• We need your support!

    We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
    Click here to Donate Now (View Announcement)

Mathematics: Post your doubts here!

Messages
21
Reaction score
9
Points
13
The complex number 2/-1+i is denoted by u.

1) Find the modulus and argument of u and u^2.

Can someone help me to solve this question? :)
 
Messages
869
Reaction score
374
Points
73
The complex number 2/-1+i is denoted by u.

1) Find the modulus and argument of u and u^2.

Can someone help me to solve this question? :)
ok, when you have a question in that form u need to change it to the form x+ yi and to do this u multiply the denominator by the conjugate (1-i) and u multiply the nominator by the conjugate so..

2/1+i = 2(1-i)/(1+i)(1-i) = 2-2i/(1^2 + 1^2) = 1-i

the mod of |1-i| = √(1^2 + 1^2) = √2
arg (z) = tan^-1 (-1/1)
u^2 = (1-i)(1-i) = -2i

that's all, hope u got it bro!
 
Messages
188
Reaction score
147
Points
53
Hey. I need some help in Finding the area bounded by a curve and the y-axis and Volume of revolution about the y-axis. I study using examsolutions.net and this topic isn't there yet. I have the CIE A-Level Mathematics Pure 1 book with a black cover and some sea shell thingy, and I cant find this topic there either. Any help would be appreciated.
 
Messages
878
Reaction score
1,474
Points
153
Hey. I need some help in Finding the area bounded by a curve and the y-axis and Volume of revolution about the y-axis. I study using examsolutions.net and this topic isn't there yet. I have the CIE A-Level Mathematics Pure 1 book with a black cover and some sea shell thingy, and I cant find this topic there either. Any help would be appreciated.

i assume u have no problem with rev about x-axis..also, I can only give a brief explanation here but i hope u get it:

volume around x-axis = pi int(y^2) dx, and then put the limits
if the question says find volume about y-axis, simply make 'x' the subject and then V= pi int(x^2) dx
Now, if you are given the limits for y, simply put the values and get the result..If you are only given the values of x, then put them in the equation and get the limits for y.

e.g
y=x^2 ; Find the volume of rev when the region between x=1 and x=5 and the curve 'y' is rotated about y-axis.

> make x subject: x= √y
> we are only given x-coordinates but to revolve around y-axis we nede y-coordinates: when x=1, y=1..when x= 5, y=25.....so our limits are 1 and 25
> square and integrate: V= pi int(√y)^2 dy =>V=pi int(y) dy ==> V=pi (y^2 /2) with limits 1 , 25.
the rest is simple calculation...hope i helped
 
Messages
164
Reaction score
111
Points
43
i assume u have no problem with rev about x-axis..also, I can only give a brief explanation here but i hope u get it:

volume around x-axis = pi int(y^2) dx, and then put the limits
if the question says find volume about y-axis, simply make 'x' the subject and then V= pi int(x^2) dx
Now, if you are given the limits for y, simply put the values and get the result..If you are only given the values of x, then put them in the equation and get the limits for y.

e.g
y=x^2 ; Find the volume of rev when the region between x=1 and x=5 and the curve 'y' is rotated about y-axis.

> make x subject: x= √y
> we are only given x-coordinates but to revolve around y-axis we nede y-coordinates: when x=1, y=1..when x= 5, y=25.....so our limits are 1 and 25
> square and integrate: V= pi int(√y)^2 dy =>V=pi int(y) dy ==> V=pi (y^2 /2) with limits 1 , 25.
the rest is simple calculation...hope i helped

I have written extensive material regarding volume of revolution on my website here:

http://www.whitegroupmaths.com/2010/02/understanding-matters-5.html

Hope it helps. Peace.
 
Messages
29
Reaction score
4
Points
13
Can someone help me out with this integration sum?

Use a substitution of the form ax+b=u to find the integral for x/(2x+3)

The answer's x/2 - 3 ln |2x+3| +k
But I got x/2 +3/4 - 3 ln |2x+3| +k

I'd very much appreciate if you showed all the steps please.
 
Messages
878
Reaction score
1,474
Points
153
Can someone help me out with this integration sum?

Use a substitution of the form ax+b=u to find the integral for x/(2x+3)

The answer's x/2 - 3 ln |2x+3| +k
But I got x/2 +3/4 - 3 ln |2x+3| +k

I'd very much appreciate if you showed all the steps please.

taking sub u=2x+3:
du/dx = 2
dx=du/2

x=(u-3)/2

insert values in original eq:
=> ⌡{u-3)/2}/u du/2
take 1/4 outside: 1/4 ⌡(u-3)/u du
simplify: 1/4 ⌡1 - (3/u) du
integrate: 1/4 [u - 3lnu] + k
replace u by x: 1/4[(2x+3) - 3ln|2x+3|] +k
simplify: x/2 + 3/4 -3/4 ln|2x+3| + k

mine includes one more term too..are u sure that answer is correct??
 
Messages
188
Reaction score
147
Points
53
i assume u have no problem with rev about x-axis..also, I can only give a brief explanation here but i hope u get it:

volume around x-axis = pi int(y^2) dx, and then put the limits
if the question says find volume about y-axis, simply make 'x' the subject and then V= pi int(x^2) dx
Now, if you are given the limits for y, simply put the values and get the result..If you are only given the values of x, then put them in the equation and get the limits for y.

e.g
y=x^2 ; Find the volume of rev when the region between x=1 and x=5 and the curve 'y' is rotated about y-axis.

> make x subject: x= √y
> we are only given x-coordinates but to revolve around y-axis we nede y-coordinates: when x=1, y=1..when x= 5, y=25.....so our limits are 1 and 25
> square and integrate: V= pi int(√y)^2 dy =>V=pi int(y) dy ==> V=pi (y^2 /2) with limits 1 , 25.
the rest is simple calculation...hope i helped
Thank you very much. Can you please tell me about the AREA bounded by a curve and the y-axis too?
 
Messages
878
Reaction score
1,474
Points
153
Thank you very much. Can you please tell me about the AREA bounded by a curve and the y-axis too?
Its the same thing except that you don't have to square and don't need to put pi....but there are certain cases where u have to be careful e.g the area bounded by the sine curve and x-axis..u have to calculate the negative and positive part separately..u will have to do the same for y-axis
 
Messages
188
Reaction score
147
Points
53
i assume u have no problem with rev about x-axis..also, I can only give a brief explanation here but i hope u get it:

volume around x-axis = pi int(y^2) dx, and then put the limits
if the question says find volume about y-axis, simply make 'x' the subject and then V= pi int(x^2) dx
Now, if you are given the limits for y, simply put the values and get the result..If you are only given the values of x, then put them in the equation and get the limits for y.

e.g
y=x^2 ; Find the volume of rev when the region between x=1 and x=5 and the curve 'y' is rotated about y-axis.

> make x subject: x= √y
> we are only given x-coordinates but to revolve around y-axis we nede y-coordinates: when x=1, y=1..when x= 5, y=25.....so our limits are 1 and 25
> square and integrate: V= pi int(√y)^2 dy =>V=pi int(y) dy ==> V=pi (y^2 /2) with limits 1 , 25.
the rest is simple calculation...hope i helped
What's funny is that ( y ) turned into (y)
 
Messages
188
Reaction score
147
Points
53
Its the same thing except that you don't have to square and don't need to put pi....but there are certain cases where u have to be careful e.g the area bounded by the sine curve and x-axis..u have to calculate the negative and positive part separately..u will have to do the same for y-axis
Can you do an example too?
 
Messages
21
Reaction score
9
Points
13
ok, when you have a question in that form u need to change it to the form x+ yi and to do this u multiply the denominator by the conjugate (1-i) and u multiply the nominator by the conjugate so..

2/1+i = 2(1-i)/(1+i)(1-i) = 2-2i/(1^2 + 1^2) = 1-i

the mod of |1-i| = √(1^2 + 1^2) = √2
arg (z) = tan^-1 (-1/1)
u^2 = (1-i)(1-i) = -2i

that's all, hope u got it bro!
thanks!
 
Top