I am trying to find for which choices of $\theta$ for which the theta method absolutely stable but I am having a lot of trouble solving the resulting inequality. It is straight forward to produce the function

$$f(z,\theta) = \frac{1+z\theta}{1-z(1-\theta)}$$

but finding the values of $\theta$ such that

$$|f(z,\theta)| < 1$$

for all $z\in \{z\in \mathbb{C}:Re(z) < 1\}$ is proving to be beyond my ability. After some searching, it looks like this has already been answered here absolute stability / inequality, but the proposed solution does not satisfy me because I don't think that will work for a complex valued function (although some of the literature I have read seems to suggest it is enough to consider the case when $z$ is real, but I cannot figure out why).

To list some things I have tried:

I tried writing this function in polar form, using the triangle inequality and every other standard manipulation.

I observed that this looks like a moebius transformation, and spent a few hours studying moebius transformations and linear fractional transformations hoping for a useful result.

I tried equating this function to 1, and then later $e^{i\rho}$ to try to calculate the boundary, but the result was still a rational function of $z$ and $\theta$ that was not making it any easier.

Finally, I have not tried to plot the contour lines, but I would like to see an answer without doing it numerically.

  • $\begingroup$ This is worked out in this answer. $\endgroup$ Mar 26, 2015 at 14:32
  • $\begingroup$ @GodricSeer Thank you (that answer is already linked in my question). I've actually eventually solved the problem on my own by using the following theorem: Let $r$ be an arbitrary non-constant rational function. Then $\lvert r(z) \rvert <1$ for all $z\in \mathbb{C}^{-}$ if and only if all poles of $r$ have positive real parts and $\lvert r(it)\rvert \leq 1$ for all $t\in \mathbb{R}$. $\endgroup$
    – JessicaK
    Mar 26, 2015 at 14:41
  • $\begingroup$ Yes, I assumed you had found an answer over the last year, but for others who find your question prior to the other (such as myself) I thought a link may be useful. $\endgroup$ Mar 26, 2015 at 14:44
  • 1
    $\begingroup$ @GodricSeer The solution you linked (which I raised a complaint about when asking my question) doesn't actually address the problem I had. The proposed solution there assumes that by squaring both sides, there is meaning to the symbol "<", but we cannot totally order the complex numbers in that manner. It is fine if you accept that $h\lambda \in \mathbb{R}$, but that method does not work for $h\lambda \in \mathbb{C}$. $\endgroup$
    – JessicaK
    Mar 26, 2015 at 14:47
  • $\begingroup$ I guess I need to learn to actually read at some point in my education.... $\endgroup$ Mar 26, 2015 at 14:52

1 Answer 1


An answer that just uses algebra of the complex numbers and does not make reference to theorems.

The problem is to find $\theta\in[0,1]$ such that $|1+\theta z|<|1-(1-\theta)z|$ for all $z\in\mathbb{C}^-$.

Let $z=a+bi$, $a<0$. Then the inequality is \begin{equation} |1+\theta a + \theta bi|<|1-(1-\theta)a -(1-\theta)bi|. \ \ \ \ \ \ (1) \end{equation}


  • $\theta<\frac12$: Will show that the absolute values of the real and imaginary components of the left hand side of (1) are strictly less than the real and imaginary components of the right hand side.

    Real Parts: Two cases: $1+\theta a \le 0$ and $1+\theta a >0$. (Note that $1-(1-\theta)a$ is always positive so $1-(1-\theta)a=|1-(1-\theta)a|$.)

    $\le 0$: Since $\theta<1/2$, \begin{align*} 0 &> (2\theta - 1) \\ 0 &< (2\theta -1)a \\ 0 &< 2 + \theta a + (\theta-1) a\\ -1-\theta a &< 1 - (1-\theta)a \\ |1+\theta a| &< |1-(1-\theta)a|. \end{align*} $>0$: \begin{align*} a &<0 \\ (\theta + (1-\theta))a &< 0 \\ \theta a &< -(1-\theta) a \\ 1+\theta a &< 1 - (1-\theta) a \\ |1+\theta a| &< |1 - (1-\theta) a|. \end{align*}Imaginary Parts: $0\le\theta<1-\theta$ so \begin{align*} \theta |b| &\le (1-\theta)|b|\\ |\theta b| &\le |-(1-\theta)b|. \end{align*} So the absolute value of the imaginary part of the left hand side is less than or equal to the absolute value of the imaginary part of the right hand side, but the corresponding real parts are strictly less than in magnitude. Thus equation (1) holds when $\theta<\frac12$ so A-stable for $\theta<1/2$.

  • $\theta=\frac12$: Now $\theta = (1-\theta)$ so that the imaginary parts of both sides of (1) are the same. The same inequalities corresponding to the real part of the $\theta<\frac12$ section still hold (for $\le 0$ the first two lines are equalities but the third line is strict inequality). So A-stable for $\theta=1/2$.

  • $\theta>\frac12$: Here we just find a particular $z\in\mathbb{C}^-$ that gives equality in (1). Since $\theta\in(\frac12,1]$, write $\theta=\frac12 + \epsilon$, where $\frac12\ge\epsilon>0$. Let $z=-\frac1\epsilon$. The the magnitudes of left and right sides are both $\frac{1}{2\epsilon}$. So the method is not A-stable for $\theta>\frac12$.

So the final answer is the method is A-Stable if and only if $\theta\le\frac12$.

  • $\begingroup$ Thanks for the answer. I was wondering how the stability region looks like in the case $\theta > 0.5$? $\endgroup$
    – Arthur
    Feb 23, 2020 at 16:58

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