Dear Michael Rogers. You gave 2 false solutions. It costs someone a fortune. Therefore, false solutions cannot be given. Let Wolfram check his decisions himself.

when x≈-1.61094633143513 + 1.24647431055970 i x^5-3x^(2+0.3i)-32=-2.2382096176443156e-13 - 1.1102230246251565e-13 i

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

when x≈0.41631317642439 - 1.80284912785277 i x^5-3x^(2+0.3i)-32=-10,468522626439754 - 7,7696708390686755 i

when x≈0.54874748513938 + 1.82722607360372 i x^5-3x^(2+0.3i)-32=1.5631940186722204e-13 + 1.3011813848606835e-13 i

when x≈2.14480223951730 + 0.03341204516261 i x^5-3x^(2+0.3i)-32=3.410605131648481e-13 + 3.3173463975799677e-13 i

I ask again. How can Wolfram get all 5 solutions of the equation x^5-3x^(2+0.3i)-32=0. But be sure not to make false decisions. Only true solutions are required.

You can set the precision to at least 100 digits. But these 2 solutions when x≈-1.46272237563773 - 1.44560206980216 i x^5-3x^(2+0.3i)-32=-4,439712865010737 + 21,40303899299964 i

when x≈0.41631317642439 - 1.80284912785277 i x^5-3x^(2+0.3i)-32=-10,468522626439754 - 7,7696708390686755 i

they will never give 0. These decisions are false. True solutions and a graph with points in the complex plane are shown in the book on the ultraradical. Take a look and see for yourself. I am attaching these graphs and a table of true solutions to this post.

Attachments:

ultraradical_Bring.pdf

Dear Neil Singer? спасибо за ваше исследования. To understand who is wrong, look at your calculator for what it is (-1,46272237563773 - 1,44560206980216 i)^(2+0.3i)

A: =-1.74483229995554 + 8.41151553899432*i

OR B: = -0,2649280115795637 + 1,2771692076895944 i

From the point of view of the operation of raising to a complex power, a complex number, they are both correct. But from the point of view of the equation, there are only 5 solutions. One method produces some solutions, and the other method gives different solutions. So which one is correct? How the calculator should raise to a complex power: (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... My calculator takes n=0 Your calculator takes n=-1. Both your calculator and mine do not violate the rule of raising to a complex power, but they give different consequences. So that the consequences do not contradict, we must take n=0. Why does your calculator take n=-1? If he doesn't take n=0, then let him take n=-5, +27... How to be in this situation? Whose calculator is set up correctly? Only in fairness, not in favor of authority, with a strange choice of n=-1

Equation solutions x^5-3x^(2+0,3i)-32 Total 5. My calculator showed some roots, yours others. Who to believe?

Sorry, I meant

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

Sergei,

Every time I’ve seen discrepancies like this it usually stems from a definition or branch cut in a function (probably related to the issue Gianluca pointed out). You have surpassed my knowledge of mathematics on this one! I would ask someone with deeper understanding of mathematics and Mathematica such as @Daniel Lichtblau. He may have some insight and suggestions for you.

Regards

Neil

@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

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?!!!

Thanks for the clear answer. It turns out that mathematics depends on programmers for all resources. I know of only 2 resources that are oriented in branching after the complex logarithm. But their programmers can't work with big numbers. They use number type = number. It has an accuracy of only 2^52. And Wolfram's accuracy is unlimited. For the sake of mathematicians who need it, ask your programmers to work with ramifications after the complex logarithm.

It is not difficult to create this order. It is enough to focus on the value of the value "n" in the complex logarithm formula. They can use the existing experience of these resources. Everything is simple there. Each root, each branch on its shelf. The shelf shows who was received and how, and this is enough to know with whom and how to work. I'm attaching to this post a research report on ultra-radical function. See how easy it is to navigate through different roots, different branches. Any student will understand this.

Attachments:

ultraradical_Bring.pdf

Thanks Daniel Lichtblau

x ≈ 0.4163131764243939365848417988905193285865 - 1.802849127852765001086475519941363599262 i when (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=-1

Yes. if n=-1 when x^(2+3i/10) (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...