Solving Complex Equations: Does f(z) = 0 imply [f(z)]* = 0?

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Discussion Overview

The discussion revolves around the implications of solving complex equations, specifically whether the condition f(z) = 0 implies that [f(z)]* = 0 and f(z*) = 0. Participants explore the relationships between these equations, particularly in the context of functions with real coefficients.

Discussion Character

  • Exploratory, Technical explanation, Homework-related

Main Points Raised

  • One participant questions whether solving [f(z)]* = 0 provides solutions for f(z) = 0, suggesting that it should hold true.
  • Another participant clarifies that f(z)* = 0 if and only if f(z) = 0, but notes that f(z*) = 0 does not necessarily follow, providing an example with f(z) = i + z.
  • It is proposed that if f is a polynomial with real coefficients, then f(z*) = f(z)* may hold, allowing for solutions to be derived from f(z*) = 0.
  • Participants confirm that [exp(z)]* = exp(z*) is correct, with one explaining how this can be shown using power series and properties of complex conjugation.
  • A participant acknowledges a mistake in their reasoning, revealing that there was an imaginary coefficient involved in their specific case.

Areas of Agreement / Disagreement

Participants generally agree on the relationship between f(z) and its conjugate, but there is no consensus on the implications for f(z*) without additional conditions. The discussion remains unresolved regarding the broader applicability of these relationships.

Contextual Notes

Limitations include the dependence on the nature of the coefficients in the function f(z) and the specific conditions under which the relationships hold. The discussion does not resolve the implications for all types of functions.

praharmitra
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If i am solving for a complex equation f(z) = 0. can i assume that if i solve [f(z)]* = 0, I'll get the solution to the first one??

I mean, does f(z) = 0 imply [f(z)]* = f(z*) = 0?

I think it should be, but its giving me vague answers for a question I'm trying to solve. pls help
 
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Well, obviously f(z)* = 0 if and only if f(z) = 0, since 0* = 0. On the other hand, it is not necessarily true that f(z*) = 0. For instance, if f(z) = i+z, then f(i*) = f(-i) = 0, but f(i) = 2i ≠ 0.

Now, it may be the case that you can prove that f(z*) = f(z)* for every z -- for instance, if f was a polynomial in z with real coefficients, this equation would hold, and then solving for f(z*) = 0 would get you a solution to the original equation. But it's not true in general.
 
ya...i had realized that...but the question i am trying to solve, has real coefficients...and also...

[exp(z)]* = exp(z*) right?

I don't know why i am getting stuck... neway. i'll post it in the homework section if i can't do it
 
praharmitra said:
[exp(z)]* = exp(z*) right?

Yes, that is correct.
 
praharmitra said:
[exp(z)]* = exp(z*) right?
Yes, you can expand the exp function into powerseries. Plugging the properties a*+b*=(a+b)* and a*∙b*=(a∙b)* into these powerseries will lead you pretty quickly to [exp(z)]* = exp(z*)

This work for any holomorphic function you can expand into powerseries with real coefficients.
 
ok... I found my mistake. I can't do what I have been trying to do. Turns our there's an imaginary coefficient after all.

Thanks anyway.
 

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