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Question Number 148068 by tabata last updated on 25/Jul/21

	Answered by Olaf_Thorendsen last updated on 25/Jul/21
	$(B) Let f(z) = z^7 −4z^3 +z−1 = 0, ∣z∣ = 1 and g(z) = −4z^3 • f has 0 pole : P_f = 0 • g has 3 roots (trivial) and 0 pole Z_g = 3 and P_g = 0 ∣f(z)−g(z)∣ = ∣z^7 +z−1∣ ∣f(z)∣ ≤ ∣z∣^7 +∣z∣+1 = 3 ∣g(z)∣ = 4∣z∣^3 = 4 • Rouche′s theorem : ∀z\∣z∣ = 1, ∣f(z)−g(z)∣ ≤ ∣g(z)∣ then Z_f −P_f = Z_g −P_g Z_f −0 = 4−0 Z_f = 3 The number of roots of f interior to ∣z∣ = 1 is at least 3.$
	$(B)$ $Let f (z) = z^{7} - 4 z^{3} + z - 1 = 0, ∣ z ∣ = 1$ $and g (z) = - 4 z^{3}$ $∙ f has 0 pole : P_{f} = 0$ $∙ g has 3 roots (trivial) and 0 pole$ $Z_{g} = 3 and P_{g} = 0$ $∣ f (z) - g (z) ∣ = ∣ z^{7} + z - 1 ∣$ $∣ f (z) ∣ ⩽ ∣ z ∣^{7} + ∣ z ∣ + 1 = 3$ $∣ g (z) ∣ = 4 ∣ z ∣^{3} = 4$ $∙ {Rouche}^{'} s theorem :$ $\forall z ∖ ∣ z ∣ = 1, ∣ f (z) - g (z) ∣ ⩽ ∣ g (z) ∣$ $then Z_{f} - P_{f} = Z_{g} - P_{g}$ $Z_{f} - 0 = 4 - 0$ $Z_{f} = 3$ $The number of roots of f interior to$ $∣ z ∣ = 1 is at least 3 .$
	Commented by tabata last updated on 25/Jul/21

	$m s r p l e a s h e l p m e i n q u e s t i o n 148082 p l e a s e a j u s t t h i s$
	Answered by Olaf_Thorendsen last updated on 25/Jul/21

	$(A)$ $f (z) = \frac{8 - z}{z (4 - z)}$ $Res (f, - 4) = 2 (\frac{8 - 4}{4}) = 1$ $Res (f, 0) = (\frac{8 - 0}{4 - 0}) = 2$ $\int_{C} f (z) d z = 2 i π Σ {Ind}_{C} (z_{k}) Res (f, z_{k})$ $\int_{C} f (z) d z = 2 i π (- 1 - 2) = - 6 i π$
	Answered by mathmax by abdo last updated on 26/Jul/21
	$f(z)=((1−cosz)/z^2 ) Res(f,o) (pole double) ⇒ Res(f,o)=lim_(z→0) (1/((2−1)!)){z^2 f(z)}^((1)) =lim_(z→0) {1−cosz}^((1)) =lim_(z→0) sinz=0 let use laurent serie we have cosz=Σ_(n=0) ^∞ (((−1)^n z^(2n) )/((2n)!))=1−(z^2 /2)+(z^4 /(4!))−(z^6 /(6!))+... ⇒ f(z)=(1/z^2 )(1−(z^2 /(2!))+(z^4 /(4!))−(z^6 /(6!)))=(1/z^2 )−(1/(2!))+(z^2 /(4!))−(z^4 /(6!))+....$
	$f (z) = \frac{1 - cosz}{z^{2}} Res (f, o) (pole double) \Rightarrow$ $Res (f, o) = \lim_{z \to 0} \frac{1}{(2 - 1)!} {z^{2} f (z)}^{(1)} = \lim_{z \to 0} {1 - cosz}^{(1)}$ $= \lim_{z \to 0} sinz = 0$ $let use laurent serie we have$ $cosz = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} z^{2 n}}{(2 n)!} = 1 - \frac{z^{2}}{2} + \frac{z^{4}}{4!} - \frac{z^{6}}{6!} + . . . \Rightarrow$ $f (z) = \frac{1}{z^{2}} (1 - \frac{z^{2}}{2!} + \frac{z^{4}}{4!} - \frac{z^{6}}{6!}) = \frac{1}{z^{2}} - \frac{1}{2!} + \frac{z^{2}}{4!} - \frac{z^{4}}{6!} + . . . .$
	Commented by mathmax by abdo last updated on 26/Jul/21

	$Res (f, o) = coefficient de \frac{1}{z} = 0$
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