Evaluate $\int_0^ie^z\ dz$.

I have never encountered an integral with limits in Complex Analysis yet. I am only familiar with the symbol $\int_C$ where $C$ is some path. What do the limits mean in this problem?

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    $\begingroup$ Here the integral does not depends on path. You can integrate along the segment $\gamma(t) = it$ for example. $\endgroup$ – user171326 Feb 19 '15 at 0:06
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    $\begingroup$ it means that the path starts at $0$ and ends at $i$ and as @N.H.mentioned, the result does not depend on the actual path itself $\endgroup$ – benji Feb 19 '15 at 0:10

If the function you are integrating is entire (analytic everywhere), then the integral does not depend on path and you can integrate just as you do for real functions, $$\int_\alpha^\beta f(z)\,dz=F(\beta)-F(\alpha)\ ,$$ where $F'(z)=f(z)$. In your case, $f(z)=e^z$ so $F(z)=e^z$ and $$\int_0^i e^z\,dz=e^i-e^0=\cos1+i\sin1-1\ .$$ There is more that can be said about this kind of problem, I am sure there will be theorems about it in your course.


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