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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!
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!
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