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Mathematics: Post your doubts here!

Rutzaba

Rutzaba

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as far as i think there will be twom circles made one with centre point 0,0 and modz = z-(0 +0i)
the second circle would be mid z-2 take it in general form ul get z- (2+0i) so 2 and zero as centre point.
this must be satisfied that the shaded region wud be in accordance with the in equality
the second thing is argument ... the region wch will be shaded will be between 0 and pi/4...
i no thisisnt a satisfactory explanation... but i have forgotten this chater completely ...:(
 
applepie1996

applepie1996

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Rutzaba said:
as far as i think there will be twom circles made one with centre point 0,0 and modz = z-(0 +0i)
the second circle would be mid z-2 take it in general form ul get z- (2+0i) so 2 and zero as centre point.
this must be satisfied that the shaded region wud be in accordance with the in equality
the second thing is argument ... the region wch will be shaded will be between 0 and pi/4...
i no thisisnt a satisfactory explanation... but i have forgotten this chater completely ...:(
Click to expand...
its okay :D
atleast you tried :)
thanks :D
 
H

Hassi123

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applepie1996 said:
you first plot points A and B on the diagram and i am assuming you know how to do that
next you find the mid point between A and B
you get -1/2-1/2i
this will be the centre of the circle
no you have to find the radius of the circle
find the distance between the two points using sqrt[(x2-x1)^2+(y2-y1)^2 and you get an answer
square root of answer will give you the radius :)
Click to expand...
Ohh, i get all of it, but im still confused in one small thing, how do you find the point w^2 ( I know how to find mod and arg of w^2 but not the point )
 
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PhyZac

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minie23 said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_12.pdf

Hey there ! I need a lil help for no. 4 ! Please help :p
Click to expand...
find the gradient of line AB
(3-1)/(-1-3)
= -1/2

now find the gradient of perpendicular line ( tht is L2)
2

find the line equation
(y-1) = 2(x-3)
y= 2x -5

find gradient of OB ( which is similar to L1)
(1-0)/(3-0)
= 1/3

find equation of L1
(y-3) = 1/3 (x-(-1))
3y - 9 = x + 1
3y = x + 10

Find intersection between L1 and L2
by simultaneous equation.

sub the yellow to blue
3(2x-5)=x +10
6x - 15 = x + 10
5x = 25
x = 5

now sub this to yellow
y=2(5) -5
y=5

C is (5,5)
 
Rutzaba

Rutzaba

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PhyZac said:
find the gradient of line AB
(3-1)/(-1-3)
= -1/2

now find the gradient of perpendicular line ( tht is L2)
2

find the line equation
(y-1) = 2(x-3)
y= 2x -5

find gradient of OB ( which is similar to L1)
(1-0)/(3-0)
= 1/3

find equation of L1
(y-3) = 1/3 (x-(-1))
3y - 9 = x + 1
3y = x + 10

Find intersection between L1 and L2
by simultaneous equation.

sub the yellow to blue
3(2x-5)=x +10
6x - 15 = x + 10
5x = 25
x = 5

now sub this to yellow
y=2(5) -5
y=5

C is (5,5)
Click to expand...
lol much? :p
 
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