# Questions from holiday revision...Watch

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#1
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
0
13 years ago
#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
0
#3
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?
0
13 years ago
#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
0
13 years ago
#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
Yup you can do it this way aswell, I just prefer to do it the other way, just a matter of preference
0
#6
(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
Ahh...makes sense... Thanks
0
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