At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

and 5/2 is a constant too :)

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"

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

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)

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

yes! :-)

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

oh forget mu'(t)...

late

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

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

yes, have that...

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

Agreed?

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

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

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.

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

I got some like terms when I distributed so combined them

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...

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???

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! :-)