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

- anonymous

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

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.

- anonymous

and 5/2 is a constant too :)

- 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?

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## More answers

- anonymous

Can you write out the question using the equation editor?

- anonymous

oh, how do I pull that up?

- anonymous

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

- anonymous

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

- anonymous

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

- 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?

- 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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

and x = t

- anonymous

Yes, that's what I was going to suggest

- 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

- 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)

- anonymous

could that be right for the distributive part??

- anonymous

just let me see

- anonymous

it's past my bedtime

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

yes! :-)

- 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?

- anonymous

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

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

- anonymous

oh forget mu'(t)...

- anonymous

late

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

yes, have that...

- anonymous

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

- anonymous

Agreed?

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

That should be mistake-free

- 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.

- anonymous

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

- anonymous

Happy so far?

- anonymous

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

- anonymous

ok

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- 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.

- anonymous

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

- 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?

- anonymous

I got some like terms when I distributed so combined them

- 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.

- anonymous

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

- anonymous

I'm double-checking my expansion.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

So it's all good to go...

- 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?

- anonymous

That looks right.

- anonymous

Now sub your limits.

- anonymous

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

- anonymous

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

- anonymous

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

- 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.

- anonymous

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

- anonymous

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

- anonymous

sorry ^ 1/3

- anonymous

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

- anonymous

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

- anonymous

So you're fine now?

- anonymous

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

- anonymous

yes

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

Yes

- anonymous

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

- anonymous

thanks, I will! :-)

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