I'm having a bit of confusion here, i think this may be right but if someone could just quickly check it and let me know where I've gone wrong, that would be greatly appreciated
here we go:
Evaluate the following Integrals.
$$\int_{\gamma} \frac{dz}{z}$$ where $\gamma(t) = cos(t) + 2i sin(t),~t\in [\frac{\pi}{2},\frac{3\pi}{2}]$
From my understanding then, we can write $$\gamma_{0}(t) = cos(t) + 2i sin(t)$$ define $\gamma_{1}(t) = 2(cost + sin(t))$ and then show that these two are homotopic via $H(s,t)=(1-s)\gamma_{0}+s\gamma_{1}$
H(t,0) = $\gamma_{1}$, H(t,1) = $\gamma_{2}$, with $\gamma_1(\frac{\pi}{2}) = \gamma_2(\frac{\pi}{2})$ and $\gamma_1(\frac{3\pi}{2}) = \gamma_2(\frac{3\pi}{2})$
to show these two paths are homotopic, then via the deformation theorem we have for any $A \subseteq_{open} \mathbb{C}$ with f analyic on A, if $\gamma_1$,$\gamma_2$ are two piecewise smooth curves in A that ar ehomotpic to each other then $$\int_{\gamma_{1}} f(z)dz = \int_{\gamma_{2}}f(z)dz$$ so then for f as defined above we have
$$\int_{cos(t)+2i sin(t)}\frac{dz}{z} = \int_{2(cost+isint)}\frac{dz}{z} = \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \frac{i cost+sint}{cost+isint} dt$$ from the contour chain rule then $$\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}\frac{i cost+sint}{cost+isint} dt = [it]^{3\pi/2}_{\pi/2} = \pi i $$
does this look right? anyone see any holes? thanks in advance. Sincerely
**edit: to check if the homotopy is with in $\{z \neq 0\}$
so from my understanding all i need to do is make sure that $H(s,t) \neq 0$, $\forall s \in [0,1],t \in [\pi/2,3\pi/2]$, then the above integral is fine.
Expanding $\gamma_1, \gamma_2$ gives $H(s,t) = (1+s) cos(t)+i 2sin(t) \neq 0$
**edit 2: incidentially i just noticed i should have used this homotopoy in the first place....im stupid.