I'm trying to solve

$$\int _{C} e^zdz $$

where C is the polygonal path consisting of the line segments from $z = 0$ to $z = 2$ and from $z = 2$ to $z = 1 + \pi it$

so I'm having difficulty finding the parametrization for $z = 2$ to $z = 1 + \pi it$

if I use the formula for a line segment $$z(t)=2(1-t)+(1+i\pi)t = 2-t+i\pi, \quad 0\le t\le1$$

but I tried to use the point-slope formula $$y-y_1=m(x-x_1)$$

For $z_1=2$, $x_1=2, y_1=0$ so $(2,0) $

and $z_2=1+i\pi$, $x_2=1, y_2=\pi$ so $(1,\pi)$, am i right?

so $y-\pi=-\pi(x-1)$

gives $$ y = -\pi x + 2\pi $$

Replace $t=x$ then

$$ y = -\pi t+2\pi $$

which gives coordinate form $(t, -\pi t+2\pi)$

Since the orientation does not match, I changed $t$ to $-t$ and got $(-t,t\pi+2 \pi)$

so $$z(t)= -t+\pi(t+2)i, \quad -2 \le t \le -1$$

I compared the result using first and second method, if my second approach is correct, it should give the same answer as the first one, but it doesn't.

So is my second approach wrong?
$$z(t)= -t+\pi ti + 2\pi i, \quad 2-t+i\pi t$$

If I am wrong can you tell me where is the wrong part? Thanks.

  • $\begingroup$ Why are you overusing <\br> ? Why even use it ? $\endgroup$ – A---B Nov 7 '17 at 10:05
  • $\begingroup$ The first is $2-t+i\pi$ or $2-t+i\pi t$ ? $\endgroup$ – Nosrati Nov 7 '17 at 10:16
  • $\begingroup$ @A---B to make it easier to read(?) $\endgroup$ – fiksx Nov 7 '17 at 10:19
  • $\begingroup$ @MyGlasses sorry typo, yes $2-t+i \pi t$ $\endgroup$ – fiksx Nov 7 '17 at 10:19
  • $\begingroup$ the first with $t=u+2$ gives $y=-u+\pi u i+2\pi i$ where $-2\leq u\leq-1$ is the same as you found in the second! $\endgroup$ – Nosrati Nov 7 '17 at 10:24

The two answers you have are the equivalent. The only difference is in the domain. The first has $t \in [0,1]$ and the second has $t \in [-2,-1]$

If you Let $t = u-2$ in the second answer, you get $$ z(u) = -(u-2) + \pi ui, -2 \le u-2 \le -1 $$


$$ z(u) = 2-u + \pi ui, \quad 0 \le u \le 1 $$

which is equal to the first answer.

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  • $\begingroup$ Thankyou so much!! So its the same right? $\endgroup$ – fiksx Nov 8 '17 at 4:37

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