Complex conjugation of a complex function
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I have a function with a constant term upfront and was wondering how to obtain the complex conjugate. The function is say:
$$(g(z))^* = (frac{-iGamma}{2pi}times f(z))^* $$
Would the conjugation be conducted 'distributively' such that:
$$ (g(z))^* = (frac{-iGamma}{2pi})^*times f(z)^* $$
$$ rightarrow(g(z))^* = frac{iGamma}{2pi}times f(z)^* $$
also in general for a complex function if a complex function is contained within another how is the conjugate computed, like this?
$$g(z) = h(f(z)) $$
$$ rightarrow g(z)^* = (h(f(z)))^* $$
$$ rightarrow g(z)^* = h^*(f^*(z)) $$
Sorry for the dual question , but thank you all for your time!
complex-analysis complex-numbers
asked Nov 21 at 16:43
QuantumPanda
848
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up vote
0
down vote
favorite
I have a function with a constant term upfront and was wondering how to obtain the complex conjugate. The function is say:
$$(g(z))^* = (frac{-iGamma}{2pi}times f(z))^* $$
Would the conjugation be conducted 'distributively' such that:
$$ (g(z))^* = (frac{-iGamma}{2pi})^*times f(z)^* $$
$$ rightarrow(g(z))^* = frac{iGamma}{2pi}times f(z)^* $$
also in general for a complex function if a complex function is contained within another how is the conjugate computed, like this?
$$g(z) = h(f(z)) $$
$$ rightarrow g(z)^* = (h(f(z)))^* $$
$$ rightarrow g(z)^* = h^*(f^*(z)) $$
Sorry for the dual question , but thank you all for your time!
complex-analysis complex-numbers
asked Nov 21 at 16:43
QuantumPanda
848
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have a function with a constant term upfront and was wondering how to obtain the complex conjugate. The function is say:
$$(g(z))^* = (frac{-iGamma}{2pi}times f(z))^* $$
Would the conjugation be conducted 'distributively' such that:
$$ (g(z))^* = (frac{-iGamma}{2pi})^*times f(z)^* $$
$$ rightarrow(g(z))^* = frac{iGamma}{2pi}times f(z)^* $$
also in general for a complex function if a complex function is contained within another how is the conjugate computed, like this?
$$g(z) = h(f(z)) $$
$$ rightarrow g(z)^* = (h(f(z)))^* $$
$$ rightarrow g(z)^* = h^*(f^*(z)) $$
Sorry for the dual question , but thank you all for your time!
complex-analysis complex-numbers
asked Nov 21 at 16:43
QuantumPanda
848
I have a function with a constant term upfront and was wondering how to obtain the complex conjugate. The function is say:
$$(g(z))^* = (frac{-iGamma}{2pi}times f(z))^* $$
Would the conjugation be conducted 'distributively' such that:
$$ (g(z))^* = (frac{-iGamma}{2pi})^*times f(z)^* $$
$$ rightarrow(g(z))^* = frac{iGamma}{2pi}times f(z)^* $$
also in general for a complex function if a complex function is contained within another how is the conjugate computed, like this?
$$g(z) = h(f(z)) $$
$$ rightarrow g(z)^* = (h(f(z)))^* $$
$$ rightarrow g(z)^* = h^*(f^*(z)) $$
Sorry for the dual question , but thank you all for your time!
complex-analysis complex-numbers
complex-analysis complex-numbers
asked Nov 21 at 16:43
QuantumPanda
848
asked Nov 21 at 16:43
QuantumPanda
848
asked Nov 21 at 16:43
QuantumPanda
848
asked Nov 21 at 16:43
QuantumPanda
848
asked Nov 21 at 16:43
QuantumPanda
848
848
add a comment |
add a comment |
1 Answer
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Concerning your first question, you are right. This follow from the equality $overline{ztimes w}=overline ztimesoverline w$.
Now, if $f$ is a function from $mathbb C$ into itself, let us define $overline f(z)$ as $overline{f(z)}$. Then $overline{hcirc f}=overline hcirc f$, which, in general, is not the same thing as $overline hcircoverline f$.
answered Nov 21 at 16:49
José Carlos Santos
142k20111207
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Concerning your first question, you are right. This follow from the equality $overline{ztimes w}=overline ztimesoverline w$.
Now, if $f$ is a function from $mathbb C$ into itself, let us define $overline f(z)$ as $overline{f(z)}$. Then $overline{hcirc f}=overline hcirc f$, which, in general, is not the same thing as $overline hcircoverline f$.
answered Nov 21 at 16:49
José Carlos Santos
142k20111207
add a comment |
up vote
1
down vote
accepted
Concerning your first question, you are right. This follow from the equality $overline{ztimes w}=overline ztimesoverline w$.
Now, if $f$ is a function from $mathbb C$ into itself, let us define $overline f(z)$ as $overline{f(z)}$. Then $overline{hcirc f}=overline hcirc f$, which, in general, is not the same thing as $overline hcircoverline f$.
answered Nov 21 at 16:49
José Carlos Santos
142k20111207
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Concerning your first question, you are right. This follow from the equality $overline{ztimes w}=overline ztimesoverline w$.
Now, if $f$ is a function from $mathbb C$ into itself, let us define $overline f(z)$ as $overline{f(z)}$. Then $overline{hcirc f}=overline hcirc f$, which, in general, is not the same thing as $overline hcircoverline f$.
answered Nov 21 at 16:49
José Carlos Santos
142k20111207
Concerning your first question, you are right. This follow from the equality $overline{ztimes w}=overline ztimesoverline w$.
Now, if $f$ is a function from $mathbb C$ into itself, let us define $overline f(z)$ as $overline{f(z)}$. Then $overline{hcirc f}=overline hcirc f$, which, in general, is not the same thing as $overline hcircoverline f$.
answered Nov 21 at 16:49
José Carlos Santos
142k20111207
answered Nov 21 at 16:49
José Carlos Santos
142k20111207
answered Nov 21 at 16:49
José Carlos Santos
142k20111207
answered Nov 21 at 16:49
José Carlos Santos
142k20111207
142k20111207
add a comment |
add a comment |
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