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Patel, Patel... I want PATEL SOLUTIONS Plz, they have more q's. I especially want them for complex numbers.

... I thought the first one was out of the syllabus :) or should i change to a :( ?

... k thanx all . Thats pretty interesting too(ill remember that). Im talking about the argument thing. hahahahahahahahahahhaa, that questions too easy. I apologise for asking that question.

the note on the bottom of page 66 in cambridge , says: "A is excluded from the locus in each case , since z# 1+2i as arg0 is undefined" z=0 + 0i argz= arctan(0/0) = pi/2 isnt it? or when they say undefined are they meaning [Like for figure 2.48 b) ] , arg0 # pi/6 ( that is when z= 1+2i)...

... Sorry i wasnt clear cscx = cosecx

.... Ah sorry i wasnt clear, i wanted help lol. U CANT USE TRIG SUBSTITUTION AT ALL. Lazurus u could do that if it was a definite integral.

...... Is this for Motors and Generators or Ideas to Implementations? I dont remember that in motors.

Integrate (x^2+1)^1/2 Integrate (4-x^2)^1/2 Integrate x^2(1+x^2)^1/2 You have to do these by parts. U cant use substitution and for the first q ,the book tells u a hint. Hint: [(x^2+1)^1/2]dx= [(x^2+1)^1/2 * 1 ]dx u cant use x=tanu. Also for the first one, u get to a stage when...

haha.. haha k. Spicegirl- Today i saw something cool . LOL. I saw a Double Integral and Directional derivative stuff. Have u done that stuff?

haha haha so why would we want u'se as tutors , if we can get ur tutors as tutors lol. JK

Okay I solved it.. First delete my other post about prove xyz= a^3 + b ^3 , ive posted too much in it. In that topic i asked Integrate xcsc^2 x partially. I meant for to ask it in new topic. Anyway i solved it. I rearanged d/dx(u/v) = (u'v - v'u)v^2 = 1/v * u' - u/v^2 * v' when u...

Can some Intergrate xcsc^2 x by parts? :) - Someone can

:) cool, thanx alll. I get it now. :) thats a cool way spicegirl , but i think the other way will come to me quicker. And Ill remember that. Maybe it will be usefull for another q.

Ooops. My understanding is u can let z equal something as long as it doesnt have a and b from the z= a + bw^2. I meant since z is already expressed in terms of a and b. btw sorry for posting too much. Delete this after if u want.

... I still dont get why u would let z= -a/b. I know it gives the desired result .... so are u saying u can let z= (a^2+1)/(b+1) or something if u wanted to , but it wont get u anywhere? My understanding is u can let z equal something as long as it doesnt have a and b from the z= a + bw^2...

oh yeah.. oh yeah i was lost on that q, so i just did what Mc Lake did lol. HAHA Do u think ill get any marks out of 4 for that q? Kane- I got Hammered completly - I want my Revenge.

thanks... But why does z= -a/b ? howd u get it? I cant think today. And did u read the request :) ? Hope im not asking too much

Given w, w^2 are the two complex roots of Unity. Prove xyz = a^3 + b^3 if x = a + b y= a +bw z= a +bw^2 Im lost. I got screwed in todays test :( ahhhhh A Request: Spice Girl and OldMan: Could u'se plz pick hard q's from cambridge or patel and solve them on the board...

Ur not really showing it, are u? I see what u have done but the problem is: show that { [ (1-sin2x) over (1+ sin2x)] dx = { (tan square x)dx. what u did is u found a=pi/4 , now how does this help to show [ (1-sin2x) over (1+ sin2x)]dx =tan^2xdx ? guesswho ,note:(1-cos2x /(1+ cos2x)...

Heres something lol. P=I^2*R (since P=VI , V=IR) P=Et , E is Energy , P is power therefore E=P/t = I^2*R/t And Resistance is Proportional to L(As length of wire increases , Resistance Increases). Let us say this greater resistance in the longer wire in comparison(or can u say...