Test a conjecture

Consider the family

\[ f_a(z)=z+az^2, \qquad a\geq0. \]

For which values of \(a\) is \(f_a\) starlike in the unit disk? This example starts with a failed case, finds the exact reason, and then solves the whole family.

Run it

python examples/research_workflows/conjecture_counterexample.py \
  --a 1 \
  --output /tmp/gfa-conjecture

What you will see

The result marks \(a=1\) as non-starlike and gives the witness \(z=-3/4\). At that point,

\[ \frac{zf_a'(z)}{f_a(z)}=-2, \]

so its real part is negative. The SVG shows the same logarithmic derivative, and the JSON records the exact witness.

Solve the whole family

Away from the removable value at zero,

\[ q_a(z)=\frac{zf_a'(z)}{f_a(z)} =\frac{1+2az}{1+az}. \]

If \(0<a\leq1/2\), then

\[ \operatorname{Re}q_a(z) >\frac{1-2a}{1-a}\geq0 \]

for every \(|z|<1\). The endpoint \(a=1/2\) is included because the inequality is strict inside the disk.

If \(a>1/2\), the derivative vanishes at \(z=-1/(2a)\) inside the disk. Such a function cannot be univalent, and therefore cannot be starlike.

The corrected statement is

\[ f_a\text{ is starlike exactly when }0\leq a\leq\frac12. \]

Try the endpoint:

python examples/research_workflows/conjecture_counterexample.py \
  --a 1/2 \
  --output /tmp/gfa-conjecture-endpoint

A failed sufficient test is not a counterexample

Some coefficient criteria are sufficient but not necessary. If such a test fails, it has learned nothing about functions outside the criterion. The Koebe dilation \(z/(1-z/4)^2\) is starlike even though the coefficient sum used by one of the package's simple tests exceeds its threshold.

By contrast, the negative value of \(\operatorname{Re}(zf'/f)\) at an interior point is a direct counterexample. The workflow keeps those two outcomes separate.