# Is f(z) =(1+z)/(1-z) a real function?

• patric44
In summary: The reflection principle states that a real valued function exists on a closed curve in the complex plane if and only if the function's real part is reflected in the real line at every point on the curve. This is definitely not always the case, but it is always the case for certain types of functions.In summary, the function is real, but not in the sense that the conjugate equals the ratio of conjugates.
patric44
Homework Statement
f(z) =(1+z)/(1-z)
Relevant Equations
f(z)*=f(z*)
hi guys
i found this problem in a set of lecture notes I have in complex analysis, is the following function real:
$$f(z)=\frac{1+z}{1-z}\;\;, z=x+iy$$
simple enough we get
$$f=\frac{1+x+iy}{1-x-iy}=$$
after multiplying by the complex conjugate of the denominator and simplification
$$f=\frac{1-x^{2}-y^{2}}{(1-x)^{2}+y^{2}}+\frac{i2y}{(1-x)^{2}+y^{2}}$$
clearly the function has an imaginary part, so i assumed its not real!?, but i found in the notes the its a real function by the following proof
$$f^{*}(z)=\frac{1+z^{*}}{1-z^{*}}=f(z^{*})$$
how is that a proof that its real?!
i will appreciate any help

Are you given constraints on the domain of definition?
Edit: I believe the argument presented( more like intended), is that a value is Real if it's equal to its conjugate.

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Could also be that f(z) must be differentiable, i.e. satisfy Caucy-Riemann?

The "f(z*) = f*(z)" condition is to show a function is analytic, not real.

Could you provide a screenshot or something from your lecture notes?

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patric44, Delta2, dextercioby and 1 other person
How do the notes define "real function"? There appears to be no standard definition of that term (unlike "real-valued function", which this isn't as its codomain is not a subset of the reals unless its domain is restricted to a subset of the reals, in which case $f^{*}(z) = f(z^{*})$ is trivially true).

Real valued Complex functions are nowhere-Analytic. They may be differentiable at some points, but not Analytic ( i.e. differentiable in an open ball about a point).
I believe the problem is the assumption that the conjugate of the ratio equals the ratio of conjugates. I believe this is false. Will double check asap.

patric44 and malawi_glenn
WWGD said:
I believe the problem is the assumption that the conjugate of the ratio equals the ratio of conjugates.
This is true. Representing the numerator and denominator in polar coordinates makes it fairly easy to see.

patric44 and WWGD
WWGD said:
Are you given constraints on the domain of definition?
Edit: The argument presented is that a value is Real if it's equal to its conjugate.
the problem was in a hand written lecture notes with no other information!

WWGD
and those were written by you, or some other?

could the question in the lecture notes be mistaken by another, I mean could he meat to ask , is f(z) analytic in which this conditions proves it?, the problem was asking first for the real part of f(z) then asked, is f(z) a real function?! clearly it wasn't real since it contained an imaginary part!

drmalawi said:
and those were written by you, or some other?
by another person

I hope that you understand the truth of the matter and will cross out that note. It is just confusing.

patric44 and malawi_glenn
WWGD said:
Real valued Complex functions are nowhere-Analytic.
Well … ##f(z) = 0## is a real valued complex function …

mathwonk
For z not 1,
$$\frac{1+z}{1-z}=\frac{(1+z)(1-z^*)}{(1-z)(1-z^*)}=\frac{1-zz^*+z-z^*}{(1-z)(1-z^*)}$$
z-z* is imaginary for any complex z. The relation
$$f(z)^*=f(z^*)$$ seems to have nothing to do with "real", e.g.
$$f(z)=z$$
is not real function in my sense.

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Delta2
anuttarasammyak said:
For z not 1,
$$\frac{1+z}{1-z}=\frac{(1+z)(1-z^*)}{(1-z)(1-z^*)}=\frac{1-zz^*+z-z^*}{(1-z)(1-z^*)}$$
z-z* is imaginary for any complex z. The relation
$$f(z)^*=f(z^*)$$ seems to have nothing to do with "real", e.g.
$$f(z)=z$$
is not real function in my sense.
Yes, my bad. Let me edit. I meant the equality is true for Reals , and that the intent was to show the value was equal to its conjugate, but miswrote.

the property, f(z)* = f(z*), of commuting with taking conjugates seems to be true of holomorphic functions that take real values on real arguments. perhaps that is the meaning of a "real" function?

patric44 and Maarten Havinga
Consider a holomorphic function, defined on a connected domain D in the complex plane which is symmetric about the real axis, i.e. invariant under conjugation, e.g. any open disc about the origin, or the entire complex plane. Then that function commutes with conjugation on that complex domain, f(z)* = f(z*) for all z in D, if and only if its restriction to that part of the real axis meeting that domain takes real values, i.e. f(z) is real for all z in DmeetR. (See the "reflection principle".)

Thus the word "real" in this context seems to refer to the coefficients in the power series expansion of the function, or in this example, to the coefficients in the rational function.

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patric44
mathwonk said:
Consider a holomorphic function, defined on a connected domain in the complex plane which is symmetric about the real axis, i.e. invariant under conjugation. Then that function commutes with conjugation on that complex domain, if and only if its restriction to part of the real axis meeting that domain takes real values. (See the "reflection principle".)
Yes, my bad, I misread that the function as a whole was Real-Valued. I remember the Schwarz reflection principle you're quoting.

It's clear what f(z)*=f(z*) proves to me: if * leaves something unchanged, it's a real number. Thus if z is a real number, f(z)*=f(z*)=f(z) meaning the image of f of the reals is real too.

patric44 and mathwonk

## 1. Is f(z) an algebraic function?

Yes, f(z) is an algebraic function as it can be expressed as a ratio of two polynomials.

## 2. What is the domain of f(z)?

The domain of f(z) is all complex numbers except for z = 1, as this would result in a division by zero.

## 3. Is f(z) a continuous function?

Yes, f(z) is a continuous function as it is defined for all values in its domain and there are no breaks or jumps in its graph.

## 4. What is the range of f(z)?

The range of f(z) is all complex numbers except for z = -1, as this would result in a division by zero.

## 5. Is f(z) a differentiable function?

Yes, f(z) is a differentiable function as it is a ratio of two differentiable functions and the denominator is never equal to zero in its domain.

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