Mixing
Does swapping $i to -i$ in $f(x)$ always give $overline{f(x)}$?
$begingroup$
I am a physics student. Often we wish to find the complex conjugate of a function , for example
$$ f: mathbb{R}to mathbb{C}, f(x)=e^{ix},$$
To do this, we just replace all $i to -i$. In the special case $f(x)=a(x)+b(x)i$ it is very clear that this is equivalent to conjugation. And for a particular function, like the one above, it is also easy to prove, so that is not my question. But is there a way to see that for arbitrary $f(x)$ or even $f(z)$, making the replacement $i to -i$ always gives the complex conjugate?
Most valuable to me would be a "proof sketch" wherein I understand the reasoning, but maybe not every epsilon.
complex-analysis complex-numbers
asked Jan 8 at 9:58
user27084user27084
576218
$endgroup$
add a comment |
$begingroup$
I am a physics student. Often we wish to find the complex conjugate of a function , for example
$$ f: mathbb{R}to mathbb{C}, f(x)=e^{ix},$$
To do this, we just replace all $i to -i$. In the special case $f(x)=a(x)+b(x)i$ it is very clear that this is equivalent to conjugation. And for a particular function, like the one above, it is also easy to prove, so that is not my question. But is there a way to see that for arbitrary $f(x)$ or even $f(z)$, making the replacement $i to -i$ always gives the complex conjugate?
Most valuable to me would be a "proof sketch" wherein I understand the reasoning, but maybe not every epsilon.
complex-analysis complex-numbers
asked Jan 8 at 9:58
user27084user27084
576218
$endgroup$
1
$begingroup$
the replacement is not well-defined, where is the $i$ in $f(x):=sqrt x$? or in $f(x):=ln x$ for $sqrt{cdot}$ and $ln{cdot}$ expanded in some way on $Bbb C$ when $x<0$? In these cases you would need to write the functions in some canonical form where the $i$ appear. This doesn't seems feasible for an arbitrary functions.
$endgroup$
– Masacroso
Jan 8 at 10:10
$begingroup$
@Masacroso Interesting. So since it does not work for these functions, it seems one might just have to really prove it one function at a time, or look at a restricted class of functions. It seems something should be provable though, since I've seen this replacement done on a large variety of functions and successfully give the conjugate.
$endgroup$
– user27084
Jan 8 at 18:07
add a comment |
$begingroup$
I am a physics student. Often we wish to find the complex conjugate of a function , for example
$$ f: mathbb{R}to mathbb{C}, f(x)=e^{ix},$$
To do this, we just replace all $i to -i$. In the special case $f(x)=a(x)+b(x)i$ it is very clear that this is equivalent to conjugation. And for a particular function, like the one above, it is also easy to prove, so that is not my question. But is there a way to see that for arbitrary $f(x)$ or even $f(z)$, making the replacement $i to -i$ always gives the complex conjugate?
Most valuable to me would be a "proof sketch" wherein I understand the reasoning, but maybe not every epsilon.
complex-analysis complex-numbers
asked Jan 8 at 9:58
user27084user27084
576218
$endgroup$
I am a physics student. Often we wish to find the complex conjugate of a function , for example
$$ f: mathbb{R}to mathbb{C}, f(x)=e^{ix},$$
To do this, we just replace all $i to -i$. In the special case $f(x)=a(x)+b(x)i$ it is very clear that this is equivalent to conjugation. And for a particular function, like the one above, it is also easy to prove, so that is not my question. But is there a way to see that for arbitrary $f(x)$ or even $f(z)$, making the replacement $i to -i$ always gives the complex conjugate?
Most valuable to me would be a "proof sketch" wherein I understand the reasoning, but maybe not every epsilon.
complex-analysis complex-numbers
complex-analysis complex-numbers
asked Jan 8 at 9:58
user27084user27084
576218
asked Jan 8 at 9:58
user27084user27084
576218
asked Jan 8 at 9:58
user27084user27084
576218
asked Jan 8 at 9:58
user27084user27084
576218
asked Jan 8 at 9:58
user27084user27084
576218
576218
1
$begingroup$
the replacement is not well-defined, where is the $i$ in $f(x):=sqrt x$? or in $f(x):=ln x$ for $sqrt{cdot}$ and $ln{cdot}$ expanded in some way on $Bbb C$ when $x<0$? In these cases you would need to write the functions in some canonical form where the $i$ appear. This doesn't seems feasible for an arbitrary functions.
$endgroup$
– Masacroso
Jan 8 at 10:10
$begingroup$
@Masacroso Interesting. So since it does not work for these functions, it seems one might just have to really prove it one function at a time, or look at a restricted class of functions. It seems something should be provable though, since I've seen this replacement done on a large variety of functions and successfully give the conjugate.
$endgroup$
– user27084
Jan 8 at 18:07
add a comment |
1
$begingroup$
the replacement is not well-defined, where is the $i$ in $f(x):=sqrt x$? or in $f(x):=ln x$ for $sqrt{cdot}$ and $ln{cdot}$ expanded in some way on $Bbb C$ when $x<0$? In these cases you would need to write the functions in some canonical form where the $i$ appear. This doesn't seems feasible for an arbitrary functions.
$endgroup$
– Masacroso
Jan 8 at 10:10
$begingroup$
@Masacroso Interesting. So since it does not work for these functions, it seems one might just have to really prove it one function at a time, or look at a restricted class of functions. It seems something should be provable though, since I've seen this replacement done on a large variety of functions and successfully give the conjugate.
$endgroup$
– user27084
Jan 8 at 18:07
1
1
$begingroup$
the replacement is not well-defined, where is the $i$ in $f(x):=sqrt x$? or in $f(x):=ln x$ for $sqrt{cdot}$ and $ln{cdot}$ expanded in some way on $Bbb C$ when $x<0$? In these cases you would need to write the functions in some canonical form where the $i$ appear. This doesn't seems feasible for an arbitrary functions.
$endgroup$
– Masacroso
Jan 8 at 10:10
$begingroup$
the replacement is not well-defined, where is the $i$ in $f(x):=sqrt x$? or in $f(x):=ln x$ for $sqrt{cdot}$ and $ln{cdot}$ expanded in some way on $Bbb C$ when $x<0$? In these cases you would need to write the functions in some canonical form where the $i$ appear. This doesn't seems feasible for an arbitrary functions.
$endgroup$
– Masacroso
Jan 8 at 10:10
$begingroup$
@Masacroso Interesting. So since it does not work for these functions, it seems one might just have to really prove it one function at a time, or look at a restricted class of functions. It seems something should be provable though, since I've seen this replacement done on a large variety of functions and successfully give the conjugate.
$endgroup$
– user27084
Jan 8 at 18:07
$begingroup$
@Masacroso Interesting. So since it does not work for these functions, it seems one might just have to really prove it one function at a time, or look at a restricted class of functions. It seems something should be provable though, since I've seen this replacement done on a large variety of functions and successfully give the conjugate.
$endgroup$
– user27084
Jan 8 at 18:07
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It does not work for $f(z)= iz$ where changing $i$ t0 $-i$ results in $-iz$ which is not the conjugate of $iz$
Another counter example is $f(z)=e^{iz}$ where changing $i$ to $-i$ results in $e^{-iz}$ which is not the conjugate of $e^{iz}$
answered Jan 8 at 10:41
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
$endgroup$
$begingroup$
Thanks for the useful example. The remainder of the question is then just for functions of a real variable $f(x)$ which map in general to complex numbers.
$endgroup$
– user27084
Jan 8 at 18:00
$begingroup$
In case of $f(x) = u(x)+iv(x)$, changing $i$ to $-i$ results in $u(x)-iv(x)$ which is the conjugate of f(x).
$endgroup$
– Mohammad Riazi-Kermani
Jan 8 at 18:48
$begingroup$
Yes, but essentially I was wondering why it works with a large class of more complicated examples, like $f(x) = frac{e^{ix} + i sin(x)}{i(x+2i)(x-3i)}$
$endgroup$
– user27084
Jan 8 at 19:16
$begingroup$
That comes from properties of conjugation which goes through addition, subtraction, multiplication and division.
$endgroup$
– Mohammad Riazi-Kermani
Jan 8 at 19:19
add a comment |
$begingroup$
Well, if you mean $f(bar z)=overline{f(z)}$, then, yes (at least for most well-behaved functions).
To start off, define $h(z)$ (where $zinmathbb C$) as being $=bar z$ (for all $z$).
Now, suppose $u$ and $v$ are any two complex numbers with $u=a+bi$ and $v=c+di$.
Then:
$h(u+v)=h((a+bi)+(c+di))=h(a+c+(b+d)i)=a+c-(b+d)i=(a-bi)+(c-di)=h(a+bi)+h(c+di)=h(u)+h(v)$
And:
$h(uv)=h((a+bi)(c+di))=h(ac+adi+bci+bdi^2)=h(ac-bd+(ad+bc)i)=\ ac-bd-(ad+bc)i=ac-bd-adi-bci=a(c-di)-b(d+ci)=a(c-di)-bi(c-di)=(a-bi)(c-di)=h(a+bi)h(c+di)=h(u)h(v)$
So we have, for any $u,v inmathbb C$: $quad h(u+v)=h(u)+h(v)quad$ & $quad h(uv)=h(u)h(v)$
i.e. $overline{u+v}=bar u +bar vquad$ & $quadoverline{uv}=bar ubar v$
So, suppose we have some polynomial $p(z)=a_0+a_1z+a_2z^2+...+a_nz^n$ where $a_0,a_1,...,a_n$ are all real (so that $overline {a_i}=a_i$ for all i).
Based on what we have shown,
$p(bar z)=a_0+a_1bar z+a_2bar z^2+...+a_nbar z^n=a_0+a_1bar z+a_2overline {z^2}+...+a_noverline {z^n}=overline {a_0}+overline {a_1z}+overline {a_2z^2}+...+overline{a_nz^n}=overline {a_0+a_1z+a_2z^2+...+a_nz^n}=overline {p(z)}$
Now, no part of this argument excluded $f$ from having an 'infinite' degree, so to speak. So you could apply this to any function with a Taylor Series expansion (at least, over the domain where the series was valid).
Edit: $f$ restricted to $mathbb R$ must map back to $mathbb R$ so that the coefficients of its Taylor series are real
answered Jan 8 at 11:04
Cardioid_Ass_22Cardioid_Ass_22
32614
$endgroup$
$begingroup$
It is not exactly the same question I was asking, but a related one and certainly helped my understanding, thank you. At least if $f(x)$ can be written as $f(iy)$, then if $f$ can be written as a convergent series this works.
$endgroup$
– user27084
Jan 8 at 18:11
add a comment |
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2 Answers
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2 Answers
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$begingroup$
It does not work for $f(z)= iz$ where changing $i$ t0 $-i$ results in $-iz$ which is not the conjugate of $iz$
Another counter example is $f(z)=e^{iz}$ where changing $i$ to $-i$ results in $e^{-iz}$ which is not the conjugate of $e^{iz}$
answered Jan 8 at 10:41
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
$endgroup$
$begingroup$
Thanks for the useful example. The remainder of the question is then just for functions of a real variable $f(x)$ which map in general to complex numbers.
$endgroup$
– user27084
Jan 8 at 18:00
$begingroup$
In case of $f(x) = u(x)+iv(x)$, changing $i$ to $-i$ results in $u(x)-iv(x)$ which is the conjugate of f(x).
$endgroup$
– Mohammad Riazi-Kermani
Jan 8 at 18:48
$begingroup$
Yes, but essentially I was wondering why it works with a large class of more complicated examples, like $f(x) = frac{e^{ix} + i sin(x)}{i(x+2i)(x-3i)}$
$endgroup$
– user27084
Jan 8 at 19:16
$begingroup$
That comes from properties of conjugation which goes through addition, subtraction, multiplication and division.
$endgroup$
– Mohammad Riazi-Kermani
Jan 8 at 19:19
add a comment |
$begingroup$
It does not work for $f(z)= iz$ where changing $i$ t0 $-i$ results in $-iz$ which is not the conjugate of $iz$
Another counter example is $f(z)=e^{iz}$ where changing $i$ to $-i$ results in $e^{-iz}$ which is not the conjugate of $e^{iz}$
answered Jan 8 at 10:41
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
$endgroup$
$begingroup$
Thanks for the useful example. The remainder of the question is then just for functions of a real variable $f(x)$ which map in general to complex numbers.
$endgroup$
– user27084
Jan 8 at 18:00
$begingroup$
In case of $f(x) = u(x)+iv(x)$, changing $i$ to $-i$ results in $u(x)-iv(x)$ which is the conjugate of f(x).
$endgroup$
– Mohammad Riazi-Kermani
Jan 8 at 18:48
$begingroup$
Yes, but essentially I was wondering why it works with a large class of more complicated examples, like $f(x) = frac{e^{ix} + i sin(x)}{i(x+2i)(x-3i)}$
$endgroup$
– user27084
Jan 8 at 19:16
$begingroup$
That comes from properties of conjugation which goes through addition, subtraction, multiplication and division.
$endgroup$
– Mohammad Riazi-Kermani
Jan 8 at 19:19
add a comment |
$begingroup$
It does not work for $f(z)= iz$ where changing $i$ t0 $-i$ results in $-iz$ which is not the conjugate of $iz$
Another counter example is $f(z)=e^{iz}$ where changing $i$ to $-i$ results in $e^{-iz}$ which is not the conjugate of $e^{iz}$
answered Jan 8 at 10:41
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
$endgroup$
It does not work for $f(z)= iz$ where changing $i$ t0 $-i$ results in $-iz$ which is not the conjugate of $iz$
Another counter example is $f(z)=e^{iz}$ where changing $i$ to $-i$ results in $e^{-iz}$ which is not the conjugate of $e^{iz}$
answered Jan 8 at 10:41
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
answered Jan 8 at 10:41
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
answered Jan 8 at 10:41
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
answered Jan 8 at 10:41
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
41.5k42061
$begingroup$
Thanks for the useful example. The remainder of the question is then just for functions of a real variable $f(x)$ which map in general to complex numbers.
$endgroup$
– user27084
Jan 8 at 18:00
$begingroup$
In case of $f(x) = u(x)+iv(x)$, changing $i$ to $-i$ results in $u(x)-iv(x)$ which is the conjugate of f(x).
$endgroup$
– Mohammad Riazi-Kermani
Jan 8 at 18:48
$begingroup$
Yes, but essentially I was wondering why it works with a large class of more complicated examples, like $f(x) = frac{e^{ix} + i sin(x)}{i(x+2i)(x-3i)}$
$endgroup$
– user27084
Jan 8 at 19:16
$begingroup$
That comes from properties of conjugation which goes through addition, subtraction, multiplication and division.
$endgroup$
– Mohammad Riazi-Kermani
Jan 8 at 19:19
add a comment |
$begingroup$
Thanks for the useful example. The remainder of the question is then just for functions of a real variable $f(x)$ which map in general to complex numbers.
$endgroup$
– user27084
Jan 8 at 18:00
$begingroup$
In case of $f(x) = u(x)+iv(x)$, changing $i$ to $-i$ results in $u(x)-iv(x)$ which is the conjugate of f(x).
$endgroup$
– Mohammad Riazi-Kermani
Jan 8 at 18:48
$begingroup$
Yes, but essentially I was wondering why it works with a large class of more complicated examples, like $f(x) = frac{e^{ix} + i sin(x)}{i(x+2i)(x-3i)}$
$endgroup$
– user27084
Jan 8 at 19:16
$begingroup$
That comes from properties of conjugation which goes through addition, subtraction, multiplication and division.
$endgroup$
– Mohammad Riazi-Kermani
Jan 8 at 19:19
$begingroup$
Thanks for the useful example. The remainder of the question is then just for functions of a real variable $f(x)$ which map in general to complex numbers.
$endgroup$
– user27084
Jan 8 at 18:00
$begingroup$
Thanks for the useful example. The remainder of the question is then just for functions of a real variable $f(x)$ which map in general to complex numbers.
$endgroup$
– user27084
Jan 8 at 18:00
$begingroup$
In case of $f(x) = u(x)+iv(x)$, changing $i$ to $-i$ results in $u(x)-iv(x)$ which is the conjugate of f(x).
$endgroup$
– Mohammad Riazi-Kermani
Jan 8 at 18:48
$begingroup$
In case of $f(x) = u(x)+iv(x)$, changing $i$ to $-i$ results in $u(x)-iv(x)$ which is the conjugate of f(x).
$endgroup$
– Mohammad Riazi-Kermani
Jan 8 at 18:48
$begingroup$
Yes, but essentially I was wondering why it works with a large class of more complicated examples, like $f(x) = frac{e^{ix} + i sin(x)}{i(x+2i)(x-3i)}$
$endgroup$
– user27084
Jan 8 at 19:16
$begingroup$
Yes, but essentially I was wondering why it works with a large class of more complicated examples, like $f(x) = frac{e^{ix} + i sin(x)}{i(x+2i)(x-3i)}$
$endgroup$
– user27084
Jan 8 at 19:16
$begingroup$
That comes from properties of conjugation which goes through addition, subtraction, multiplication and division.
$endgroup$
– Mohammad Riazi-Kermani
Jan 8 at 19:19
$begingroup$
That comes from properties of conjugation which goes through addition, subtraction, multiplication and division.
$endgroup$
– Mohammad Riazi-Kermani
Jan 8 at 19:19
add a comment |
$begingroup$
Well, if you mean $f(bar z)=overline{f(z)}$, then, yes (at least for most well-behaved functions).
To start off, define $h(z)$ (where $zinmathbb C$) as being $=bar z$ (for all $z$).
Now, suppose $u$ and $v$ are any two complex numbers with $u=a+bi$ and $v=c+di$.
Then:
$h(u+v)=h((a+bi)+(c+di))=h(a+c+(b+d)i)=a+c-(b+d)i=(a-bi)+(c-di)=h(a+bi)+h(c+di)=h(u)+h(v)$
And:
$h(uv)=h((a+bi)(c+di))=h(ac+adi+bci+bdi^2)=h(ac-bd+(ad+bc)i)=\ ac-bd-(ad+bc)i=ac-bd-adi-bci=a(c-di)-b(d+ci)=a(c-di)-bi(c-di)=(a-bi)(c-di)=h(a+bi)h(c+di)=h(u)h(v)$
So we have, for any $u,v inmathbb C$: $quad h(u+v)=h(u)+h(v)quad$ & $quad h(uv)=h(u)h(v)$
i.e. $overline{u+v}=bar u +bar vquad$ & $quadoverline{uv}=bar ubar v$
So, suppose we have some polynomial $p(z)=a_0+a_1z+a_2z^2+...+a_nz^n$ where $a_0,a_1,...,a_n$ are all real (so that $overline {a_i}=a_i$ for all i).
Based on what we have shown,
$p(bar z)=a_0+a_1bar z+a_2bar z^2+...+a_nbar z^n=a_0+a_1bar z+a_2overline {z^2}+...+a_noverline {z^n}=overline {a_0}+overline {a_1z}+overline {a_2z^2}+...+overline{a_nz^n}=overline {a_0+a_1z+a_2z^2+...+a_nz^n}=overline {p(z)}$
Now, no part of this argument excluded $f$ from having an 'infinite' degree, so to speak. So you could apply this to any function with a Taylor Series expansion (at least, over the domain where the series was valid).
Edit: $f$ restricted to $mathbb R$ must map back to $mathbb R$ so that the coefficients of its Taylor series are real
answered Jan 8 at 11:04
Cardioid_Ass_22Cardioid_Ass_22
32614
$endgroup$
$begingroup$
It is not exactly the same question I was asking, but a related one and certainly helped my understanding, thank you. At least if $f(x)$ can be written as $f(iy)$, then if $f$ can be written as a convergent series this works.
$endgroup$
– user27084
Jan 8 at 18:11
add a comment |
$begingroup$
Well, if you mean $f(bar z)=overline{f(z)}$, then, yes (at least for most well-behaved functions).
To start off, define $h(z)$ (where $zinmathbb C$) as being $=bar z$ (for all $z$).
Now, suppose $u$ and $v$ are any two complex numbers with $u=a+bi$ and $v=c+di$.
Then:
$h(u+v)=h((a+bi)+(c+di))=h(a+c+(b+d)i)=a+c-(b+d)i=(a-bi)+(c-di)=h(a+bi)+h(c+di)=h(u)+h(v)$
And:
$h(uv)=h((a+bi)(c+di))=h(ac+adi+bci+bdi^2)=h(ac-bd+(ad+bc)i)=\ ac-bd-(ad+bc)i=ac-bd-adi-bci=a(c-di)-b(d+ci)=a(c-di)-bi(c-di)=(a-bi)(c-di)=h(a+bi)h(c+di)=h(u)h(v)$
So we have, for any $u,v inmathbb C$: $quad h(u+v)=h(u)+h(v)quad$ & $quad h(uv)=h(u)h(v)$
i.e. $overline{u+v}=bar u +bar vquad$ & $quadoverline{uv}=bar ubar v$
So, suppose we have some polynomial $p(z)=a_0+a_1z+a_2z^2+...+a_nz^n$ where $a_0,a_1,...,a_n$ are all real (so that $overline {a_i}=a_i$ for all i).
Based on what we have shown,
$p(bar z)=a_0+a_1bar z+a_2bar z^2+...+a_nbar z^n=a_0+a_1bar z+a_2overline {z^2}+...+a_noverline {z^n}=overline {a_0}+overline {a_1z}+overline {a_2z^2}+...+overline{a_nz^n}=overline {a_0+a_1z+a_2z^2+...+a_nz^n}=overline {p(z)}$
Now, no part of this argument excluded $f$ from having an 'infinite' degree, so to speak. So you could apply this to any function with a Taylor Series expansion (at least, over the domain where the series was valid).
Edit: $f$ restricted to $mathbb R$ must map back to $mathbb R$ so that the coefficients of its Taylor series are real
answered Jan 8 at 11:04
Cardioid_Ass_22Cardioid_Ass_22
32614
$endgroup$
$begingroup$
It is not exactly the same question I was asking, but a related one and certainly helped my understanding, thank you. At least if $f(x)$ can be written as $f(iy)$, then if $f$ can be written as a convergent series this works.
$endgroup$
– user27084
Jan 8 at 18:11
add a comment |
$begingroup$
Well, if you mean $f(bar z)=overline{f(z)}$, then, yes (at least for most well-behaved functions).
To start off, define $h(z)$ (where $zinmathbb C$) as being $=bar z$ (for all $z$).
Now, suppose $u$ and $v$ are any two complex numbers with $u=a+bi$ and $v=c+di$.
Then:
$h(u+v)=h((a+bi)+(c+di))=h(a+c+(b+d)i)=a+c-(b+d)i=(a-bi)+(c-di)=h(a+bi)+h(c+di)=h(u)+h(v)$
And:
$h(uv)=h((a+bi)(c+di))=h(ac+adi+bci+bdi^2)=h(ac-bd+(ad+bc)i)=\ ac-bd-(ad+bc)i=ac-bd-adi-bci=a(c-di)-b(d+ci)=a(c-di)-bi(c-di)=(a-bi)(c-di)=h(a+bi)h(c+di)=h(u)h(v)$
So we have, for any $u,v inmathbb C$: $quad h(u+v)=h(u)+h(v)quad$ & $quad h(uv)=h(u)h(v)$
i.e. $overline{u+v}=bar u +bar vquad$ & $quadoverline{uv}=bar ubar v$
So, suppose we have some polynomial $p(z)=a_0+a_1z+a_2z^2+...+a_nz^n$ where $a_0,a_1,...,a_n$ are all real (so that $overline {a_i}=a_i$ for all i).
Based on what we have shown,
$p(bar z)=a_0+a_1bar z+a_2bar z^2+...+a_nbar z^n=a_0+a_1bar z+a_2overline {z^2}+...+a_noverline {z^n}=overline {a_0}+overline {a_1z}+overline {a_2z^2}+...+overline{a_nz^n}=overline {a_0+a_1z+a_2z^2+...+a_nz^n}=overline {p(z)}$
Now, no part of this argument excluded $f$ from having an 'infinite' degree, so to speak. So you could apply this to any function with a Taylor Series expansion (at least, over the domain where the series was valid).
Edit: $f$ restricted to $mathbb R$ must map back to $mathbb R$ so that the coefficients of its Taylor series are real
answered Jan 8 at 11:04
Cardioid_Ass_22Cardioid_Ass_22
32614
$endgroup$
Well, if you mean $f(bar z)=overline{f(z)}$, then, yes (at least for most well-behaved functions).
To start off, define $h(z)$ (where $zinmathbb C$) as being $=bar z$ (for all $z$).
Now, suppose $u$ and $v$ are any two complex numbers with $u=a+bi$ and $v=c+di$.
Then:
$h(u+v)=h((a+bi)+(c+di))=h(a+c+(b+d)i)=a+c-(b+d)i=(a-bi)+(c-di)=h(a+bi)+h(c+di)=h(u)+h(v)$
And:
$h(uv)=h((a+bi)(c+di))=h(ac+adi+bci+bdi^2)=h(ac-bd+(ad+bc)i)=\ ac-bd-(ad+bc)i=ac-bd-adi-bci=a(c-di)-b(d+ci)=a(c-di)-bi(c-di)=(a-bi)(c-di)=h(a+bi)h(c+di)=h(u)h(v)$
So we have, for any $u,v inmathbb C$: $quad h(u+v)=h(u)+h(v)quad$ & $quad h(uv)=h(u)h(v)$
i.e. $overline{u+v}=bar u +bar vquad$ & $quadoverline{uv}=bar ubar v$
So, suppose we have some polynomial $p(z)=a_0+a_1z+a_2z^2+...+a_nz^n$ where $a_0,a_1,...,a_n$ are all real (so that $overline {a_i}=a_i$ for all i).
Based on what we have shown,
$p(bar z)=a_0+a_1bar z+a_2bar z^2+...+a_nbar z^n=a_0+a_1bar z+a_2overline {z^2}+...+a_noverline {z^n}=overline {a_0}+overline {a_1z}+overline {a_2z^2}+...+overline{a_nz^n}=overline {a_0+a_1z+a_2z^2+...+a_nz^n}=overline {p(z)}$
Now, no part of this argument excluded $f$ from having an 'infinite' degree, so to speak. So you could apply this to any function with a Taylor Series expansion (at least, over the domain where the series was valid).
Edit: $f$ restricted to $mathbb R$ must map back to $mathbb R$ so that the coefficients of its Taylor series are real
answered Jan 8 at 11:04
Cardioid_Ass_22Cardioid_Ass_22
32614
edited Jan 8 at 11:31
answered Jan 8 at 11:04
Cardioid_Ass_22Cardioid_Ass_22
32614
answered Jan 8 at 11:04
Cardioid_Ass_22Cardioid_Ass_22
32614
answered Jan 8 at 11:04
Cardioid_Ass_22Cardioid_Ass_22
32614
32614
$begingroup$
It is not exactly the same question I was asking, but a related one and certainly helped my understanding, thank you. At least if $f(x)$ can be written as $f(iy)$, then if $f$ can be written as a convergent series this works.
$endgroup$
– user27084
Jan 8 at 18:11
add a comment |
$begingroup$
It is not exactly the same question I was asking, but a related one and certainly helped my understanding, thank you. At least if $f(x)$ can be written as $f(iy)$, then if $f$ can be written as a convergent series this works.
$endgroup$
– user27084
Jan 8 at 18:11
$begingroup$
It is not exactly the same question I was asking, but a related one and certainly helped my understanding, thank you. At least if $f(x)$ can be written as $f(iy)$, then if $f$ can be written as a convergent series this works.
$endgroup$
– user27084
Jan 8 at 18:11
$begingroup$
It is not exactly the same question I was asking, but a related one and certainly helped my understanding, thank you. At least if $f(x)$ can be written as $f(iy)$, then if $f$ can be written as a convergent series this works.
$endgroup$
– user27084
Jan 8 at 18:11
add a comment |
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$begingroup$
the replacement is not well-defined, where is the $i$ in $f(x):=sqrt x$? or in $f(x):=ln x$ for $sqrt{cdot}$ and $ln{cdot}$ expanded in some way on $Bbb C$ when $x<0$? In these cases you would need to write the functions in some canonical form where the $i$ appear. This doesn't seems feasible for an arbitrary functions.
$endgroup$
– Masacroso
Jan 8 at 10:10
$begingroup$
@Masacroso Interesting. So since it does not work for these functions, it seems one might just have to really prove it one function at a time, or look at a restricted class of functions. It seems something should be provable though, since I've seen this replacement done on a large variety of functions and successfully give the conjugate.
$endgroup$
– user27084
Jan 8 at 18:07