Mate Avanzada

Post on 14-Apr-2016

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Transcript of Mate Avanzada

3√8 i

√ 14 i

√2−3 i

3√8

4√54

l n (1+2 i )

t an(2π+i ln 2¿)¿

θe2xθ e−2 x

θex θsen x2

θ sen ( x )θsen (2x )

θ cos ( x )θe−x

θ cos ( x )θcos( x2 )

∮l

❑ ln z dz( z2+162 )2

; l :|z|=10

∮l

❑ cos z dz( z2+3 ) ( z−5 )3

; l :|z|=10

∮l

❑ (2−3 z+z3 )dz( z+3 )3 ( z−4 )2

; l :|z|=10

∮l

❑ cos5 x dx(x2+49 ) (x2+16 )

∮l

❑ sin x2dx

(x2+16 ) (x2+9 )