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Solve the following equation in $ \mathbb{C} $: $ |z - |z + 1|| = |z + |z - 1|| $

I started it but I don't know how to finish it. Here is what I did so far:

$ |z - |z + 1||^2 = (z - |z + 1|)(\bar{z} - \overline{|z + 1|}) = |z|^2 - z \cdot \overline{|z + 1|} - \bar{z} \cdot |z + 1| + |z + 1|^2 $ $ |z + |z - 1||^2 = (z + |z - 1|)(\bar{z} + \overline{|z - 1|}) = |z|^2 + z \cdot \overline{|z - 1|} + \bar{z} \cdot |z - 1| + |z - 1|^2 $

Now, $ |z - |z + 1|| = |z + |z - 1|| \Rightarrow |z - |z + 1||^2 = |z + |z - 1||^2 $

So we have: $ |z|^2 - z \cdot \overline{|z + 1|} - \bar{z} \cdot |z + 1| + |z + 1|^2 = |z|^2 + z \cdot \overline{|z - 1|} + \bar{z} \cdot |z - 1| + |z - 1|^2 $

$ z(\overline{|z + 1| + |z - 1|}) + \bar{z}(|z + 1| + |z - 1|) = (|z + 1| + |z - 1|)(|z + 1| - |z - 1|) $

We divide all by |z + 1| + |z - 1| and we get: $ z + \frac{||z + 1| + |z - 1||^2}{(|z + 1| + |z - 1|)^2} + \bar{z} = |z + 1| - |z - 1|$

But $ |z + 1| + |z - 1| $ is a positive number so $ ||z + 1| + |z - 1|| = |z + 1| + |z - 1| $

So $ z + \bar{z} = |z + 1| - |z - 1| $.

And now I don't know what to do with this equation.

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The division actually obscured the result. If you undo it by multiplying your final result by $|z+1|+|z-1|$ you get with $\Re(z)=x$

$$2x(|z+1|+|z-1|)=|z+1|^2-|z-1|^2\tag{1}$$

The term on the right hand side of (1) can be simplified:

$$|z+1|^2-|z-1|^2=|z|^2+2x+1-|z|^2+2x-1=4x\tag{2}$$

Combining (1) and (2) and dividing by $2x$ gives

$$|z+1|+|z-1|=2\tag{3}$$

which is the equation of a degenerate ellipse in the complex plane. Note that the eccentricity equals $1$, so you end up with a line segment on the real axis in the range $x\in[-1,1]$. Note that because we divided by $x$, Eq. (3) disregards the solution $x=0$, i.e. the whole imaginary axis. So, in sum, the regions satisfying your original equation are

$$x\in[-1,1],\;y=0\quad\text{and}\quad x=0,\;y\in\mathbf{R}$$

i.e. a line segment on the real axis, and the whole imaginary axis.

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  • $\begingroup$ And the imaginary axis. $\endgroup$ – Empy2 Feb 21 '15 at 12:18
  • $\begingroup$ @Michael: What do you mean? $\endgroup$ – Matt L. Feb 21 '15 at 12:25
  • $\begingroup$ You cancelled $x$ from both sides. Check that $z=iy$ satisfies the equation. $\endgroup$ – Empy2 Feb 21 '15 at 12:28
  • $\begingroup$ But the ellipse is just the real axis between $\pm1$. $\endgroup$ – Empy2 Feb 21 '15 at 12:36
  • $\begingroup$ @Michael: Yes, you're right, it's a degenerate ellipse with eccentricity=1. Thanks for pointing that out! I've edited my answer. $\endgroup$ – Matt L. Feb 21 '15 at 13:01
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$z+\overline{z}=|z+1|-|z-1|$
You might square twice
$$(z+\overline{z})^2=2z\overline{z}+2-2|z+1||z-1|\\ 4(z+1)(\overline{z}+1)(z-1)(\overline{z}-1)=(2-z^2-\overline{z}^2)^2\\ 4(z^2-1)(\overline{z}^2-1)=4-4z^2-4\overline{z}^2+z^4+2z^2\overline{z}^2+\overline{z}^4\\ 0=(z^2-\overline{z}^2)^2$$ so $z^2$ is real, so $z=x$ or $z=iy$.
If $z=iy$, then $z+\overline{z}=0=|iy+1|-|iy-1|$, so the whole imaginary axis is included.
If $z=x$, then just the interval $[-1,1]$ is included.

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Here is a much simpler geometric proof:

enter image description here

In the diagram, $z_1=z_0+1$ and $z_2=z_0-1$. Then arcs are drawn with centre at A and tip at $z_1$ and $z_2$ to meet real axis at D and C respectively. Clearly, point D will represent $|z_0+1|$ and point C will represent $-|z_0-1|.$

The length of the line segment joining $z_0$ to D will represent $|z_0 - |z_0 + 1||$ and the one joining $z_0$ to C will represent $|z_0 + |z_0 - 1||$ .

The original equation requires both these line segments to be equal, and this implies that $z_0$ must lie on the perpendicular bisector of D and C. Algebraically, this is equivalent to:

$$Re(z_0)={|z_0+1|-|z_0-1|\over 2}$$

Multiplying both sides by $|z_0+1|+|z_0-1|$, you will get the the following equation after simplification:

$$Re(z_0)(|z_0+1|+|z_0-1|)=2Re(z_0)$$

Which is representing either the imaginary axis or the points lying between -1 and 1

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