Solving Cauchy Residual Theorem for Holomorphic Function at z=2i

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The discussion focuses on solving the Cauchy Residual Theorem for a holomorphic function at z=2i. The numerator is transformed into e^i(z+3) while the denominator is factored into (z-2i)(z+2i), with only (z+2i) being relevant since the function is holomorphic at z=2i. The user attempts to calculate the residue using Res(f,2i)=g(2i), arriving at an expression involving e^(i(2i+3)/(4i). After some calculations, the user initially suspects the answer might be 0.00477463 but ultimately finds it to be -0.2. The discussion highlights the complexities of applying the theorem and the challenges in verifying the calculations.
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Alright so I posted a picture asking the exact question.

Here is my best attempt...

According to my professor's terrible notes, the numerator can magically turn into the form:

e^i(z+3)

when converted to complex. The denominator will be factored into

(z-2i)(z+2i)

but the function is only holomorphic at z=2i so only (z+2i) can be used.

From there the Res(f,2i)=g(2i) which is equal to what I believe is something like

e^(i(2i+3)/(4i)

It follows that

J=e^(-2+3i)*Pi

and sovling for the real part gives me an incorrect answer.

I might be missing some steps but I'm going off a theorem and it's really hard to relate to this problem. Help me PLEASE!
 

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Is it 0.00477463?
 
This is $$\frac{\sin (3)}{4 e^2}$$

Let me know if this is the answer. I can explain how i got it.
 
I finally calculated the answer... it turned out to be -0.2

See the picture if you're interested
 

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I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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