Integrate the following functions over the circle C0 3 a iz3

Integrate the following functions over the circle C[0, 3].

(a) i^(z-3)

(b) sinz / ((z^2+(1/2))^2)

Solution

Problem (a)

₀³i^(z3) dz

i = imaginary

₀³(1)^(z3)/2 dz

Substitute u= (z3)/2

du/dz = 1/2

= 2(1)^u du

Now solving:

(1)^u du

Apply exponential rule:

= (1)^u/ln(1)

Plug in solved integrals:

2(1)^udu

= 2(1)^u/ln(1)

Undo substitution u = (z3)/2

=2(1)^(z3)/2/ln(1) | [0,3]

= [2(1)^(33)/2/ln(1) + 2(1)^(03)/2/ln(1)]

Theoretically solution will be not possible since ln(-1) is an invalid argument.

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