# How do you graph an inequality on Real/Imaginary plane?

Suppose we have $z$ as a complex number, $z \in C$, how do you graph an inequality which has $z$ in it?

This kinds of inequalities arise when we need to graph the shape of stability region of a given numerical method:

For example, for RK2 method we can derive the stability inequality:

$|1+z+\frac{z^2}{2}| < 1$

What I could do so far was to set the inequality to zero and solve for $z$:

$|1+z+\frac{z^2}{2}| = 1$ so $z = 0, -2$. Therefore $z$ must be between $0$ and $-2$.

I also know the existence of circle formula: $(x-h)^2 + (y-k)^2 = r^2$ which sometimes helps to graph an inequality by transforming the inequality into the circle formula at origin $(h, k)$ with radius $r$.

But I can't transform the above RK2 inequality to a circle formula, and also unable to get the imaginary part properly, if I take the 'set to equality' method (though I know $-2< Re{\{z\}} < 0$).

Ok the answers provided by Eric Stucky and Morgan Rodgers both pointing at replacing $z$ with $x+iy$ is prefect. The problem is that in the exams there is no computer to plot the inequality derived in terms of $x$ and $y$.

Below is the approach I learned from http://www.math.ubc.ca/~peirce/M405_607E_Lecture%2018.pdf which I found the most suitable approach when you are in exam room:

First set the inequality to the function G such that:

$G(z) = 1 + z + \frac{z^2}{2}$

For stability we require $|G(z)| < 1$.

Solve $z$ in terms of $G$:

$z = -1 \pm \sqrt{2G-1}$

Now we can start graphing the $z$. First we assume $G$ is a circle of radius 1 at the origin of real/imaginary plane. Then we scale $G$ by two and we get $2G$, then we shift the circle to the left and we get $2G-1$ and so on unil we get $-1 \pm \sqrt{2G-1}$.

The process has been shown with step by step conformal map in http://www.math.ubc.ca/~peirce/M405_607E_Lecture%2018.pdf

I consider this as the best answer, specially in exam rooms. I will wait for further confirmation by the community and then will accept this as answer after few days.

You sketch in the boundary line, where the equation equals 0, then you shade in one side (corresponding to the inequality).

Notice that in your equation, you have $|1+z+\frac{z^{2}}{2}| = 0$, but it's not just as simple as factoring the inner equation; also for complex numbers it is not especially meaningful to say that $z$ is "between" $0$ and $2$. You actually get $$\sqrt{(1+z+\frac{z^{2}}{2})(1 + \overline{z} + \frac{\overline{z}^{2}}{2})} - 1= 0.$$ I think if you want to solve this you should view $z = x+y\mathrm{i}$ and treat it like an equation in two variables. This gives (once you get rid of the square root) a 4th degree equation in $x$ and $y$.

• Thanks. Setting $z = x + iy$ is the key. May 8, 2016 at 7:47

The beginning and end of the story here is: set $z=x+iy$ and convert to an inequality of two variables. There's no particularly good reason to believe that your domain should be anything nice like a circle.

In this case, square both sides and do the substitution to get

\begin{align*} \left|1+x+iy+\frac12x^2+ixy-\frac12y^2\right|^2<1 \\ \left(1+x+\frac{x^2-y^2}{2}\right)^2 + (y+xy)^2<1 \\ 1+x^2+\frac{x^4-2x^2y^2+y^4}{4}+2x+x(x^2-y^2)+(x^2-y^2) + y^2(1+2x+x^2)< 1 \\ 2x^2+\frac{x^4+2x^2y^2+y^4}{4}+2x+x(x^2+y^2) < 0 \\ 2x^2+\frac{|z|^4}{4}+2x+x|z|^2 < 0 \\ \end{align*}

This is a quadratic inequality in $|z|^2$ and $x$, with bounds being $-2x\pm2\sqrt{-x(x+2)}$. Since the bounds must be real, we recover your already discovered inequality on the real part.

Alternatively, you can leave it in terms of $x$ and $y$; which is probably the better form for putting it into an automated graphing device. It seems you will need this (or a lot of patience, trial, and error), since the region described is not so nice analytically.

• I appreciate your effort of solving the inequality in terms of $x$ and $y$. I think what I was missing is the key of setting $z = x + iy$ which Morgan Rodgers also mentioned. Meanwhile I found this which is less error prone and more efficient: math.ubc.ca/~peirce/M405_607E_Lecture%2018.pdf May 8, 2016 at 7:46
• I wrote down the answer in separate post below. Actually the region is very nice analytically. It is the inner bound of a circle placed in left half plane. May 8, 2016 at 14:22