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If $\gamma$ is a piecewise, smooth, positively oriented simple closed curve in $D$, then Cauchy's formula states that $f(z)=1/2\pi i\int_\gamma {f(a)\over {a-z}}$.

My textbook also stated that for $z$ is the center of the circle then $f(z)=1/2\pi \int_0^{2\pi}f(z+re^{it})dt$. There was no further justification,

I technically understand everything the way they were written, because a circle can be formulated as $re^{it}$, and since $z$ is the center, so you add $z$. And you integrate from $0\to \pi$. So that all makes sense, but where does the $i$ go, and if I want to, how do I write a formal proof for this formula?

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In the integral formula, you have

$$f(z) = \frac{1}{2\pi i}\int_\gamma \frac{f(a)}{a-z}\,da.$$

Now, with the parametrisation $\gamma(t) = z + re^{it}$ and $a\: \hat{=}\: \gamma(t)$, the differential $da$ becomes

$$da = \gamma'(t)\,dt = ire^{it}\,dt = i(\gamma(t)-z)\,dt,$$

or, in other words,

$$ \frac{da}{i(a-z)} \:\hat{=}\: dt.$$

The $i$ is cancelled by (part of) the derivative of $\gamma$.

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  • $\begingroup$ Beat me to it! I was halfway through the solution when yours popped up.. Although might I inquire about this hat-equals symbol? I've seen it once or twice in the past but I have no idea what it means. $\endgroup$ Mar 23, 2014 at 21:28
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    $\begingroup$ "Corresponds to". Not quite equal, strictly speaking. In this case, we have $dt$ the pull-back of $\frac{da}{i(a-z)}$ by $\gamma$, the forms live on different spaces. $\endgroup$ Mar 23, 2014 at 21:31
  • $\begingroup$ Right. Makes sense. Thanks! $\endgroup$ Mar 23, 2014 at 21:34

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