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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 $n=-1$)

Little Rookie
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  • $z'(t)=2ir{}{}$ – Nosrati Oct 01 '17 at 10:39
  • pull out $r$ and let $1+i(2t-1)=u$ – Nosrati Oct 01 '17 at 10:48
  • 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. – Did Oct 04 '17 at 07:55

1 Answers1

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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}

Nosrati
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