How can i know that this equation is a ellipse?

[Fabio010]

Sketch the following region:

| z+i | + | z-1 | =2




I know that it is a ellipse, but i am trying to replace the z for x +iy.



I reached the following equation.

3x^2+3y^2 = 4x-4y+2xy


now how can a simplify it in order to have the ellipse equation: (x^2/a^2) + (y^2/b^2) = 1

 

[arildno]

Supposing you have done this correctly:

1. There is no reason that the axes of the ellipse should lie parallell to the x and y-axes.
That this is NOT the case, is given by the mixed quadratic term 2xy.
You need to rotate your coordinate system by some fixed angle "a", by introducing new, general coordinates (u,v), for example:
x=u*cos(a)-v*sin(a), y=u*sin(a)+v*cos(a)
We then determine the angle "a" so that the cross term in "u" and "v" vanishes.
Here, "a" is the angle between the positive x-axis and the positive u-axis, so that if "a"=0 yields x=u, y=v, while "a"=pi/2 in radians, x=-v and y=u.

2. Furthermore: The origin of the ellipse will NOT lie at the origin. Complete the squares in u and v to get (after a while) an equation of the form:
(u-u')^2/K^2+(v-v')^2/L^2=1

where u', v', K and L are constants to be determined.

 

[arildno]

Now, following 1, you'll get:
3(u^2+v^2)=4(u(cos(a)-sin(a))-v(cos(a)+sin(a)))+2((u^2-v^2)cos(a)sin(a)+uv(cos(2a)).

Choosing a=pi/4, then, yields:
3(u^2+v^2)=-4sqrt(2)*v+u^2-v^2, that is:
2u^2+4(v^2+sqrt(2)v)=0

Now, you need to complete the square in v, according to 2.

This gives you:
2u^2+4(v+1/sqrt(2))^2=2,

That is:
u^2/K^2+(v+1/sqrt(2))^2/L^2=1, (**)
where u'=0, v'=-1/sqrt(2), K=1, L=1/sqrt(2)

(**) is your equation for the ellipse, where the u-axis makes pi/4 radians angle to the x-axis

The origin of the ellipse is therefore at (u',v')=(0,-1/sqrt(2)), which works out to be at (x',y')=(1/2,-1/2)

In order to draw it, you ought to find the extremals along the (u,v) axis pairs, and then work out the (x,y) coordinate pairs.
For example,
(u,v)=(0,0) works out as..(x,y)=(0,0)

 

[HallsofIvy]

The geometric definition of 'ellipse' is "a plane figure having two points, called foci, such that the total distances from any point on the ellipse to the two foci is a constant".

In the complex plane, geometrically, |a- b| is the distance between point a and point b.

so |z+ i|+ |z- 1|= 2 says exactly that the total distance from any point, z, on the figure to -i and 1, is a constant, 2. That tells us it is an ellipse, having -i and 1 as foci.

 

[Fabio010]

People thanks for the help!

The problem is not supposed to be that hard.

My teacher just said that it was an ellipse. I tried to prove it by the equation.