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- Homework Statement
- if ## \gamma (t):= i+3e^{2it } , t \in \left[0,4\pi \right] , then \int_0^{4\pi } \frac {dz} {z} \ ##

- Relevant Equations
- complex numbers

if ## \gamma (t):= i+3e^{2it } , t \in \left[0,4\pi \right] , then \int_0^{4\pi} \frac {dz} {z} ##

in order to solve such integral i substitute z with ##\gamma(t)## and i multiply by ##\gamma'(t)##

that is:

##\int_0^{4 \pi} \frac {6e^{2it}}{i+3e^{2it}}dt=\left.log(i+3e^{2it}) \right|_0^{4 \pi}=##

##= log\left (i+3e^{i8 \pi }\right) - log\left (i+3\right)=##

##= log\left (i+3\right)-log\left (i+3\right)##

there must be something wrong but i don't see where I'm making the mistake. Because in such way the result will be zero but shouldn't be so.

in order to solve such integral i substitute z with ##\gamma(t)## and i multiply by ##\gamma'(t)##

that is:

##\int_0^{4 \pi} \frac {6e^{2it}}{i+3e^{2it}}dt=\left.log(i+3e^{2it}) \right|_0^{4 \pi}=##

##= log\left (i+3e^{i8 \pi }\right) - log\left (i+3\right)=##

##= log\left (i+3\right)-log\left (i+3\right)##

there must be something wrong but i don't see where I'm making the mistake. Because in such way the result will be zero but shouldn't be so.

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