Question

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

Answer 1

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.

Answer 2

and 5/2 is a constant too :)

Answer 3

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?

Answer 4

Can you write out the question using the equation editor?

Answer 5

oh, how do I pull that up?

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

and x = t

Answer 15

Yes, that's what I was going to suggest

Answer 16

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

Answer 17

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)

Answer 18

could that be right for the distributive part??

Answer 19

just let me see

Answer 20

it's past my bedtime

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

yes! :-)

Answer 25

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?

Answer 26

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

Answer 27

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,\[\mu (t)=t^2+t+2i t^{3/2} \rightarrow \mu'(t)=2t+1+3 i t^{1/2} \]

Answer 28

oh forget mu'(t)...

Answer 29

late

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

yes, have that...

Answer 34

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

Answer 35

Agreed?

Answer 36

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

Answer 37

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

Answer 38

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

Answer 39

That should be mistake-free

Answer 40

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.

Answer 41

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

Answer 42

Happy so far?

Answer 43

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

Answer 44

ok

Answer 45

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

Answer 46

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

Answer 47

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

Answer 48

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

Answer 49

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.

Answer 50

actually, not bad... now can't find it, just had it

Answer 51

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?

Answer 52

I got some like terms when I distributed so combined them

Answer 53

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.

Answer 54

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

Answer 55

I'm double-checking my expansion.

Answer 56

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

Answer 57

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

Answer 58

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

Answer 59

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

Answer 60

So it's all good to go...

Answer 61

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?

Answer 62

That looks right.

Answer 63

Now sub your limits.

Answer 64

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

Answer 65

\[\frac{1}{3}+i \frac{4}{3}\]

Answer 66

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

Answer 67

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.

Answer 68

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

Answer 69

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

Answer 70

sorry ^ 1/3

Answer 71

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

Answer 72

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

Answer 73

So you're fine now?

Answer 74

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

Answer 75

yes

Answer 76

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

Answer 77

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

Answer 78

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

Answer 79

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

Answer 80

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

Answer 81

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

Answer 82

Yes

Answer 83

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

Answer 84

thanks, I will! :-)