Evaluation of Sc z e^(z^2) dz on the complex plane

[lastlydreaming]

Hello, is this correct:

Evaluate the integral on the complex plan of Sc z e^(z^2) dz from (1+i) to (3+2i):

From 0<=t<=1, I find z=(1+i)(1-t) +(3+2i)(t)
to be z=(2t+1)+ i(t+1).
Then dz is (2+i) dt.
Do a substitution where u = z^2, du = 2z dz

Then

(2+i)/2 S e^ u du (from 0 to 1) = (2+1)/2 e^ u ]0 to 1 = (2+i)/2 (e-1)?

Thanks.

 

[pka]

\(\displaystyle \L
\int\limits_{1 + i}^{3 + 2i} {ze^{z^2 } dz} = \left. {\frac{{e^{z^2 } }}{2}} \right|_{1 + i}^{3 + 2i} = \frac{{e^{\left( {3 + 2i} \right)^2 } }}{2} - \frac{{e^{\left( {1 + i} \right)^2 } }}{2}.\)

Entire functions have integrals that are path-wise independent.
Therefore, they can be handled exactly as you did in beginning calculus.

 

[lastlydreaming]

Can you leave your sol'n in that form?

The "(1/2)e^((3+2i)^2) - (1/2)e^((1+i)^2)" form?

 

[pka]

Can you leave your sol'n in that form?

That is up to your instructor to say.

 

[lastlydreaming]

How would you simplify that?