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

• praharmitra
In summary, the conversation discusses whether solving for [f(z)]* = 0 can provide a solution to f(z) = 0. It is stated that while f(z)* = 0 if and only if f(z) = 0, it is not always true that f(z*) = 0. However, for functions with real coefficients, this equation can hold and lead to a solution for the original equation. It is also mentioned that the same applies for the exponential function. Ultimately, it is concluded that this approach may not work for all functions, as seen in the example where there is an imaginary coefficient.
praharmitra
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

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.

## 1. What is the significance of the complex function f(z) = 0?

The complex function f(z) = 0 represents the set of complex numbers that satisfy the equation. This means that when z is substituted into the function, it will result in a value of 0. It is commonly used in solving complex equations to find the roots or solutions.

## 2. How does the statement [f(z)]* = 0 relate to f(z) = 0?

The symbol [f(z)]* represents the complex conjugate of f(z), which means that the imaginary part of the function is multiplied by -1. In other words, [f(z)]* = a - bi, where f(z) = a + bi. Therefore, if f(z) = 0, then [f(z)]* = 0 as well.

## 3. Can the statement [f(z)]* = 0 be used to solve complex equations?

Yes, the statement [f(z)]* = 0 can be used to solve complex equations. By taking the complex conjugate of both sides, we can often simplify the equation and solve for the variable z. However, this method may not work for all equations and should be used with caution.

## 4. Is f(z) = 0 the only solution to complex equations?

No, f(z) = 0 is not the only solution to complex equations. There can be multiple solutions or roots to a complex equation, and f(z) = 0 is just one of them. Other solutions can be found by manipulating the equation or using different methods such as graphing or numerical approximation.

## 5. Are there any exceptions to the statement [f(z)]* = 0 if f(z) = 0?

Yes, there can be exceptions to the statement [f(z)]* = 0 if f(z) = 0. One exception is when z is a purely imaginary number, meaning it has no real component. In this case, the complex conjugate of f(z) will not equal 0. Additionally, if the function f(z) is undefined at z = 0, then [f(z)]* = 0 may not hold true. It is important to consider the specific function and its domain when using this statement.

• General Math
Replies
2
Views
1K
• General Math
Replies
1
Views
884
• General Math
Replies
13
Views
4K
• General Math
Replies
1
Views
925
• General Math
Replies
13
Views
2K
• Calculus and Beyond Homework Help
Replies
3
Views
434
• Calculus
Replies
3
Views
456
• Calculus and Beyond Homework Help
Replies
17
Views
1K
• General Math
Replies
4
Views
1K
• Topology and Analysis
Replies
7
Views
2K