## anonymous 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. anonymous

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. anonymous

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

3. anonymous

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

4. amistre64

this appears to be the same question that was posted earlier

5. anonymous

@experimentX dude i could not do that :/

6. anonymous

and today is the last day :/

7. experimentX

which one??

8. anonymous

they are all connected

9. experimentX

i think i already did the first one.

10. anonymous

you found sqrt2-i(pi)/4

11. anonymous

i am not sure it is true or not

12. experimentX

what's the answer supposed to be??

13. anonymous

14. experimentX

for question no 1??

15. anonymous

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. anonymous

:) okey bro thanks again

23. anonymous

@mukushla @richyw @eliassaab any idea ?

24. anonymous

Have you had the residue theorem?

25. anonymous

yes i guess

26. anonymous

but i did not learn well :/

27. anonymous

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

28. anonymous

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. anonymous

Mimic my post above for 2 remaining lines.

30. anonymous

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

31. anonymous

@eliassaab ?

32. anonymous

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

33. anonymous

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

34. anonymous

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. anonymous

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

36. anonymous

ok i understand i guess thank u :)

37. anonymous

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

38. anonymous

log=ln

39. anonymous

thank u so much :)

40. anonymous

yw