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

[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

 

[WWGD]

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.

 

[malawi_glenn]

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?

 

[pasmith]

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).

 

[WWGD]

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.

 

[FactChecker]

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]

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!

 

[malawi_glenn]

and those were written by you, or some other?

 

[patric44]

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!

 

[patric44]

and those were written by you, or some other?

by another person

 

[FactChecker]

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

 

[Orodruin]

Real valued Complex functions are nowhere-Analytic.

Well … $f(z) = 0$ is a real valued complex function …

 

[anuttarasammyak]

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.

 

[WWGD]

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.

 

[mathwonk]

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?

 

[mathwonk]

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.

 

[WWGD]

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.

 

[Maarten Havinga]

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.