Complex No. (1 Viewer)

[scardizzle]

http://i.imgur.com/rXlgT.jpg
Comes from 2002 SBHS trial have no clue how to solve part ii

 

[ohexploitable]

Didn't link properly...

 

[Affinity]

think about what arg((a-b)/arg(c-b)) is

 

[scardizzle]

think about what arg((a-b)/arg(c-b)) is

sorry man, i'm still not seeing it... arg(a-b)/arg(c-b) = arg(a-b) -arg(c-b) do you have to use the result that opposite angles of a cyclic quad are supplementary? I still cant put the pieces together.

 

[Affinity]

almost there arg(a-b) -arg(c-b) is the angle at b

 

[Sakeeee]

concyclic means opposite angles add to 180!!!

Thus

arg(z1-z2) - arg(z3-z2) + arg(z3-z4) - arg(z1-z4) = 180

.'. arg [(z1-z2)(z3-z4)]/[(z1-z4)(z3-z2)]=180 (using the rules of arg which are so fking similar to log)

(z1-z2)(z3-z4)]/[(z1-z4)(z3-z2)= a real number (when given an angle = 180 you will know for sure it's a positive real number on the complex plane.)


Now if you still don't get it that's okay, im kind of lost too so lets work this out.

z1-z2 just means the line from z2 to z1 (learn this in the vectors topic)
It is imperative that you realise that the z2 after the minus sign is where the line begins because z3-z2 would make an angle at z2 with z1-z2.

Hence im gonna call that S(2-1) [S stands for a s of the quadrilateral, (2-1) stands for line from point 2 to point 1]

so would you agree that the angle made by S(2-1) and S(2-3) can be written as
arg(z1-z2) - arg(z3-z2) [ arg changes the line into an angle just like m=tan@ does]

therefore we have the two angles opposite of a cyclic quad and they must add to 180.

Man that was confusing wasn't it BUT KEEP FUCKING TRYING or u will never get to bang that nerdy chick that sits in the corner D=!!!!

 

[scardizzle]

Thanks for the help guys but i dont see how the angle at z1 made by z4 and z2 = arg(z1-z2) - arg (z1 - z4) isnt the argument take from the origin? I guess what im try to say is how can you find an argument of a vector?

 

[Sakeeee]

Thanks for the help guys but i dont see how the angle at z1 made by z4 and z2 = arg(z1-z2) - arg (z1 - z4) isnt the argument take from the origin?

so z1-z2 is a straight line right. now arg (z1-z2) is the angle it makes with the x axis

say this angle is 130 degrees.

now arg(z1-z4) is the angle made by the line z4z1 and the angle made by this is say 100 degrees against the x-axis. DRAW THESE TWO LINES MAKING SURE ONE IS 130 DEGREES AND OTHER IS 100DEGREES.

now you will find that the corner angle made by these two lines is 30 degrees.


OKAY IMPORTANT THING IS IN COMPLEX arg(z1-z2) - arg (z1 - z4) WILL NOT GIVE AN ANGLE. like i said "THE SECOND BIT OF THE z1-z2 IS THE BASE. hence you can't have angle at two places i.e z2 z4. so you can only do arg(z1-z2) - arg (z3-z2) THIS MEANS TALKING ABOUT THE CORNER LOCATED AT z2.

PLEASE GO THROUGH UR VECTORS AGAIN OR YOU WILL NEVER REALLY UNDERSTAND COMPLEX

 

[murraysp]

think about that thing with the two arguments equaling the the angle.

like bakes put on the board

this is tough

if u figure it out gimme a yell

i think...

i got an idea

ok so possible like maybe equal roots on the circle

if lets say we a talking about z^4 then you get complex conjugates

if u get the conjugates then the imaginery parts cancel i think

i think...