By default fsolve will compute all the real roots of a univariate polynomial, in a single call.

Calling allvalues of a forced RootOf wastes time. At higher working precision it can also allow (very small) imaginary artefacts to appear unnecessarily in the result.

One can simply compute like,

    fsolve(simplify(ChebyshevT(N, x)))

and get pretty good efficiency (including at higher working precision) without unnecessary imaginary artefacts (at higher working precision).

I'll point out that the OP's mentioned target expression,

   I*ln((-I*z+b)/(I*z+b))

is not equal to his mentioned input expression,

   I*(ln(-I*z+b)-ln(I*z+b))

for all z and b. That's is what member vv has demonstrated, by example.

My point is that it's not the case that combine(...,symbolic) is misbehaving. The difference primarily lies in the fact that the two expressions (input, and target) mentioned by the OP are not equal everywhere. The demonstrated difference (for some b and z) is inherent to the expressions themselves.

There are various ways to force the combining of the ln terms in the original expression, ignoring branch distinctions. If you perform manipulations involving ln(... exp(something) ...) or combine(expr,symbolic) then be aware of possible consequences.

I'm stating all this primarily for the benefit of the OP.

@C_R If you really wanted you could pass additional options to g, so that those were passed on to the Int call.

An option to force the specific (coarser) rounding may be part of that. (Additional cruft digits may actually be detrimental here.)

For what it might be worth to you,

Download comb_ln_alt.mw

By the way, if you want to spawn off a related conversation you could use the "Branch" button, instead of starting a wholly distint thread. That automatically inserts URL links between parent and child threads, which helps keep the related content easily accessible. The "Branch" button appears at the bottom of an item, eg. next to the Reply Button.

@janhardo Without extras,

Download combine_symbolic_ex.mw

Why is it not direct enough for you?

You might be confusing that terse approach with additional explanation, manipulation (or leading sign, etc...), and a slightly different starting expression, as shown above.

By the way, in the Reply above you have a typo, so that the expression doesn't agree with your target reformulation. You claim that you want,
    I*ln ((-z*I + b)/(z*I + b))
but you start with,
    I*ln(-z*I + b) - ln(z*I + b)
That's a typo on your part. Presumably you intended to write,
   I*(ln(-z*I + b) - ln(z*I + b))

@janhardo You could also attempt numerical validation.

You could do that for a specific pair of values for b and z.

Or you could plot it, testing many values of b and z.

The following computes for many (here, purely real) values of both b and z. It has some numeric roundoff error but the difference is clearly very nearly zero. That means that there's no sign difference (which concerned you). Of course the value b=0 is problematic, but we already knew that.

plot3d([Re,Im](arctan(z/b)/b-ans),b=-4..4,z=-4..4,
       color=[red,blue],grid=[100,100],view=-1e-14..1e-14);

@Carl Love A starting point...

@Ronan We saw that you might now be using that as a Start Page.

I suppose that you could also (right-click) toggle off that Table's borders, just prior to exporting.

@janhardo Your call diff(%) is invalid syntax, and makes no sense. You've forgotten to tell Maple the name with which to differentiate.

Perhaps you meant to differentiate with respect to x.

simplify(exp(x*alpha)*cos(x*beta)*alpha/(alpha^2 + beta^2)
         + exp(x*alpha)*sin(x*beta)*beta/(alpha^2 + beta^2));

diff(%, x);

simplify(%);

@Christopher2222 Yes, there have been several postings over the years in which it's been shown that `simplify` does not try all (speculative) flavours of calls to `collect` and so can miss the most compact forms.

That is now quite well "known".

There is a balance between power and time, in simplification. Of course it's easy to say that some trivial example (like this one) ought to be handled. But one of the problems is in drawing a line in the sand, because some examples get slowed down tremendously while others are easy. Where a line is drawn will always be unsatisfactory to some people.

The subtleties here are more than most people realize. To make effective progress requires a huge collection of examples, to guard against severe behavioral regression.

However, `collect` is tangential here since it's already been mentioned that internally `simplify` actually produces the desired, partially factored forms under discussion. The central issue seems to be why its not always returned.

@mmcdara We're discussing how it decides which to accept. In both cases the partially factored result is generated and -- somehow -- analyzed.

That fully factored form doesn't get generated by the code we're discussing, so doesn't enter into the specifics of the computation path under discussion. The fact that simplify doesn't always call factor is also interesting, but tangential here. (It also doesn't always call collect.These are not a obscure facts, but are tangential in this quite specific context.)

@Carl Love I'd agree that sieving out nonreal results is a good last step.

But I'm not sure that option `real` to `solve` is best here. It can even interfere in some nonpolynomial cases.

The Jave GUI's 3D plot driver can finicky, especially w.r.t. operating sysyem graphic drivers.

Precisely which version of ubuntu are you using? Is it in the list of 64bit Linux versions officially supported by Maple 2019?

What's the video card and what (if any 3rd party) graphics driver might you have installed for it?

@vv You're now refering to a different procedure than you mentioned before, `simplify\size\size\object`.

But that's beside my main point. The original Question asked why t^2*x^2+t^2*y^2 partially factored while t^2*x^2-t^2*y^2 did not. Your answer of why the former happens does not explain why the latter does not.

Both of those respectively have a smaller measure for the corresponding partially factored form, according to `simplify\size\size`. or even `simplify/size/size/object` for that matter. Your Answer omits this additional information for the other expression,

`simplify/size/size`~([ t^2*x^2+t^2*y^2, t^2*(x^2+y^2) ]);

                [56, 38]

`simplify/size/size`~([ t^2*x^2-t^2*y^2, t^2*(x^2-y^2) ]);

                [59, 46]

Or,

`simplify/size/size/object`~([ t^2*x^2+t^2*y^2, t^2*(x^2+y^2) ]);

                [16, 12]

`simplify/size/size/object`~([ t^2*x^2-t^2*y^2, t^2*(x^2-y^2) ]);

                [17, 13]

So those measures' pairwise straight comparisons do not explain why one of them does get partially factored and one of them does not.

There must be some other key difference (be it structural, or that the measure comparison is graded, etc.)

@vv I don't see how `simplify/size/size` relates to whether t^2*x^2+t^2*y^2
 gets factored, or why t^2*x^2-t^2*y^2 does not get partially factored. I'll explain what I mean:

As far as I can see, a measure by `simplify/size/size` is not compared against another when dealing with t^2*x^2+t^2*y^2, since `simplify/size/size` gets called only once. So that part of the explanation -- that results from that procedure are compared -- seems inaccurate.

More importantly, it doesn't address the important aspect of the actual Question, which is the key point that t^2*x^2-t^2*y^2 does not get factored wrt t even though a similar measurement -- were it even made twice -- would provide the same kind of straight relative comparison.

I don't understand all the up-votes, as this doesn't seem to be a correct explanation. It doesn't rely on the measurement of t^2*(x^2-y^2).