# Math Help - Stuck trying to sketch this complex graph

1. ## Stuck trying to sketch this complex graph

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?

2. ## Re: Stuck trying to sketch this complex graph

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.

3. ## Re: Stuck trying to sketch this complex graph

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.

4. ## Re: Stuck trying to sketch this complex graph

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

5. ## Re: Stuck trying to sketch this complex graph

It's supposedly the right side of a hyperbola.

6. ## Re: Stuck trying to sketch this complex graph

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.

7. ## Re: Stuck trying to sketch this complex graph

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 $\displaystyle (x, y) = \left(\frac{1}{2}, 0\right)$.

8. ## Re: Stuck trying to sketch this complex graph

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.

9. ## Re: Stuck trying to sketch this complex graph

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.