MX2 Marathon (2 Viewers)

HazzRat

H̊ͯaͤz͠z̬̼iẻͩ̊͏̖͈̪
Joined
Aug 29, 2021
Messages
1,248
Gender
Male
HSC
2024
what bs is this. Can someone plz explain to me how these r equal
rh534tgerhtrg4r.PNG
 
Last edited:

liamkk112

Well-Known Member
Joined
Mar 26, 2022
Messages
1,054
Gender
Female
HSC
2023
what bs is this. Can someone plz explain to be how these r equal
View attachment 42231
if z-1/z+1 is purely imaginary, then the real part is zero, aka that x(x+1) + y(y-1) = 0
=> x^2 + x + y^2 - y = 0
=> x^2 + x + 1/4 + y^2 - y + 1/4 = 1/4 + 1/4 (completing the square)
=> (x+1/2)^2 + (y-1/2)^2 = 1/2
so basically it should be = 1/2, u can see they made a mistake because they noted that radius is 1/sqrt(2) which makes sense if
(x+1/2)^2 + (y-1/2)^2 = 1/2
 

liamkk112

Well-Known Member
Joined
Mar 26, 2022
Messages
1,054
Gender
Female
HSC
2023
if z-1/z+1 is purely imaginary, then the real part is zero, aka that x(x+1) + y(y-1) = 0
=> x^2 + x + y^2 - y = 0
=> x^2 + x + 1/4 + y^2 - y + 1/4 = 1/4 + 1/4 (completing the square)
=> (x+1/2)^2 + (y-1/2)^2 = 1/2
so basically it should be = 1/2, u can see they made a mistake because they noted that radius is 1/sqrt(2) which makes sense if
(x+1/2)^2 + (y-1/2)^2 = 1/2
also there should be an open circle at (-1, 0) as this would make the denominator 0
 

Luukas.2

Well-Known Member
Joined
Sep 21, 2023
Messages
443
Gender
Male
HSC
2023
also there should be an open circle at (-1, 0) as this would make the denominator 0
And is also impossible, as it makes

which is not purely imaginary - so two open circles, at opposite ends of a diameter.
 

Luukas.2

Well-Known Member
Joined
Sep 21, 2023
Messages
443
Gender
Male
HSC
2023
what bs is this. Can someone plz explain to me how these r equal
View attachment 42231
Note that a purely geometric approach would provide the graph, with the points and excluded, and is quicker:


Geometrically, this statement tells us that the angle between the vector from to and the vector from to is a right angle.

Applying the converse of the angle in a semicircle theorem, it follows that lies on one of the two semi-circles whose diameter is the interval joining and .

The two end points of the diameter, and , are excluded from the locus as each results in one of the two vectors being the zero vector, and hence one of the two arguments being and thus undefined.

---

Taking the algebraic approach, the two constraints that lead to points being excluded are that and . The point (-1, 0) violates both constraints, whilst the point (0, 1) violates only the second constraint.
 

HazzRat

H̊ͯaͤz͠z̬̼iẻͩ̊͏̖͈̪
Joined
Aug 29, 2021
Messages
1,248
Gender
Male
HSC
2024
Can someone plz explain to me the second last line of reasoning? That is: z1^2 + z2^2 = z1^2*(1+a^2). How did they get that?
jakaksjfj.PNG
 

ExtremelyBoredUser

Bored Uni Student
Joined
Jan 11, 2021
Messages
2,479
Location
m
Gender
Male
HSC
2022
nah i'm a queen 💅
Interdice, this is a man. Yes he is asian but he is not japanese or korean, I can confirm from personal experience, you might be a dwarf in comparison however like a pitbull I am worried you might try to slober all over him. He is a hard working person and has a family to take care of, please do not s** assault this man when you get on UNSW campus, he is simply making a very ironic and funny joke which juxtaposes my claim that he is a king to seem zesty, this does not warrant any abuse.

Much appreciated,
A worried friend of Lith_30
 

HazzRat

H̊ͯaͤz͠z̬̼iẻͩ̊͏̖͈̪
Joined
Aug 29, 2021
Messages
1,248
Gender
Male
HSC
2024
gfaew.PNG

for question (iii) here, where did they get the first statement of the proof from? I could solve the question after getting that statement but like idk where they got it
jnjanwfjwan.PNG
 

liamkk112

Well-Known Member
Joined
Mar 26, 2022
Messages
1,054
Gender
Female
HSC
2023
View attachment 42394

for question (iii) here, where did they get the first statement of the proof from? I could solve the question after getting that statement but like idk where they got it
View attachment 42395
in ii) let a^3 = (a^3/1+a^3), b^3 = (b^3 /1+b^3), c^3 = (1/1+c^3) for the first one.
then taking cube roots to get a,b and c u see how the rhs comes about
 

liamkk112

Well-Known Member
Joined
Mar 26, 2022
Messages
1,054
Gender
Female
HSC
2023
though i'm not sure what the motivation is for thinking of that specific substitution
 

HazzRat

H̊ͯaͤz͠z̬̼iẻͩ̊͏̖͈̪
Joined
Aug 29, 2021
Messages
1,248
Gender
Male
HSC
2024
gefqwd.PNG
For question (ii) here, y is the angle ∠OAC a right angle? I kinda assumed ∠OCA to be the right angle

brgefqwd.PNG
 

Luukas.2

Well-Known Member
Joined
Sep 21, 2023
Messages
443
Gender
Male
HSC
2023
View attachment 42402
For question (ii) here, y is the angle ∠OAC a right angle? I kinda assumed ∠OCA to be the right angle

View attachment 42403
The right angle is definitely OAC as the tangent to a circle is always perpendicular to the radius at the point of contact.

If OCA was a right angle, then the tangent at A would be parallel to OC and thus OA would not be a tangent, but rather would cross the circle at some point B between O and A... in which case, there must be points on the circle between A and B with a larger principal argument.

 

HazzRat

H̊ͯaͤz͠z̬̼iẻͩ̊͏̖͈̪
Joined
Aug 29, 2021
Messages
1,248
Gender
Male
HSC
2024
nfbdsfa.PNG
solid induction question from my school's 2020 test
 

Users Who Are Viewing This Thread (Users: 0, Guests: 2)

Top