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

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

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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.
and 5/2 is a constant too :)
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?

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Can you write out the question using the equation editor?
oh, how do I pull that up?
this is a new laptop and not sure what is on it exactly!!
(1/3) t^3 + (2/3) i t^(5/2) + i t^(3/2)
The equation editor is on the bottom of the message box. It says, "Equation"
\[\frac{1}{3}t^3+\frac{2}{3} i t^{5/2}+i t ^{3/2}\]...is that the thing you have to integrate or is that an answer to something?
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
ok, I hit the eqn button but my tyoibng didnt come out like yours!
actually, the "contour" ...this is in Complex Analysis.
Yeah, I guessed. The path you have to integrate over is \[y=\sqrt{x}\]right?
and x = t
Yes, that's what I was going to suggest
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
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)
could that be right for the distributive part??
just let me see
it's past my bedtime
oh, sorry...it's only past 10am here (east coast)
all of these fractional exponents and fractional constants are givng me a headache and making my eyes cross!
Were you actually given the function to integrate as\[\mu (z)=x^2+y^2+2i x y \]?
yes! :-)
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?
yep! that's sort of what the prof wrote...
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,\[\mu (t)=t^2+t+2i t^{3/2} \rightarrow \mu'(t)=2t+1+3 i t^{1/2} \]
oh forget mu'(t)...
late
\[z'(t)=1+i \frac{1}{2}t\]
woah....where did the 2 i t^ (3/2) come from?
So your integral becomes,\[\int\limits_{0}^{1}\left( t^2+t+2\it^{3/2} \right)\left( 1+i \frac{1}{2}t^{-1/2} \right) dt\]
yes, have that...
2it^(3/2) from subbing in x=t, y=t^(1/2) into 2ixy
Agreed?
got you, you were just wrtiting the original problem for me...thought htat had come from all the distributing...
The "i t" didn't show up for some reason in the integral above in the first bracket.
\[\int\limits_{0}^{1}\left( t^2+t+2 i t^{2/3} \right)\left( 1+i \frac{1}{2}t^{-1/2} \right)dt\]
That should be mistake-free
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.
\[\int\limits_{0}^{1}u(t)dt+i \int\limits_{0}^{1}v(t)dt\]
Happy so far?
yeah, I can finally type again...froze up or something
ok
ok, my question is, did I do the distributive math correctly?
You can check that with wolframalpha - have you heard of that?
way back up at 43 minutes ago... no, but I'll check it out...
Don't be surprised/put off if you distribute and get something rotten.
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.
actually, not bad... now can't find it, just had it
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?
I got some like terms when I distributed so combined them
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.
You could come up with something like that before you integrate.
I'm double-checking my expansion.
sorry, didn't catch that...I hacethe exponent as 3/2
\[t^2+i \left( \frac{5}{2}t^{3/2}+\frac{1}{2}t^{1/2} \right)\]
You get 3/2, where I get 5/2
yeah! that's what I have !!!!!!! (just saw my mistake, yes the 3/2 should be a 5/2
So it's all good to go...
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?
That looks right.
Now sub your limits.
wahoo!@! so just plug in the 1 and 0 and evaluate as usual?
\[\frac{1}{3}+i \frac{4}{3}\]
so (1/3) + 2i???
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.
oos! wherer did your i (4/3) come from
Look at the imaginary part of your result. It's \[t^{5/2}+\frac{1}{2}t^{3/2}\]
sorry ^ 1/3
(1)^{5/2}+(1/3)*(1)^{3/2} = 3/3 + 1/3 = 4/3
fool thing froze again!! I saw my mistake while you were typing! duh, forgot to multiply in the 1/3
So you're fine now?
so I had (1/3)+ i + (1/3)i to get the two terms (1/3) + (4/3) i
yes
and when t=0, the result for the other limit is 0.
Sure am!! Now just have to do the same thing for contours y=x and y=x^2 and compare all 3 answers.
Well, good luck. It's the same deal :P
sorry, computer delayed again! THANK YOU soooooo MUCH FOR HELPING ME!! ;- )
No probs. I'm hoping you fanned me ;)
sure did!!!!! :-) and this site is free????
Yes
Okay, I'm off for sleep...have fun with complex analysis ;)
thanks, I will! :-)

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