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 $n=-1$)

  • $\begingroup$ $z'(t)=2ir{}{}$ $\endgroup$ – 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

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}


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.