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

**SyNtHeSiS** · October 27th 2010, 03:18 AM

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

Attempt:

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

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

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

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

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

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

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

It looks like I have reached a dead end

**Plato** · October 27th 2010, 03:42 AM

$|z-1|$ is the distance of z to 1.
$|z+i|$ is the distance of z to -i.
So you have the sum of two is less than or equal 2.
Think ellipse.

**Also sprach Zarathustra** · October 27th 2010, 03:48 AM

Or: -12x^2+8xy+16 x-12y^2-16y <0

Which is an ellipse!

**HallsofIvy** · October 27th 2010, 08:44 AM

Originally Posted by SyNtHeSiS

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

Attempt:

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

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

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

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

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

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

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

It looks like I have reached a dead end

The boundary of that set is given by $4x^2 + 16x + 16y + 8xy + 16 + 4y^2 = 16(x^2 + y^2 + 2y + 1)$
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 "<").

**SyNtHeSiS** · October 27th 2010, 10:07 AM

Thanks. How would you change

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

into the standard form on an ellipse:

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ ?

**Plato** · October 27th 2010, 10:14 AM

The quick answer is: one does not.
The axes are not vertical/horizontal.
The foci are (1,0) and (0,-1)

**SyNtHeSiS** · October 27th 2010, 10:39 AM

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

**Plato** · October 27th 2010, 10:47 AM

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.

**Archie Meade** · October 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|.

**SyNtHeSiS** · October 28th 2010, 03:04 AM

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?

**Plato** · October 28th 2010, 03:22 AM

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

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

**mr fantastic** · October 28th 2010, 05:08 AM

Originally Posted by Plato

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

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

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