 # Complex antiderivative I am confused on a couple things: 1.) Why is it that an integral of a comp kolutastmr 2022-07-02 Answered
Complex antiderivative
I am confused on a couple things:
1.) Why is it that an integral of a complex valued function of a complex variable exists if f(z(t)) is piecewise continuous (and/or piecewise continuous on $\mathbb{C}$) and not continuous, like a real?
2.) Why is it that one cannot make use of an antiderivative to evaluate an integral of a function like 1/z, on a contour of something like $z=2{e}^{i\theta }$ positively oriented with $-\pi \le \theta \le \pi$? That is, because ${F}^{\prime }\left(z\right)=1/z$ is undefined at 0, it is disqualified (I think). But why?
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Explanation:
If the function f(z) has an antiderivative F(z) on a domain that includes a contour Γ that goes from point a to point b, then . The problem is that in many cases such an antiderivative can't exist. A case in point is your example, ${\int }_{\mathrm{\Gamma }}\frac{dz}{z}$, where $\mathrm{\Gamma }$ is the counterclockwise circle of radius 2 centred at 0. You'd like to say that an antiderivative of $f\left(z\right)=1/z$ is $\mathrm{log}\left(z\right)$, but there is no version of log(z) that is continuous on that whole circle. Indeed that must be so, because since $b=a$ (this contour ends in the same place where it starts), you'd have ${\int }_{\mathrm{\Gamma }}\frac{dz}{z}=F\left(a\right)-F\left(a\right)=0$. But, parametrizing the contour as $z\left(t\right)=2{e}^{it}$,

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