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IV 1 5 1 111 Show that 23 PL K KKK 138 Thus e 11 1 1 Solution We begin with parametrizize the circle z1 1 as z 1 eit 1 cost i sint where 0 t 2 For the nominator in the integrand e z 1 we have dz 2 Im e eRez i Imz eRezeilmz eRez e since Re z 2 on the circle And for the denominator in the integrad we have 1 z 2 cost isint Re 2 cost isint 2 cost 1 M 8 1541 2 1 0 e e Becauce the integrationcontour is a circle with radius 1 we have that L 2 ML estimate gives

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