Complex numbers: how do you figure out circular graphs? (1 Viewer)

[shimmerz_777]

this question turned up in an exam

RE [z-4/z] = 0

i really have no understanding of how the RE bit affects the circle, and i have trouble figuring them out. if some one could do the working and show some reasoning behind some bits ill really appreciate it. thanks

 

[SonyHK]

this question turned up in an exam

RE [z-4/z] = 0

i really have no understanding of how the RE bit affects the circle, and i have trouble figuring them out. if some one could do the working and show some reasoning behind some bits ill really appreciate it. thanks

let z=x+iy
RE [z-4/z] = 0
[z-4/z]=(x+iy)-4\(x+iy)=(x+iy)-4(x-iy)\(x²+y²)
.: Re [z-4/z]=x-4x\(x²+y²)=0
1-4\(x²+y²)=0
x²+y²=4
so circle is centre origin radius 2, assuming no errors.

 

[shimmerz_777]

let z=x+iy
RE [z-4/z] = 0
[z-4/z]=(x+iy)-4\(x+iy)=(x+iy)-4(x-iy)\(x²+y²)
.: Re [z-4/z]=x-4x\(x²+y²)=0
1-4\(x²+y²)=0
x²+y²=4
so circle is centre origin radius 2, assuming no errors.

i have a solution for it, although that isnt the answer, it says "locus of z is a circle center (2,0), radius 2, excluding the point (0,0)

its confusing because i dont know what the RE does, i thought it meant that the equasion had a line fixed at x=0 or something, but i have no idea really

 

[pLuvia]

The Re(z) means the real part of z, Im(z) means the imaginary part of z

Let z=x+iy
Re(z)=x
Im(z)=y

For your question

(x+yi-4)/(x+yi)=0
[(x+yi)-4][x-yi]/(x+yi)(x-yi)
x²+y²-4(x-yi) / x²+y²

Re[z-4/z] = x²+y²-4x / x²+y²=0
=x²-4x+y²=0
=(x-2)²+y²=4

The locus of z is a circle centre (2,0) and radius 2

 

[SonyHK]

Oh well you didn't put a bracket where it was appropriate=P

You said "RE [z-4/z] = 0"
The correct answer is using Re [(z-4)\z]=0 ...

 

[shimmerz_777]

The Re(z) means the real part of z, Im(z) means the imaginary part of z

Let z=x+iy
Re(z)=x
Im(z)=y

For your question

(x+yi-4)/(x+yi)=0
[(x+yi)-4][x-yi]/(x+yi)(x-yi)
x²+y²-4(x-yi) / x²+y²

Re[z-4/z] = x²+y²-4x / x²+y²=0
=x²-4x+y²=0
=(x-2)²+y²=4

The locus of z is a circle centre (2,0) and radius 2

ok, i knew that it refered to the real part of z, which i believed to be the x value, i just dont get what would happen if the Re was not apart of the equasion. i probably should work it out but i keep doubting myself when i get an answer. ill figure it out eventually. and sorry about the bracket i forgot.

ps, you forgot to add x^2 + y^2 cannot = 0 as it is dividing part of the equasion, thus you have to exclude the point (0,0)

 

[pLuvia]

ps, you forgot to add x^2 + y^2 cannot = 0 as it is dividing part of the equasion, thus you have to exclude the point (0,0)

Yep that too

 

[Riviet]

(z-4)/z = 0, where z is an element of the complex numbers.
z-4=0, z =/= 0
z=4 or x+iy=4, where z=x+iy

So I think it's just that point (4,0)

 

[shimmerz_777]

ok pretty much what i was hoping to find out from this thread was a general way in which you can solve questions like the one i posted, what ive figured out is you multiply by (bar z)/(bar z) and go from there, but also, i still am yet to understand how the Re bit affects it, as well as if Im were to be placed infront of it, how it would change the question. does the Re bit mean you have to find your answer in terms of real numbers or something? anyway thanks for helping me get the first bit.

 

[243_robbo]

stuie the "Re" bit actually makes it easier cos it means you can disregard any thing with i in it when you are finding your locus. but rerember i^2 is -1 so that it is also part of the real part