Understanding $\frac{1}{2}(\sin(z))^2$ and its Solution Using the Attached Image

[nacho-man]

Referring to the attached image.

i seem to have forgotten this material and am trying to revise,

how/why does it become $\frac{1}{2}$ $(\sin(z))^2 $' ?

and does the path $C$ that we are given come into play in the solution?

Sorry for the noobness, any help is appreciated!

 

[Prove It]

It doesn't. \displaystyle \begin{align*} \sin{(z)}\cos{(z)} = \frac{1}{2}\sin{(2z)} \end{align*}, not \displaystyle \begin{align*} \frac{1}{2}\sin^2{(z)} \end{align*}...

 

[nacho-man]

I should clarify I meant how did they go from question to step 1, then step 2, step 3.

what is being done?

 

[Prove It]

Couldn't be more wrong I'm afraid. First of all, this is a LINE INTEGRAL. The so-called "Solution" has not taken that into account at all. In fact, there has not been any integrating done at all either...

 

[chisigma]

Referring to the attached image.

i seem to have forgotten this material and am trying to revise,

how/why does it become $\frac{1}{2}$ $(\sin(z))^2 $' ?

and does the path $C$ that we are given come into play in the solution?

Sorry for the noobness, any help is appreciated!

The function $\displaystyle f(z) = \frac{1}{2}\ \sin 2 z $ is analytic over rhe entire complex plane, so that the line integral is independent from the path connecting the point 0 to the point 1 + i. That means that is...

$\displaystyle \frac{1}{2} \int_{c} \sin 2 z\ d z = \frac{1}{2}\ (\int_{0}^{1} \sin 2 x\ d x + i\ \int_{0}^{1} \sin 2\ (1 + i y)\ d y )\ (1)$

Kind regards

$\chi$ $\sigma$

 

[nacho-man]

thanks fellas!