Thanks Daniel Lichtblau

Daniel Lichtblau. Understand that not only your branch is eligible to apply. Here is one solution to this equation. How will you use it if you don't know what branch it is from?!!! (-1.5940218425876456 -1.2172363174830494i)^(2+0.3i) 0,07430109274278866+1,2867625442131887i, n=0 0,48935160069992917+8,474697847462767i, n=-1 0,011281566004640335+0,19537662285926974, n=1 (X+xi)^(Y+yi)=e^((Y+yi)*Ln(X+xi)) Ln(X+xi)=ln((X^2+x^2)^(1/2))+i(f+2Pn), tg(f)=x/X, n=0,1,2... It's pointless. The branch must be specified. Otherwise, you will not know what to do with this decision.

How will you use it if you don't know what branch it is from?!!!

@Daniel Lichtblau. The problem is visible due to the lack of a link to the branch. I will give you the root x= -1.5940218425876456 -1.2172363174830494i But I won't tell you what branch it is from. Then you won't be able to use it. The number of branches is infinite. For example, you want to know if x^5-3x^(2+0.3i)-32=0 is equal or not. You need to find a branch like a needle in a haystack. And if I tell another person that the branch is from n=0 He gets it right the first time. (X+xi)^(Y+yi)=e^((Y+yi)(ln((X^2+x^2)^(1/2))+i(arctg(x/X)+2Pn)))

What's the use of knowing this solution if you don't know which branch it's from? At such roots, a branch must be indicated. I attached an example where you can see how short, but it is clear which branch has which roots. Without such a signature, the roots of the unknown branch are meaningless.

Attachments:

ultraradical_Bring.pdf

@NeilSinger Not much I can add. The five solutions all validate just fine. The claim that two are incorrect is just wrong. Mathematica uses the (standard) principal branch of the logarithm, and this is then used in the definition of powers. So possibly there is just a misunderstanding on the part of the original poster about the definition of x^(2+.3*I).

I will note that FindRoot replicates the two in question (given suitable starting points). So there is general agreement between different methodologies.

Dear Gianluca Gorni. You're right. We are the problem with this. Prudent calculators suggest choosing what n is equal to. To know how to process it later and why such a result. When not chosen, by default, n=0, so that different researchers who solve the same problem with different methods do not get disappointed in each other. We need Wolfram to do the same. It is rarely important, but when it is important, it is very important.

Raising a complex number to another complex number is an operation fraught with trouble. It has infinitely many different values. Take for example I^I. Mathematica assumes that this means Exp[I Log[I]], where Log[I] is the principal value of the Logarithm, which is an arbitrary choice. Other values for the Log give different results for I^I:

generalLogI = Solve[E^x == I, x][[1]]
threeSampleValuesForLogI = 
 logI /. {{C[1] -> 0}, {C[1] -> 1}, {C[1] -> -1}}
E^(I x) /. threeSampleValuesForLogI

I am not sure, but the disagreement of the results may stem from different choices for the complex logarithm.

As you saw above, I separately verified the roots are correct by substituting them back into the original equation. It seems highly improbable that Solve would get the answer wrong and then Mathematica would suddenly be unable to compute a polynomial numerically (and in a way that happens to exactly match an error in the Solve routine). It seems most likely that your reference has an error since it appears to have been derived manually (or there was a typographical error). For completeness, I tried this expression in an independent complex number calculator and I do not get your answer.

x≈-1.46272237563773 - 1.44560206980216 i x^5-3x^(2+0.3i)-32=-4,439712865010737 + 21,40303899299964 I

I get 0.

S = Solve[x^5 - 3 x^(2 + 3/10* I) - 32 == 0 && -10 < Re[x] < 10 && -10 < Im[x] < 10, x]
x^5 - 3 x^(2 + 3/10* I) - 32 == 0 /. S // FullSimplify

(*{True, True, True, True, True}*)

Exact solution.

x^5 - 3 x^(2 + 3/10*I) - 32 == 0 /. 
 x -> 0.4163131764243939365848417988905193285865 - 
   I*1.802849127852765001086475519941363599262
(*True*)

Sergey,

Michael's answer seems correct to me to machine precision. You get a warning about machine precision and an answer. The warning is that you are only getting answers to machine precision and no better. You can get better precision if you use 3/10 instead of 0.3 and you can them specify any precision you want. For example:

In[22]:= ans = 
 Solve[x^5 - 3 x^(2 + 3 I/10) - 32 == 0 && -10 < Re[x] < 10 && -10 < 
    Im[x] < 10, x, WorkingPrecision -> 50]

Out[22]= {{x -> -1.6109463314351326392988566553712692531598610618242 \
+ 1.2464743105597007675161723821942332067829191674769 I}, {x -> \
-1.4627223756377303297165335908679364247281052953771 - 
    1.4456020698021564339153511956933357083440670693014 I}, {x -> 
   0.4163131764243939365848417988905193285864645894657 - 
    1.8028491278527650010864755199413635992617259467944 I}, {x -> 
   0.5487474851393810132123387637492439707377603539200 + 
    1.8272260736037174136816701325357997471978574779219 I}, {x -> 
   2.1448022395172964092246332752422095752340324284946 + 
    0.0334120451626065251059523893164733974520934509475 I}}

You can confirm your answer with:

In[23]:= x^5 - 3 x^(2 + 3 I/10) - 32 /. ans

Out[23]= {0.*10^-48 + 0.*10^-48 I, 0.*10^-48 + 0.*10^-48 I, 
 0.*10^-48 + 0.*10^-48 I, 0.*10^-48 + 0.*10^-48 I, 
 0.*10^-48 + 0.*10^-48 I}

So you are getting close to 50 digits of precision here.

Michael's answer was correct to machine precision (15 digits on my machine) with is usually all anyone ever needs.

Regards,

Neil

I don't know which Wolfram product you wish to use but this Alpha query shows five solutions.

This WL command does too:

Solve[x^5 - 3 x^(2 + 0.3 I) - 32 == 0 && 
  -10 < Re[x] < 10 && -10 < Im[x] < 10, x]

For a transcendental equation (which this is because of x^(2 + 0.3 i)), one often needs to specify a bounded domain.