Back
to top
Questions from holiday revision... Watch
Announcements
Page 1 of 1
Go to first unread
Skip to page:
I was lucky enough to get 6 pages of holiday revision on Wednesday!
I'm having trouble with these..
*Show that tanA+cotA=2cosecA for all possible values of A.
*Find z if 3z-i=2+iz
I subbed in z=x+iy and got down to;
2x+y=i(-2y+x+1)
I don't know what to do now..
*Sketch and describe the locus of |z-1+i| is smaller or equal to 3.
I expanded that out and got;
|x+iy-1+i|<3
|(x-1)+i(y+1)|<3
and I don't know what to do...
*AB, Ac are chords of a circle such that AB=AC. P and Q are two points on the chord BC, and AP, AQ meet the circle again at R and S. Let angle ABC=a and angle BCS=b. Prove that the points P, Q, S and R are concyclic.
Here's what it looks like..
Any help would be great
I'm having trouble with these..
*Show that tanA+cotA=2cosecA for all possible values of A.
*Find z if 3z-i=2+iz
I subbed in z=x+iy and got down to;
2x+y=i(-2y+x+1)
I don't know what to do now..
*Sketch and describe the locus of |z-1+i| is smaller or equal to 3.
I expanded that out and got;
|x+iy-1+i|<3
|(x-1)+i(y+1)|<3
and I don't know what to do...
*AB, Ac are chords of a circle such that AB=AC. P and Q are two points on the chord BC, and AP, AQ meet the circle again at R and S. Let angle ABC=a and angle BCS=b. Prove that the points P, Q, S and R are concyclic.
Here's what it looks like..
Any help would be great
0
reply
Report
#2
1)
tanA + cotA
= sinA/cosA + cosA/sinA
=(sin^2A + coz^2A)/cosAsinA
= 1/cosAsinA
=1/(0.5sinA)
=2cosec2A
So i think you may have typed the question out wrong or i made a stupid mistake
2)Now what you do is equate real and imaginary parts:
Equating real parts: 2x + y = 0
Equating imaginary parts: -2y + x + 1 = 0
Giving two simultaneous equations which you can solve
(this is assuming that you have subbed in and rearranged correctly)
3)|(x-1)+i(y+1)|<3
sqrt((x-1)^2 + (y+1)^2) < 3
(x-1)^2 + (y+1)^2 < 9
Therefore it is a circle with centre (1,-i) and radius 3
Soz Im a bit tired and am going to go to bed now so i cant do the last one right now. Im not sure if you have rearanged your eqaution from qu 2 correctly, but Im to tired to check Havent been to sleep in about 38 hours 
. Iapologise for any errors
tanA + cotA
= sinA/cosA + cosA/sinA
=(sin^2A + coz^2A)/cosAsinA
= 1/cosAsinA
=1/(0.5sinA)
=2cosec2A
So i think you may have typed the question out wrong or i made a stupid mistake
2)Now what you do is equate real and imaginary parts:
Equating real parts: 2x + y = 0
Equating imaginary parts: -2y + x + 1 = 0
Giving two simultaneous equations which you can solve
(this is assuming that you have subbed in and rearranged correctly)
3)|(x-1)+i(y+1)|<3
sqrt((x-1)^2 + (y+1)^2) < 3
(x-1)^2 + (y+1)^2 < 9
Therefore it is a circle with centre (1,-i) and radius 3
Soz Im a bit tired and am going to go to bed now so i cant do the last one right now. Im not sure if you have rearanged your eqaution from qu 2 correctly, but Im to tired to check
Havent been to sleep in about 38 hours 
. Iapologise for any errors
0
reply
Thanks for the help rpotter!
I need to remember to make my Zs look like Zs and not the number 2! I just realised that I wrote one of the questions down wrong becasue of that!
For 3z-i=2+iz, is this correct?
3(x+iy)-i=2+i(x+iy)
3x+3iy-i=2+ix-y
3x-2+y=ix-3iy+i
3x-2+y=i(x-3y+1)
3x-2+y=0 equation 1
x-3y+1=0 so, x=3y-1 equation2
sub 2 into 1
3(3y-1)-2+y=0
9y-3-2+y=0
y=0.5
sub y=0.5 into equation 1,
3x-2+0.5=0
3x=1.5
x=0.5
and then z=0.5x+0.5iy?
I need to remember to make my Zs look like Zs and not the number 2! I just realised that I wrote one of the questions down wrong becasue of that!
For 3z-i=2+iz, is this correct?
3(x+iy)-i=2+i(x+iy)
3x+3iy-i=2+ix-y
3x-2+y=ix-3iy+i
3x-2+y=i(x-3y+1)
3x-2+y=0 equation 1
x-3y+1=0 so, x=3y-1 equation2
sub 2 into 1
3(3y-1)-2+y=0
9y-3-2+y=0
y=0.5
sub y=0.5 into equation 1,
3x-2+0.5=0
3x=1.5
x=0.5
and then z=0.5x+0.5iy?
0
reply
Report
#4
You don't need to look at real and imaginary parts. We have
3z - i = 2 + iz
z(3 - i) = 2 + i
z = (2 + i)/(3 - i)
Now to get this into the form a+bi multiply the top and bottom by the complex conjugate of the denominator, namely 3+i. We do this because z multiplied by it's conjugate is |z|^2, which is real. So we get
z = [(2 + i)(3 + i)]/(3^2 + 1^2) = (6 + 2i + 3i - 1)/10 = (1+i)/2
3z - i = 2 + iz
z(3 - i) = 2 + i
z = (2 + i)/(3 - i)
Now to get this into the form a+bi multiply the top and bottom by the complex conjugate of the denominator, namely 3+i. We do this because z multiplied by it's conjugate is |z|^2, which is real. So we get
z = [(2 + i)(3 + i)]/(3^2 + 1^2) = (6 + 2i + 3i - 1)/10 = (1+i)/2
0
reply
Report
#5
(Original post by mikesgt2)
You don't need to look at real and imaginary parts. We have
3z - i = 2 + iz
z(3 - i) = 2 + i
z = (2 + i)/(3 - i)
Now to get this into the form a+bi multiply the top and bottom by the complex conjugate of the denominator, namely 3+i. We do this because z multiplied by it's conjugate is |z|^2, which is real. So we get
z = [(2 + i)(3 + i)]/(3^2 + 1^2) = (6 + 2i + 3i - 1)/10 = (1+i)/2
You don't need to look at real and imaginary parts. We have
3z - i = 2 + iz
z(3 - i) = 2 + i
z = (2 + i)/(3 - i)
Now to get this into the form a+bi multiply the top and bottom by the complex conjugate of the denominator, namely 3+i. We do this because z multiplied by it's conjugate is |z|^2, which is real. So we get
z = [(2 + i)(3 + i)]/(3^2 + 1^2) = (6 + 2i + 3i - 1)/10 = (1+i)/2
0
reply
(Original post by mikesgt2)
You don't need to look at real and imaginary parts. We have
3z - i = 2 + iz
z(3 - i) = 2 + i
z = (2 + i)/(3 - i)
Now to get this into the form a+bi multiply the top and bottom by the complex conjugate of the denominator, namely 3+i. We do this because z multiplied by it's conjugate is |z|^2, which is real. So we get
z = [(2 + i)(3 + i)]/(3^2 + 1^2) = (6 + 2i + 3i - 1)/10 = (1+i)/2
You don't need to look at real and imaginary parts. We have
3z - i = 2 + iz
z(3 - i) = 2 + i
z = (2 + i)/(3 - i)
Now to get this into the form a+bi multiply the top and bottom by the complex conjugate of the denominator, namely 3+i. We do this because z multiplied by it's conjugate is |z|^2, which is real. So we get
z = [(2 + i)(3 + i)]/(3^2 + 1^2) = (6 + 2i + 3i - 1)/10 = (1+i)/2
0
reply
X
Page 1 of 1
Go to first unread
Skip to page:
Quick Reply
