## SkykhanFalcon 2 years ago $\int\limits_{C}^{}dz/z$ where c consists of three line segments: From z = 1 to z = 1-i From z = 1-i to z = -1-i From z = -1-i to z = -1

1. agentx5

At first I though this wasn't going to be so hard, but then I read the sub-instructions >_< @amistre64 , @TuringTest , any insight on this one guys? PS: You did post ALL the contextual information for this one this time right @SkykhanFalcon ? ;-) (/me remembers your last super-challenging differentials question)

2. agentx5

There's no special initial conditions, restrictions, or anything like that right?

3. SkykhanFalcon

:) yeah i remember, for now there is not :)

4. amistre64

this appears to be the same question that was posted earlier

5. SkykhanFalcon

@experimentX dude i could not do that :/

6. SkykhanFalcon

and today is the last day :/

7. experimentX

which one??

8. SkykhanFalcon

they are all connected

9. experimentX

i think i already did the first one.

10. SkykhanFalcon

you found sqrt2-i(pi)/4

11. SkykhanFalcon

i am not sure it is true or not

12. experimentX

what's the answer supposed to be??

13. SkykhanFalcon

final answer should be -(pi)*i

14. experimentX

for question no 1??

15. SkykhanFalcon

no they are connected each other book says only an answer and it is -(pi)*i

16. experimentX

O ... let's see for 2

17. experimentX

|dw:1343076384065:dw|

18. experimentX

|dw:1343076480604:dw|

19. experimentX

|dw:1343076552517:dw|

20. experimentX

|dw:1343076614227:dw|

21. experimentX

|dw:1343076688932:dw| sorry man ... i couldn't get any better

22. SkykhanFalcon

:) okey bro thanks again

23. SkykhanFalcon

@mukushla @richyw @eliassaab any idea ?

24. eliassaab

Have you had the residue theorem?

25. SkykhanFalcon

yes i guess

26. SkykhanFalcon

but i did not learn well :/

27. eliassaab

If you know the Residue Theorem you can cut your work from computing three line integrals to one.

28. eliassaab

Your line integral on the first line is z = 1 +i y dz = i dy $\int_0^{-1} \frac{i}{1+i y} \, dy=\frac{1}{4} (\ln(4)-i \pi )$

29. eliassaab

Mimic my post above for 2 remaining lines.

30. SkykhanFalcon

hmm then will i continue with 1/4(ln4.... or i always use the i/1+iy

31. SkykhanFalcon

@eliassaab ?

32. eliassaab

for the second integral z= x -I dz = dx $\int_1^{-1} \frac{1}{x-i} \, dx=-\frac{1}{2} (i \pi )$

33. SkykhanFalcon

how can i understand the integral part for ex. why the second one become 1/x-i

34. eliassaab

How much is y on the second line? -1 How much is x on the seconfd line? x from 1 to -1 z= x -i

35. eliassaab

On the third line z= -1 + i y dz= i dy $\int_{-1}^0 \frac{i}{-1+i y} \, dy$

36. SkykhanFalcon

ok i understand i guess thank u :)

37. eliassaab

$\int_{-1}^0 \frac{i}{-1+i y} \, dy=-\frac{1}{4} i (\pi -2 i \log (2))$

38. eliassaab

log=ln

39. SkykhanFalcon

thank u so much :)

40. eliassaab

yw