Thanks Daniel Lichtblau

@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.

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.

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.