Stuck trying to sketch this complex graph

**terrorsquid** · September 14th 2011, 07:44 AM

Solve $|z+1|-|z-2|=1$ and sketch its graph.

First I notice that I'm finding the complex numbers where the distance between z and -1 = the distance between z and 2 +1 (not sure how the +1 affects things).

$|x+iy+1|-|x+iy-2|=1$

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

here's where I'm not sure where to go because I can't just square both sides to remove the roots with the 1 there. How do I proceed?

**piscoau** · September 14th 2011, 07:51 AM

the square root of sqrt((x+1)^2+y^2)-sqrt((x-2)^2+y^2) = 1 can be removed by take both side square more than once.

**HallsofIvy** · September 14th 2011, 11:24 AM

If I recall correctly, the figure such that the difference of the distances for each point on the figure to two fixed points is the hyperbola having those fixed points as foci.

**terrorsquid** · September 14th 2011, 07:01 PM

Originally Posted by piscoau

the square root of sqrt((x+1)^2+y^2)-sqrt((x-2)^2+y^2) = 1 can be removed by take both side square more than once.

Let m $= \sqrt{(x+1)^2+y^2}$

let j $= \sqrt{(x-2)^2+y^2}$

$\sqrt{m}-\sqrt{j} = 1$

squaring both sides:

$m-2\sqrt{m}\sqrt{j}+j = 1$

squaring again:

$m^2-4m\sqrt{m}\sqrt{j}+6mj-4j\sqrt{m}\sqrt{j}+j^2=1$

can't seriously expect to keep going :S

**terrorsquid** · September 14th 2011, 07:01 PM

It's supposedly the right side of a hyperbola.

**HallsofIvy** · September 15th 2011, 03:41 AM

I misread your equation. I have edited what I wrote before. The set of points such that the difference of the distances from two fixed point (either d1- d2 or d2- d1) is a constant is indeed a hyperbola. For only d2- d1 a constant, you have one branch of the hyperbola.

**Prove It** · September 15th 2011, 04:39 AM

Originally Posted by terrorsquid

Solve $|z+1|-|z-2|=1$ and sketch its graph.

First I notice that I'm finding the complex numbers where the distance between z and -1 = the distance between z and 2 +1 (not sure how the +1 affects things).

$|x+iy+1|-|x+iy-2|=1$

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

here's where I'm not sure where to go because I can't just square both sides to remove the roots with the 1 there. How do I proceed?

$\displaystyle \begin{align*} \sqrt{(x + 1)^2 + y^2} - \sqrt{(x - 2)^2 + y^2} &= 1 \\ \left(\sqrt{(x + 1)^2 + y^2} - \sqrt{(x - 2)^2 + y^2}\right)^2 &= 1^2 \\ (x + 1)^2 + y^2 - 2\sqrt{(x + 1)^2 + y^2}\sqrt{(x - 2)^2 + y^2} + (x - 2)^2 + y^2 &= 1 \\ (x + 1)^2 + (x - 2)^2 + 2y^2 - 1 &= 2\sqrt{(x + 1)^2 + y^2}\sqrt{(x - 2)^2 + y^2} \\ x^2 + 2x + 1 + x^2 - 4x + 4 + 2y^2 - 1 &= 2\sqrt{(x+1)^2 + y^2}\sqrt{(x-2)^2 + y^2} \\ 2x^2 - 2x + 4 + 2y^2 &= 2\sqrt{(x + 1)^2 + y^2}\sqrt{(x - 2)^2 + y^2} \\ x^2 - x + 2 + y^2 &= \sqrt{(x + 1)^2 + y^2}\sqrt{(x - 2)^2 + y^2} \\ \left(x^2 - x + 2 + y^2\right)^2 &= \left(\sqrt{(x + 1)^2 + y^2}\sqrt{(x - 2)^2 + y^2}\right)^2 \\ x^4-2x^3+2x^2y^2+5x^2-2xy^2-4x+y^4+4y^2+4 &= \left[(x + 1)^2 + y^2\right]\left[(x - 2)^2 + y^2\right] \\ x^4-2x^3+2x^2y^2+5x^2-2xy^2-4x+y^4+4y^2+4 &= x^4-2x^3+2x^2y^2-3x^2-2xy^2+4x+y^4+5y^2+4 \\ 5x^2 - 4x + 4y^2 &= -3x^2 + 4x + 5y^2\end{align*}$

$\displaystyle \begin{align*} 8x^2 - 8x - y^2 &= 0 \\ 8\left(x^2 - x\right) - y^2 &= 0 \\ 8\left[x^2 - x + \left(-\frac{1}{2}\right)^2 - \left(-\frac{1}{2}\right)^2\right] - y^2 &= 0 \\ 8\left[\left(x - \frac{1}{2}\right)^2 - \frac{1}{4}\right] - y^2 &= 0 \\ 8\left(x - \frac{1}{2}\right)^2 - y^2 - 2 &= 0 \\ 8\left(x - \frac{1}{2}\right)^2 - y^2 &= 2 \\ 4\left(x - \frac{1}{2}\right)^2 - \frac{y^2}{2} &= 1 \\ \frac{\left(x - \frac{1}{2}\right)^2}{\left(\frac{1}{2}\right)^2} - \frac{y^2}{\left(\sqrt{2}\right)^2} &= 1\end{align*}$

So this is a hyperbola, centred at $\displaystyle (x, y) = \left(\frac{1}{2}, 0\right)$ .

**mr fantastic** · September 15th 2011, 01:35 PM

Originally Posted by Prove It

$\displaystyle \begin{align*} \sqrt{(x + 1)^2 + y^2} - \sqrt{(x - 2)^2 + y^2} &= 1 \\ \left(\sqrt{(x + 1)^2 + y^2} - \sqrt{(x - 2)^2 + y^2}\right)^2 &= 1^2 \\ (x + 1)^2 + y^2 - 2\sqrt{(x + 1)^2 + y^2}\sqrt{(x - 2)^2 + y^2} + (x - 2)^2 + y^2 &= 1 \\ (x + 1)^2 + (x - 2)^2 + 2y^2 - 1 &= 2\sqrt{(x + 1)^2 + y^2}\sqrt{(x - 2)^2 + y^2} \\ x^2 + 2x + 1 + x^2 - 4x + 4 + 2y^2 - 1 &= 2\sqrt{(x+1)^2 + y^2}\sqrt{(x-2)^2 + y^2} \\ 2x^2 - 2x + 4 + 2y^2 &= 2\sqrt{(x + 1)^2 + y^2}\sqrt{(x - 2)^2 + y^2} \\ x^2 - x + 2 + y^2 &= \sqrt{(x + 1)^2 + y^2}\sqrt{(x - 2)^2 + y^2} \\ \left(x^2 - x + 2 + y^2\right)^2 &= \left(\sqrt{(x + 1)^2 + y^2}\sqrt{(x - 2)^2 + y^2}\right)^2 \\ x^4-2x^3+2x^2y^2+5x^2-2xy^2-4x+y^4+4y^2+4 &= \left[(x + 1)^2 + y^2\right]\left[(x - 2)^2 + y^2\right] \\ x^4-2x^3+2x^2y^2+5x^2-2xy^2-4x+y^4+4y^2+4 &= x^4-2x^3+2x^2y^2-3x^2-2xy^2+4x+y^4+5y^2+4 \\ 5x^2 - 4x + 4y^2 &= -3x^2 + 4x + 5y^2\end{align*}$

$\displaystyle \begin{align*} 8x^2 - 8x - y^2 &= 0 \\ 8\left(x^2 - x\right) - y^2 &= 0 \\ 8\left[x^2 - x + \left(-\frac{1}{2}\right)^2 - \left(-\frac{1}{2}\right)^2\right] - y^2 &= 0 \\ 8\left[\left(x - \frac{1}{2}\right)^2 - \frac{1}{4}\right] - y^2 &= 0 \\ 8\left(x - \frac{1}{2}\right)^2 - y^2 - 2 &= 0 \\ 8\left(x - \frac{1}{2}\right)^2 - y^2 &= 2 \\ 4\left(x - \frac{1}{2}\right)^2 - \frac{y^2}{2} &= 1 \\ \frac{\left(x - \frac{1}{2}\right)^2}{\left(\frac{1}{2}\right)^2} - \frac{y^2}{\left(\sqrt{2}\right)^2} &= 1\end{align*}$

So this is a hyperbola, centred at $\displaystyle (x, y) = \left(\frac{1}{2}, 0\right)$ .

There is an implied restriction on x in this working that will cause one of the branches of this hyperbola to be rejected (that branch is an extraneous solution and is introduced by the squaring process). But I don't have the time to read through it. However, it is sufficient to test a simple point on each branch and reject the branch that contains the point that fails to satisfy the original equation.

Personally, I favour the approach of moving a radical to one side first before squaring. The implied restriction is a lot more obvious. If I have time later, I might show the first few steps.

**mr fantastic** · October 1st 2011, 05:17 AM

Originally Posted by mr fantastic

There is an implied restriction on x in this working that will cause one of the branches of this hyperbola to be rejected (that branch is an extraneous solution and is introduced by the squaring process). But I don't have the time to read through it. However, it is sufficient to test a simple point on each branch and reject the branch that contains the point that fails to satisfy the original equation.

Personally, I favour the approach of moving a radical to one side first before squaring. The implied restriction is a lot more obvious. If I have time later, I might show the first few steps.

$|z + 1| - |z - 2| = 1$

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

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

Square both sides:

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

Expand and simplify:

$\Rightarrow 3x - 2 = \sqrt{(x - 2)^2 + y^2}$

Since RHS > 0, it is required that LHS > 0. Therefore there is an implied restriction:

$3x - 2 > 0 \Rightarrow x > \frac{2}{3}$ .

So the hyperbola obtained by continued algebra is restricted to the branch that lies to the right of x = 2/3.

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