1.6 Q5

[Nathan]

Question: $\int_y Re(z) dz$ where $y$ is the line segment from 1 to $i$.

I can't get the same answer as the one in the textbook.

Answer in textbook: $\frac{1}{2}(i-1)$

My answer:
$y(t)=$
$= 1 + (i - 1)t$
$= (1 - t) + it$

$y'(t) = i - 1$

$\int_y Re(z) dz=$
=$\int_1^i (1-t)(i-1)dt$
$=\int_1^i i - 1 - ti + t dt$
$=ti - t - \frac{t^2i}{2} + \frac{t^2}{2}|^i_1$
$=-1 -i + \frac{i}{2} - \frac{1}{2} - (i - 1-\frac{i}{2} + \frac{1}{2})$
$=-i - 1$

What is wrong with my answer?

[smarques]

Hi Nathan.

The mistake seems to be in your bounds of integration, not the process itself. With the way you parameterized the line, the integral should be from 0 to 1, not 0 to i.