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Suppose $z_0 \in \mathbb{C}$ , $r>0$,

Let $C$ be the line segment from $z_0+r-ir$ to $z_0+r+ir$.

A parametrization of the smooth curve $C$ is

$z(t) = z_0 + r + i(2rt-r)$, $t\in[0,1]$

and $z'(t)=2ri$.

Then, $\int_{C} (z-z_0)^n dz = \int_{0}^{1} [r+i(2rt-r)]^n 2r$ $dt$. $(n\in \mathbb{Z})$

I need help on further evaluating the integral. (for $n\neq -1$ and $n=-1$)

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  • $\begingroup$ $z'(t)=2ir{}{}$ $\endgroup$ – Nosrati Oct 1 '17 at 10:39
  • $\begingroup$ pull out $r$ and let $1+i(2t-1)=u$ $\endgroup$ – Nosrati Oct 1 '17 at 10:48
  • $\begingroup$ So much sloppiness in this... The phrase "polygonal line" is distracting when one only has a segment, the case to be treated separately is $n=-1$, not $n=1$ (and one can find distressing that two answerers do not correct this but, for no apparent reason, limit themselves to $n$ nonnegative), some factor $i$ disappears, you do not say why finding an antiderivative of a polynomial function is a problem at all... and so on. $\endgroup$ – Did Oct 4 '17 at 7:55
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For $n\in \mathbb{N}$ $$\int_{C} (z-z_0)^n dz = \int_{0}^{1} r^n[1+i(2t-1)]^n 2ridt$$ Now let $1+i(2t-1)=u$ then $2idt=du$ and \begin{align} \int_{C} (z-z_0)^n dz &= r^{n+1}\int_{1-i}^{1+i} u^n du \\ &= r^{n+1}\dfrac{(1+i)^{n+1}-(1-i)^{n+1}}{n+1} \\ &= \dfrac{2i}{n+1}(r\sqrt{2})^{n+1}\sin\dfrac{\pi(n+1)}{4} \end{align}

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The answer is trivial using the fact that the integrand in question has a complex anti-derivative. $ \def\lfrac#1#2{{\large\frac{#1}{#2}}} $

Take any natural $n$. (There is no need or reason to consider whether $n = 1$.)

Take any variable $z$ varying with real parameter $t$.

Then $\lfrac{d(\lfrac1{n+1}(z-z_0)^{n+1})}{dz} = (z-z_0)^n$ everywhere. (See this for a list of such facts!)

Thus $\int (z-z_0)^n\ dz = \lfrac1{n+1}(z-z_0)^{n+1} + k$ for some (complex) constant $k$.

Thus $\int_C (z-z_0)^n\ dz = \left[ \lfrac1{n+1}(z-z_0)^{n+1} \right]_{z=z_0+r-ir}^{z_0+r+ir} = \cdots$. (Just substitute and simplify.)

It's up to you whether you want to express in trigonometric form as MyGlasses did or not. Notice that he/she also (implicitly) used the same fact that I stated above.

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