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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and 5/2 is a constant too :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that'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?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Can you write out the question using the equation editor?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh, how do I pull that up?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this is a new laptop and not sure what is on it exactly!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(1/3) t^3 + (2/3) i t^(5/2) + i t^(3/2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The equation editor is on the bottom of the message box. It says, "Equation"

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{3}t^3+\frac{2}{3} i t^{5/2}+i t ^{3/2}\]...is that the thing you have to integrate or is that an answer to something?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0actually, 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, I hit the eqn button but my tyoibng didnt come out like yours!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0actually, the "contour" ...this is in Complex Analysis.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, I guessed. The path you have to integrate over is \[y=\sqrt{x}\]right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, that's what I was going to suggest

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0could that be right for the distributive part??

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh, sorry...it's only past 10am here (east coast)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0all of these fractional exponents and fractional constants are givng me a headache and making my eyes cross!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Were you actually given the function to integrate as\[\mu (z)=x^2+y^2+2i x y \]?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, 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?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yep! that's sort of what the prof wrote...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now, 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,\[\mu (t)=t^2+t+2i t^{3/2} \rightarrow \mu'(t)=2t+1+3 i t^{1/2} \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[z'(t)=1+i \frac{1}{2}t\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0woah....where did the 2 i t^ (3/2) come from?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So your integral becomes,\[\int\limits_{0}^{1}\left( t^2+t+2\it^{3/2} \right)\left( 1+i \frac{1}{2}t^{1/2} \right) dt\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.02it^(3/2) from subbing in x=t, y=t^(1/2) into 2ixy

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0got you, you were just wrtiting the original problem for me...thought htat had come from all the distributing...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The "i t" didn't show up for some reason in the integral above in the first bracket.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{1}\left( t^2+t+2 i t^{2/3} \right)\left( 1+i \frac{1}{2}t^{1/2} \right)dt\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That should be mistakefree

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{1}u(t)dt+i \int\limits_{0}^{1}v(t)dt\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, I can finally type again...froze up or something

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, my question is, did I do the distributive math correctly?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can check that with wolframalpha  have you heard of that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0way back up at 43 minutes ago... no, but I'll check it out...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Don't be surprised/put off if you distribute and get something rotten.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0actually, not bad... now can't find it, just had it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0could I actually come up with something like t^2 + (3/2) i t ^(3/2) + (1/2 i t^ (1/2) before I integrate?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I got some like terms when I distributed so combined them

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Before 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You could come up with something like that before you integrate.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm doublechecking my expansion.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry, didn't catch that...I hacethe exponent as 3/2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[t^2+i \left( \frac{5}{2}t^{3/2}+\frac{1}{2}t^{1/2} \right)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You get 3/2, where I get 5/2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah! that's what I have !!!!!!! (just saw my mistake, yes the 3/2 should be a 5/2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So it's all good to go...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, 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?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wahoo!@! so just plug in the 1 and 0 and evaluate as usual?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{3}+i \frac{4}{3}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oos! wherer did your i (4/3) come from

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Look at the imaginary part of your result. It's \[t^{5/2}+\frac{1}{2}t^{3/2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(1)^{5/2}+(1/3)*(1)^{3/2} = 3/3 + 1/3 = 4/3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0fool thing froze again!! I saw my mistake while you were typing! duh, forgot to multiply in the 1/3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so I had (1/3)+ i + (1/3)i to get the two terms (1/3) + (4/3) i

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and when t=0, the result for the other limit is 0.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sure am!! Now just have to do the same thing for contours y=x and y=x^2 and compare all 3 answers.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, good luck. It's the same deal :P

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry, computer delayed again! THANK YOU soooooo MUCH FOR HELPING ME!! ; )

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No probs. I'm hoping you fanned me ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sure did!!!!! :) and this site is free????

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, I'm off for sleep...have fun with complex analysis ;)
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