## SkykhanFalcon 3 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

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