## anonymous 5 years ago ok, so now how would you integrate (5/2) i t ^ (3/2)? Same way? What happens to the i?

1. anonymous

The i stays put (it's a constant), assuming i is the imaginary number and you're integrating with respect to t. So you integrate normally here.

2. anonymous

and 5/2 is a constant too :)

3. anonymous

that's what I thought....okkkkk, so can you check my work out on this one? I had a really long integral problem and think I did all the simplifying corectly but I end up with (1/3) t^3 + (2/3) i t^(5/2) + i t^(3/2) between 0 and 1...do i just plug in the 1 and then the 0 for t and the i remains in the answer?

4. anonymous

Can you write out the question using the equation editor?

5. anonymous

oh, how do I pull that up?

6. anonymous

this is a new laptop and not sure what is on it exactly!!

7. anonymous

(1/3) t^3 + (2/3) i t^(5/2) + i t^(3/2)

8. anonymous

The equation editor is on the bottom of the message box. It says, "Equation"

9. anonymous

$\frac{1}{3}t^3+\frac{2}{3} i t^{5/2}+i t ^{3/2}$...is that the thing you have to integrate or is that an answer to something?

10. anonymous

actually, I have to intergrate the curve from (0,0) to (1,1) where y = sq rt of x using m(z) dz where m(z) is x^2 +y^2 + 2ixy

11. anonymous

ok, I hit the eqn button but my tyoibng didnt come out like yours!

12. anonymous

actually, the "contour" ...this is in Complex Analysis.

13. anonymous

Yeah, I guessed. The path you have to integrate over is $y=\sqrt{x}$right?

14. anonymous

and x = t

15. anonymous

Yes, that's what I was going to suggest

16. anonymous

and I THINK I did the distributive correctly...the dz/dt at the end of the original integral changes to (dx/dt) + i (dy/dt) and and i came up with integral of [t^2 + t +2i t^(3/2)] [ 1 + i (1/(2sqrt t)] dt

17. anonymous

I double distributed and before simplifyling, came up with integral of t^2 + t + 2i t^(3/2) + i t^2 (1/(2sqrt t)) + i t (1/(2sqrt t) + 2i t^(3/2) (i) ((t^(1/2)/2)

18. anonymous

could that be right for the distributive part??

19. anonymous

just let me see

20. anonymous

it's past my bedtime

21. anonymous

oh, sorry...it's only past 10am here (east coast)

22. anonymous

all of these fractional exponents and fractional constants are givng me a headache and making my eyes cross!

23. anonymous

Were you actually given the function to integrate as$\mu (z)=x^2+y^2+2i x y$?

24. anonymous

yes! :-)

25. anonymous

Okay, you have $z=x+iy$as your variable and over this path, x and y will take values,$x=t, y=t^{1/2}$so$z=z(t)=t+i t^{1/2}$Your integral is$\int\limits_{\gamma}^{}\mu (z) dz=\int\limits_{0}^{1}\mu(z(t))z'(t)dt$where gamma is our defined path. Agree so far?

26. anonymous

yep! that's sort of what the prof wrote...

27. anonymous

Now, since mu(z) is as you defined it above in terms of x and y, and now that we have those x and y in terms of t along this contour, mu as a function of t is,$\mu (t)=t^2+t+2i t^{3/2} \rightarrow \mu'(t)=2t+1+3 i t^{1/2}$

28. anonymous

oh forget mu'(t)...

29. anonymous

late

30. anonymous

$z'(t)=1+i \frac{1}{2}t$

31. anonymous

woah....where did the 2 i t^ (3/2) come from?

32. anonymous

So your integral becomes,$\int\limits_{0}^{1}\left( t^2+t+2\it^{3/2} \right)\left( 1+i \frac{1}{2}t^{-1/2} \right) dt$

33. anonymous

yes, have that...

34. anonymous

2it^(3/2) from subbing in x=t, y=t^(1/2) into 2ixy

35. anonymous

Agreed?

36. anonymous

got you, you were just wrtiting the original problem for me...thought htat had come from all the distributing...

37. anonymous

The "i t" didn't show up for some reason in the integral above in the first bracket.

38. anonymous

$\int\limits_{0}^{1}\left( t^2+t+2 i t^{2/3} \right)\left( 1+i \frac{1}{2}t^{-1/2} \right)dt$

39. anonymous

That should be mistake-free

40. anonymous

You have to expand that out and group the terms into an integral for the real and imaginary components, and then do Riemann integration ('normal' integration) on each interval.

41. anonymous

$\int\limits_{0}^{1}u(t)dt+i \int\limits_{0}^{1}v(t)dt$

42. anonymous

Happy so far?

43. anonymous

yeah, I can finally type again...froze up or something

44. anonymous

ok

45. anonymous

ok, my question is, did I do the distributive math correctly?

46. anonymous

You can check that with wolframalpha - have you heard of that?

47. anonymous

way back up at 43 minutes ago... no, but I'll check it out...

48. anonymous

Don't be surprised/put off if you distribute and get something rotten.

49. anonymous

It happens all the time in complex analysis. It's only ever nice when the give you "Integrate z over the unit circle" or some pos.

50. anonymous

51. anonymous

could I actually come up with something like t^2 + (3/2) i t ^(3/2) + (1/2 i t^ (1/2) before I integrate?

52. anonymous

I got some like terms when I distributed so combined them

53. anonymous

Before I answer that, again, there's a mistake in the last integral where I declared, "No mistakes"...it's late and this site is awkward. The power on the 2it in the first bracket should be 3/2, not 2/3.

54. anonymous

You could come up with something like that before you integrate.

55. anonymous

I'm double-checking my expansion.

56. anonymous

sorry, didn't catch that...I hacethe exponent as 3/2

57. anonymous

$t^2+i \left( \frac{5}{2}t^{3/2}+\frac{1}{2}t^{1/2} \right)$

58. anonymous

You get 3/2, where I get 5/2

59. anonymous

yeah! that's what I have !!!!!!! (just saw my mistake, yes the 3/2 should be a 5/2

60. anonymous

So it's all good to go...

61. anonymous

yes, so when integrating , I would get (1/3) t^3 for the real part and i [(t^5/2) + (1/3) t^(3/2) for the imaginary part?

62. anonymous

That looks right.

63. anonymous

64. anonymous

wahoo!@! so just plug in the 1 and 0 and evaluate as usual?

65. anonymous

$\frac{1}{3}+i \frac{4}{3}$

66. anonymous

so (1/3) + 2i???

67. anonymous

Yes, you can evaluate as usual since you've converted a contour integral to one over an interval...so it's 'normal', for lack of a better term.

68. anonymous

oos! wherer did your i (4/3) come from

69. anonymous

Look at the imaginary part of your result. It's $t^{5/2}+\frac{1}{2}t^{3/2}$

70. anonymous

sorry ^ 1/3

71. anonymous

(1)^{5/2}+(1/3)*(1)^{3/2} = 3/3 + 1/3 = 4/3

72. anonymous

fool thing froze again!! I saw my mistake while you were typing! duh, forgot to multiply in the 1/3

73. anonymous

So you're fine now?

74. anonymous

so I had (1/3)+ i + (1/3)i to get the two terms (1/3) + (4/3) i

75. anonymous

yes

76. anonymous

and when t=0, the result for the other limit is 0.

77. anonymous

Sure am!! Now just have to do the same thing for contours y=x and y=x^2 and compare all 3 answers.

78. anonymous

Well, good luck. It's the same deal :P

79. anonymous

sorry, computer delayed again! THANK YOU soooooo MUCH FOR HELPING ME!! ;- )

80. anonymous

No probs. I'm hoping you fanned me ;)

81. anonymous

sure did!!!!! :-) and this site is free????

82. anonymous

Yes

83. anonymous

Okay, I'm off for sleep...have fun with complex analysis ;)

84. anonymous

thanks, I will! :-)