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Question Number 145946 by tabata last updated on 09/Jul/21

t h e t y p e o f s i n g u l a r p o i n t o f f (z) = \frac{c o s (π z)}{(1 - z^{3})} i s ?

Answered by mathmax by abdo last updated on 09/Jul/21

les points singuliers de f sont les residus def

z^{3} - 1 = 0 \Leftrightarrow (z - 1) (z^{2} + z + 1) = 0 \Leftrightarrow z = 1 or z^{2} + z + 1 = 0

z^{2} + z + 1 = 0 \to Δ = - 3 \Rightarrow z_{1} = \frac{- 1 + i \sqrt{3}}{2} = e^{\frac{2 i π}{3}}

z_{2} = \frac{- 1 - i \sqrt{3}}{2} = e^{- \frac{2 i π}{3}} \Rightarrow f (z) = \frac{\cos (π z)}{(1 - z) (z - z_{1}) (z - z_{2})}

Res (f, 1) = \frac{\cos (π)}{3} = - \frac{1}{3}

Res (f, z_{1}) = \frac{\cos (π z_{1})}{(1 - z_{1}) (z_{1} - z_{2})} = \frac{\cos (π e^{\frac{2 i π}{3}})}{(1 - e^{\frac{i 2 π}{3}}) (i \sqrt{3})}

Res (f, z_{2}) = \frac{\cos (π z_{2})}{(1 - z_{2}) (- i \sqrt{3})}

rest de terminer le calcul \dots .

Commented by mathmax by abdo last updated on 09/Jul/21

les points singuliers sont les poles c a dire 1, e^{\frac{2 i π}{3}} et e^{- \frac{2 i π}{3}} \dots . .