Someone please help!! (1 Viewer)

[Harimau]

This is a question from Patel...

If Z1, Z2, Z3 are points representing the complex numbers z1, z2, z3 and they lie on a circle which crosses the origin, prove that 1/Z1, 1/Z2, 1/Z3 are co-linnear

I know that there is an easy way to do this, but i dont quite understand on how to do it the conventional way... can someone please help me!

 

[ND]

http://x300-3-bne.servers.net.au/~admin8/community/showthread.php?s=&threadid=4493

Not exactly "easy".

 

[Rahul]

*shudders*......memories.....i am dumb and spent ages on this and didnt get it. *sigh* the joys of 4 u

 

[Harimau]

Well the easy solution that a tutor at Prior did was to let Z= 1/W, where W is a complex number. And then you sub it into the equation of a circle, letting the circle by the complex Number C.

[ W-X] = C
[ 1/Z-X]= C

and then solving from there and taking the real and imaginary parts...

Question, how do i put absolute value signs on VB? and also sqrt and exponents...

 

[ezzy85]

u can use |z| for absolute value. Thats the vertical bar that's shared with the \ key. for e^x u can use the html tag (sup)x(/sup), replacing the brackets with < or >

 

[Mathematician]

...

Where is Prior tutoring?

someone asked me before and i didnt know what he was talking about. LOL

 

[Mathematician]

..

oh yeah , i dont get how the question was solved the easy way.

General eq of a circle that passes through the origin?

and can u just do all the steps to solving it the easy way?

 

[Harimau]

Re: ...

Originally posted by Mathematician
Where is Prior tutoring?

someone asked me before and i didnt know what he was talking about. LOL

Prior Tutoring is at Ebley street bondi and also in one of the towers at Chatswood. Call them at 9369 4447

This is the solution, done by one of their tutors:

Let C be the center of the circle, and its equation |z-c|=|c| .

Let w=1/z

Sub

|1/w-c|=|c|

|1-wc|^2=|c|^2|w|^2

(1-wc)(1-w(bar)c(Bar))= 1-2Re(wc)+ |c|^2|w|^2

From this, we get

1-2Re(wc)=0 or Re(cw)=1/2

Sub w= x+iy and c=a+ib

And we get:

ax-by=1/2, a line with gradient a/b, and y intercept 1/2b

This guy is a damn genius no?

 

[KeypadSDM]

No, he's just done it enough to be able to regurgitate it.

Think of uni lecturers, when you hear them utter those words that seem so foreign to you, you are dazzled because you did not see it yourself. But they were once in your spot, not able to see it either.