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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 $-2< Re{\{z\}} < 0$).

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3 Answers 3

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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.

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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 $y$.

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

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  • $\begingroup$ 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 $\endgroup$
    – Ehsan
    May 8, 2016 at 7:46
  • $\begingroup$ 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. $\endgroup$
    – Ehsan
    May 8, 2016 at 14:22

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