) = - \frac{ \pi }{54} }\)