# Sketching |z - 1| + |z + i| <= 2 in complex plane

• Oct 27th 2010, 03:18 AM
SyNtHeSiS
Sketching |z - 1| + |z + i| <= 2 in complex plane
Sketch the following region in the complex plane: $\displaystyle { z \epsilon\mathbb{C} : |z -1| + |z + i| \leq 2}$

Attempt:

$\displaystyle |z - 1| + |z + i| \leq 2$

$\displaystyle \sqrt{(x - 1)^2 + y^2} + \sqrt{x^2 + (y + 1)^2} \leq 2$

$\displaystyle \sqrt{(x - 1)^2 + y^2} \leq 2 - \sqrt{x^2 + (y + 1)^2}$

$\displaystyle x^2 - 2x + 1 + y^2 \leq 4 - 4\sqrt{x^2 + y^2 + 2y + 1} + (x^2 + y^2 + 2y + 1)$

$\displaystyle -2x - 4 - 2y \leq -4\sqrt{x^2 + y^2 + 2y + 1}$

$\displaystyle (-2x - 4 - 2y)^2 \leq 16(x^2 + y^2 + 2y + 1)$

$\displaystyle 4x^2 + 16x + 16y + 8xy + 16 + 4y^2 \leq 16(x^2 + y^2 + 2y + 1)$

It looks like I have reached a dead end(Speechless)
• Oct 27th 2010, 03:42 AM
Plato
$\displaystyle |z-1|$ is the distance of z to 1.
$\displaystyle |z+i|$ is the distance of z to -i.
So you have the sum of two is less than or equal 2.
Think ellipse.
• Oct 27th 2010, 03:48 AM
Also sprach Zarathustra
Or: -12x^2+8xy+16 x-12y^2-16y <0

Which is an ellipse!
• Oct 27th 2010, 08:44 AM
HallsofIvy
Quote:

Originally Posted by SyNtHeSiS
Sketch the following region in the complex plane: $\displaystyle { z \epsilon\mathbb{C} : |z -1| + |z + i| \leq 2}$

Attempt:

$\displaystyle |z - 1| + |z + i| \leq 2$

$\displaystyle \sqrt{(x - 1)^2 + y^2} + \sqrt{x^2 + (y + 1)^2} \leq 2$

$\displaystyle \sqrt{(x - 1)^2 + y^2} \leq 2 - \sqrt{x^2 + (y + 1)^2}$

$\displaystyle x^2 - 2x + 1 + y^2 \leq 4 - 4\sqrt{x^2 + y^2 + 2y + 1} + (x^2 + y^2 + 2y + 1)$

$\displaystyle -2x - 4 - 2y \leq -4\sqrt{x^2 + y^2 + 2y + 1}$

$\displaystyle (-2x - 4 - 2y)^2 \leq 16(x^2 + y^2 + 2y + 1)$

$\displaystyle 4x^2 + 16x + 16y + 8xy + 16 + 4y^2 \leq 16(x^2 + y^2 + 2y + 1)$

It looks like I have reached a dead end(Speechless)

The boundary of that set is given by $\displaystyle 4x^2 + 16x + 16y + 8xy + 16 + 4y^2 = 16(x^2 + y^2 + 2y + 1)$
which, as Plato and also Sprach Zarathustra have told you, is an ellipse. The set you want is the interior and boundary of that set (because of the "<").
• Oct 27th 2010, 10:07 AM
SyNtHeSiS
Thanks. How would you change

$\displaystyle -12x^2 + 16x - 12y^2 - 16y + 8xy \leq 0$

into the standard form on an ellipse:

$\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ ?
• Oct 27th 2010, 10:14 AM
Plato
The quick answer is: one does not.
The axes are not vertical/horizontal.
The foci are (1,0) and (0,-1)
• Oct 27th 2010, 10:39 AM
SyNtHeSiS
In that case, how would you know that it is an ellipse, with only knowing those 2 points (1,0) and (0, -1)?
• Oct 27th 2010, 10:47 AM
Plato
Quote:

Originally Posted by SyNtHeSiS
In that case, how would you know that it is an ellipse, with only knowing those 2 points (1,0) and (0, -1)?

ELLIPSE
The path of a point which moves so that the sum of its distances from two fixed points is a constant. The two fixed points are the foci of the ellipse.
• Oct 27th 2010, 10:55 AM
The internal region is where the sum is <2.

Sorry about the typo on the sketch, should be |z+i|.
• Oct 28th 2010, 03:04 AM
SyNtHeSiS
Thanks. I am still a bit confused. I understand that the definition of an ellipse is that the sum of the distances from both foci to a point is constant, but how would you know this in this example, without testing 2 points on the ellipse, to see if the sum of their distances from both foci are equal?
• Oct 28th 2010, 03:22 AM
Plato
Quote:

Originally Posted by SyNtHeSiS
Thanks. I am still a bit confused. I understand that the definition of an ellipse is that the sum of the distances from both foci to a point is constant, but how would you know this in this example, without testing 2 points on the ellipse, to see if the sum of their distances from both foci are equal?

The notation $\displaystyle \left| {z - z_0 } \right|$ is the distance from $\displaystyle z$ to $\displaystyle z_0$.

Therefore $\displaystyle \left| {z - z_0 } \right|+\left| {z - z_1 } \right|=C$ must be an ellipse.
• Oct 28th 2010, 05:08 AM
mr fantastic
Quote:

Originally Posted by Plato
The notation $\displaystyle \left| {z - z_0 } \right|$ is the distance from $\displaystyle z$ to $\displaystyle z_0$.

Therefore $\displaystyle \left| {z - z_0 } \right|+\left| {z - z_1 } \right|=C$ must be an ellipse.

Provided $\displaystyle C > \left|z_1 - z_0\right|$.

@OP: If $\displaystyle C = \left|z_1 - z_0\right|$ then the locus is a line segment joining $\displaystyle z_0$ and $\displaystyle z_1$, and if $\displaystyle C < \left|z_1 - z_0\right|$ then there is no locus.