# MX2 Marathon (1 Viewer)

#### ExtremelyBoredUser

##### Bored Uni Student
they said that $\bg_white z_2=\alpha{z_1}$ so then $\bg_white z_1^2+z_2^2=z_1^2+(\alpha{z_1})^2=z_1^2(1+\alpha^2)$
King has spoken

#### ExtremelyBoredUser

##### Bored Uni Student
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

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

#### liamkk112

##### Well-Known Member
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
though i'm not sure what the motivation is for thinking of that specific substitution

#### HazzRat

##### H̊ͯaͤz͠z̬̼iẻͩ̊͏̖͈̪

For question (ii) here, y is the angle ∠OAC a right angle? I kinda assumed ∠OCA to be the right angle

#### Luukas.2

##### Well-Known Member
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.

$\bg_white \text{Arg}\,z \in \left[\phi - \theta,\ \phi + \theta\right]$

#### HazzRat

##### H̊ͯaͤz͠z̬̼iẻͩ̊͏̖͈̪

solid induction question from my school's 2020 test