Mixing
$log(z_1z_2)=log(z_1)+log(z_2)$ where $z_1,z_2in mathbb{C}${0}
$begingroup$
I need to prove the set identity of the complex logarithm $log(z_1z_2)=log(z_1)+log(z_2)$ where $z_1,z_2in mathbb{C}$.
$log(z_1)+log(z_2)=${$log|z_1|+log|z_2|+i(text{Arg}(z_1)+text{Arg}(z_2)+4kpi i|kin mathbb{Z}$}$implies$
{$log|z_1z_2|+i((text{Arg}(z_1)+2kpi i)+(text{Arg}(z_2)+2kpi i)|kin mathbb{Z}$}$implies$
{$log|z_1z_2|+i(text{arg}(z_1)+text{arg}(z_2))|kin mathbb{Z}$}$implies$
{$log|z_1z_2|+i(text{Arg}(z_1z_2)+2kpi i)|kin mathbb{Z}$}$implies$$log(z_1z_2)$
This is wclearly wrong I guess but this is how I tried to approach it. Would aprreciate any answers.Thanks a lot
complex-analysis analysis complex-numbers
asked Oct 20 '14 at 16:15
HeisenbergHeisenberg
1,3231741
$endgroup$
add a comment |
$begingroup$
I need to prove the set identity of the complex logarithm $log(z_1z_2)=log(z_1)+log(z_2)$ where $z_1,z_2in mathbb{C}$.
$log(z_1)+log(z_2)=${$log|z_1|+log|z_2|+i(text{Arg}(z_1)+text{Arg}(z_2)+4kpi i|kin mathbb{Z}$}$implies$
{$log|z_1z_2|+i((text{Arg}(z_1)+2kpi i)+(text{Arg}(z_2)+2kpi i)|kin mathbb{Z}$}$implies$
{$log|z_1z_2|+i(text{arg}(z_1)+text{arg}(z_2))|kin mathbb{Z}$}$implies$
{$log|z_1z_2|+i(text{Arg}(z_1z_2)+2kpi i)|kin mathbb{Z}$}$implies$$log(z_1z_2)$
This is wclearly wrong I guess but this is how I tried to approach it. Would aprreciate any answers.Thanks a lot
complex-analysis analysis complex-numbers
asked Oct 20 '14 at 16:15
HeisenbergHeisenberg
1,3231741
$endgroup$
1
$begingroup$
this identity is simply not true for all complex values $z_1,z_2$ (and especially when one of them is $0$ ...)
$endgroup$
– mercio
Oct 20 '14 at 16:26
$begingroup$
@mercio Sorry about that edited the title thanks
$endgroup$
– Heisenberg
Oct 20 '14 at 16:27
add a comment |
$begingroup$
I need to prove the set identity of the complex logarithm $log(z_1z_2)=log(z_1)+log(z_2)$ where $z_1,z_2in mathbb{C}$.
$log(z_1)+log(z_2)=${$log|z_1|+log|z_2|+i(text{Arg}(z_1)+text{Arg}(z_2)+4kpi i|kin mathbb{Z}$}$implies$
{$log|z_1z_2|+i((text{Arg}(z_1)+2kpi i)+(text{Arg}(z_2)+2kpi i)|kin mathbb{Z}$}$implies$
{$log|z_1z_2|+i(text{arg}(z_1)+text{arg}(z_2))|kin mathbb{Z}$}$implies$
{$log|z_1z_2|+i(text{Arg}(z_1z_2)+2kpi i)|kin mathbb{Z}$}$implies$$log(z_1z_2)$
This is wclearly wrong I guess but this is how I tried to approach it. Would aprreciate any answers.Thanks a lot
complex-analysis analysis complex-numbers
asked Oct 20 '14 at 16:15
HeisenbergHeisenberg
1,3231741
$endgroup$
I need to prove the set identity of the complex logarithm $log(z_1z_2)=log(z_1)+log(z_2)$ where $z_1,z_2in mathbb{C}$.
$log(z_1)+log(z_2)=${$log|z_1|+log|z_2|+i(text{Arg}(z_1)+text{Arg}(z_2)+4kpi i|kin mathbb{Z}$}$implies$
{$log|z_1z_2|+i((text{Arg}(z_1)+2kpi i)+(text{Arg}(z_2)+2kpi i)|kin mathbb{Z}$}$implies$
{$log|z_1z_2|+i(text{arg}(z_1)+text{arg}(z_2))|kin mathbb{Z}$}$implies$
{$log|z_1z_2|+i(text{Arg}(z_1z_2)+2kpi i)|kin mathbb{Z}$}$implies$$log(z_1z_2)$
This is wclearly wrong I guess but this is how I tried to approach it. Would aprreciate any answers.Thanks a lot
complex-analysis analysis complex-numbers
complex-analysis analysis complex-numbers
asked Oct 20 '14 at 16:15
HeisenbergHeisenberg
1,3231741
asked Oct 20 '14 at 16:15
HeisenbergHeisenberg
1,3231741
edited Oct 20 '14 at 16:26
Heisenberg
asked Oct 20 '14 at 16:15
HeisenbergHeisenberg
1,3231741
asked Oct 20 '14 at 16:15
HeisenbergHeisenberg
1,3231741
asked Oct 20 '14 at 16:15
HeisenbergHeisenberg
1,3231741
1,3231741
1
$begingroup$
this identity is simply not true for all complex values $z_1,z_2$ (and especially when one of them is $0$ ...)
$endgroup$
– mercio
Oct 20 '14 at 16:26
$begingroup$
@mercio Sorry about that edited the title thanks
$endgroup$
– Heisenberg
Oct 20 '14 at 16:27
add a comment |
1
$begingroup$
this identity is simply not true for all complex values $z_1,z_2$ (and especially when one of them is $0$ ...)
$endgroup$
– mercio
Oct 20 '14 at 16:26
$begingroup$
@mercio Sorry about that edited the title thanks
$endgroup$
– Heisenberg
Oct 20 '14 at 16:27
1
1
$begingroup$
this identity is simply not true for all complex values $z_1,z_2$ (and especially when one of them is $0$ ...)
$endgroup$
– mercio
Oct 20 '14 at 16:26
$begingroup$
this identity is simply not true for all complex values $z_1,z_2$ (and especially when one of them is $0$ ...)
$endgroup$
– mercio
Oct 20 '14 at 16:26
$begingroup$
@mercio Sorry about that edited the title thanks
$endgroup$
– Heisenberg
Oct 20 '14 at 16:27
$begingroup$
@mercio Sorry about that edited the title thanks
$endgroup$
– Heisenberg
Oct 20 '14 at 16:27
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Right away your first line is not correct, because $$log z_1 + log z_2 = log |z_1| + log |z_2| + i(operatorname{Arg} z_1 + operatorname{Arg} z_2) + 2pi i k_1 + 2 pi i k_2,$$ where $k_1, k_2 in mathbb Z$; that is, the branch of the two logs do not necessarily come from the same value of $k$.
answered Oct 20 '14 at 16:21
heropupheropup
64.8k764103
$endgroup$
$begingroup$
Thanks that was a big mistake. How do I proceed from here?
$endgroup$
– Heisenberg
Oct 20 '14 at 16:24
$begingroup$
Well, you know that $operatorname{Arg} z_1 + operatorname{Arg} z_2 = operatorname{Arg}(z_1 z_2)$, so that takes care of that part of the angle. We also know that $log |z_1 z_2| = log |z_1||z_2| = log |z_1| + log |z_2|$ from the corresponding identity for the positive reals, so that takes care of the magnitude. All that remains is to show that the two sets $${k_1 + k_2 : k_1, k_2 in mathbb Z}$$ and $${k_3 : k_3 in mathbb Z}$$ are equivalent.
$endgroup$
– heropup
Oct 20 '14 at 16:29
$begingroup$
$Argz_1+Argz_2=Arg(z_1z_2)$ is not true however $arg(z_1)+arg(z_2)=arg(z_1z_2 $
$endgroup$
– Heisenberg
Oct 20 '14 at 16:31
$begingroup$
I mean that is not true in general
$endgroup$
– Heisenberg
Oct 20 '14 at 16:31
$begingroup$
Indeed, you are correct. But can you see how to justify it? We are talking about sets.
$endgroup$
– heropup
Oct 20 '14 at 16:33
|
show 1 more comment
$begingroup$
It is true that
$$log(z_1z_2)=log(z_1)+log(z_2) tag1$$
when $(1)$ is interpreted as a set equivalence.
This means that any value of $log(z_1z_2)$ can be expressed as the sum of some value of $log(z_1)$ and some value of $log(z_2)$. And conversely, it means that the sum of any value of $log(z_1)$ and any value of $log(z_2)$ can be expressed as some value of $log(z_1z_2)$.
Note that $(1)$ is true since $log(|z_1z_2|)=log(|z_1|)+log(|z_2|)$ and $arg(z_1z_2)=arg(z_1)+arg(z_2)$.
To see that $arg(z_1z_2)=arg(z_1)+arg(z_2)$, we write $z_1 = |z_1| e^{iphi_1}$ and $z_2=|z_2|e^{iphi_2}$. Clearly, $z_1z_2=|z_1||z_2|e^{i(phi_1+phi_2)}$. So any argument of $z_1$ plus any argument of $z_2$ is an argument of $z_1z_2$. On the other hand consider any argument of $z_1z_2$, which is of the form $phi_1+phi_2+2npi$ for some integer $n$. We could simply take $arg(z_1)=phi_1$ and $arg(z_2)=phi_2+2npi$ and have the stated equality hold.
NOTE: It is important to understand that $(1)$ does not hold in general if $log(z)$ is taken on the Principal branch of the complex logarithm (or any other designated branch since then we lose a degree of freedom).
As an example, take $z_1=i$ and $z_2=-i$. Then, if we choose $log(i)=ipi/2$ and $log(-i)=-ipi/2$, then we must choose the branch of $log(z)$ for which $log(1)=0$. If on the other hand we choose $log(i)=-i3pi/2$ and $log(-i)=-ipi/2$, then we must choose the branch of $log(z)$ for which $log(1)=-i2pi$.
Conversely, if we choose the branch of $log(z)$ for which $log(1)=0$, then for $log(i)=-i3pi/2$, we must have $log(-i)=-ipi/2$.
answered Jan 28 at 20:02
Mark ViolaMark Viola
134k1278176
$endgroup$
$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
Jan 30 at 5:02
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
Right away your first line is not correct, because $$log z_1 + log z_2 = log |z_1| + log |z_2| + i(operatorname{Arg} z_1 + operatorname{Arg} z_2) + 2pi i k_1 + 2 pi i k_2,$$ where $k_1, k_2 in mathbb Z$; that is, the branch of the two logs do not necessarily come from the same value of $k$.
answered Oct 20 '14 at 16:21
heropupheropup
64.8k764103
$endgroup$
$begingroup$
Thanks that was a big mistake. How do I proceed from here?
$endgroup$
– Heisenberg
Oct 20 '14 at 16:24
$begingroup$
Well, you know that $operatorname{Arg} z_1 + operatorname{Arg} z_2 = operatorname{Arg}(z_1 z_2)$, so that takes care of that part of the angle. We also know that $log |z_1 z_2| = log |z_1||z_2| = log |z_1| + log |z_2|$ from the corresponding identity for the positive reals, so that takes care of the magnitude. All that remains is to show that the two sets $${k_1 + k_2 : k_1, k_2 in mathbb Z}$$ and $${k_3 : k_3 in mathbb Z}$$ are equivalent.
$endgroup$
– heropup
Oct 20 '14 at 16:29
$begingroup$
$Argz_1+Argz_2=Arg(z_1z_2)$ is not true however $arg(z_1)+arg(z_2)=arg(z_1z_2 $
$endgroup$
– Heisenberg
Oct 20 '14 at 16:31
$begingroup$
I mean that is not true in general
$endgroup$
– Heisenberg
Oct 20 '14 at 16:31
$begingroup$
Indeed, you are correct. But can you see how to justify it? We are talking about sets.
$endgroup$
– heropup
Oct 20 '14 at 16:33
|
show 1 more comment
$begingroup$
Right away your first line is not correct, because $$log z_1 + log z_2 = log |z_1| + log |z_2| + i(operatorname{Arg} z_1 + operatorname{Arg} z_2) + 2pi i k_1 + 2 pi i k_2,$$ where $k_1, k_2 in mathbb Z$; that is, the branch of the two logs do not necessarily come from the same value of $k$.
answered Oct 20 '14 at 16:21
heropupheropup
64.8k764103
$endgroup$
$begingroup$
Thanks that was a big mistake. How do I proceed from here?
$endgroup$
– Heisenberg
Oct 20 '14 at 16:24
$begingroup$
Well, you know that $operatorname{Arg} z_1 + operatorname{Arg} z_2 = operatorname{Arg}(z_1 z_2)$, so that takes care of that part of the angle. We also know that $log |z_1 z_2| = log |z_1||z_2| = log |z_1| + log |z_2|$ from the corresponding identity for the positive reals, so that takes care of the magnitude. All that remains is to show that the two sets $${k_1 + k_2 : k_1, k_2 in mathbb Z}$$ and $${k_3 : k_3 in mathbb Z}$$ are equivalent.
$endgroup$
– heropup
Oct 20 '14 at 16:29
$begingroup$
$Argz_1+Argz_2=Arg(z_1z_2)$ is not true however $arg(z_1)+arg(z_2)=arg(z_1z_2 $
$endgroup$
– Heisenberg
Oct 20 '14 at 16:31
$begingroup$
I mean that is not true in general
$endgroup$
– Heisenberg
Oct 20 '14 at 16:31
$begingroup$
Indeed, you are correct. But can you see how to justify it? We are talking about sets.
$endgroup$
– heropup
Oct 20 '14 at 16:33
|
show 1 more comment
$begingroup$
Right away your first line is not correct, because $$log z_1 + log z_2 = log |z_1| + log |z_2| + i(operatorname{Arg} z_1 + operatorname{Arg} z_2) + 2pi i k_1 + 2 pi i k_2,$$ where $k_1, k_2 in mathbb Z$; that is, the branch of the two logs do not necessarily come from the same value of $k$.
answered Oct 20 '14 at 16:21
heropupheropup
64.8k764103
$endgroup$
Right away your first line is not correct, because $$log z_1 + log z_2 = log |z_1| + log |z_2| + i(operatorname{Arg} z_1 + operatorname{Arg} z_2) + 2pi i k_1 + 2 pi i k_2,$$ where $k_1, k_2 in mathbb Z$; that is, the branch of the two logs do not necessarily come from the same value of $k$.
answered Oct 20 '14 at 16:21
heropupheropup
64.8k764103
answered Oct 20 '14 at 16:21
heropupheropup
64.8k764103
answered Oct 20 '14 at 16:21
heropupheropup
64.8k764103
answered Oct 20 '14 at 16:21
heropupheropup
64.8k764103
64.8k764103
$begingroup$
Thanks that was a big mistake. How do I proceed from here?
$endgroup$
– Heisenberg
Oct 20 '14 at 16:24
$begingroup$
Well, you know that $operatorname{Arg} z_1 + operatorname{Arg} z_2 = operatorname{Arg}(z_1 z_2)$, so that takes care of that part of the angle. We also know that $log |z_1 z_2| = log |z_1||z_2| = log |z_1| + log |z_2|$ from the corresponding identity for the positive reals, so that takes care of the magnitude. All that remains is to show that the two sets $${k_1 + k_2 : k_1, k_2 in mathbb Z}$$ and $${k_3 : k_3 in mathbb Z}$$ are equivalent.
$endgroup$
– heropup
Oct 20 '14 at 16:29
$begingroup$
$Argz_1+Argz_2=Arg(z_1z_2)$ is not true however $arg(z_1)+arg(z_2)=arg(z_1z_2 $
$endgroup$
– Heisenberg
Oct 20 '14 at 16:31
$begingroup$
I mean that is not true in general
$endgroup$
– Heisenberg
Oct 20 '14 at 16:31
$begingroup$
Indeed, you are correct. But can you see how to justify it? We are talking about sets.
$endgroup$
– heropup
Oct 20 '14 at 16:33
|
show 1 more comment
$begingroup$
Thanks that was a big mistake. How do I proceed from here?
$endgroup$
– Heisenberg
Oct 20 '14 at 16:24
$begingroup$
Well, you know that $operatorname{Arg} z_1 + operatorname{Arg} z_2 = operatorname{Arg}(z_1 z_2)$, so that takes care of that part of the angle. We also know that $log |z_1 z_2| = log |z_1||z_2| = log |z_1| + log |z_2|$ from the corresponding identity for the positive reals, so that takes care of the magnitude. All that remains is to show that the two sets $${k_1 + k_2 : k_1, k_2 in mathbb Z}$$ and $${k_3 : k_3 in mathbb Z}$$ are equivalent.
$endgroup$
– heropup
Oct 20 '14 at 16:29
$begingroup$
$Argz_1+Argz_2=Arg(z_1z_2)$ is not true however $arg(z_1)+arg(z_2)=arg(z_1z_2 $
$endgroup$
– Heisenberg
Oct 20 '14 at 16:31
$begingroup$
I mean that is not true in general
$endgroup$
– Heisenberg
Oct 20 '14 at 16:31
$begingroup$
Indeed, you are correct. But can you see how to justify it? We are talking about sets.
$endgroup$
– heropup
Oct 20 '14 at 16:33
$begingroup$
Thanks that was a big mistake. How do I proceed from here?
$endgroup$
– Heisenberg
Oct 20 '14 at 16:24
$begingroup$
Thanks that was a big mistake. How do I proceed from here?
$endgroup$
– Heisenberg
Oct 20 '14 at 16:24
$begingroup$
Well, you know that $operatorname{Arg} z_1 + operatorname{Arg} z_2 = operatorname{Arg}(z_1 z_2)$, so that takes care of that part of the angle. We also know that $log |z_1 z_2| = log |z_1||z_2| = log |z_1| + log |z_2|$ from the corresponding identity for the positive reals, so that takes care of the magnitude. All that remains is to show that the two sets $${k_1 + k_2 : k_1, k_2 in mathbb Z}$$ and $${k_3 : k_3 in mathbb Z}$$ are equivalent.
$endgroup$
– heropup
Oct 20 '14 at 16:29
$begingroup$
Well, you know that $operatorname{Arg} z_1 + operatorname{Arg} z_2 = operatorname{Arg}(z_1 z_2)$, so that takes care of that part of the angle. We also know that $log |z_1 z_2| = log |z_1||z_2| = log |z_1| + log |z_2|$ from the corresponding identity for the positive reals, so that takes care of the magnitude. All that remains is to show that the two sets $${k_1 + k_2 : k_1, k_2 in mathbb Z}$$ and $${k_3 : k_3 in mathbb Z}$$ are equivalent.
$endgroup$
– heropup
Oct 20 '14 at 16:29
$begingroup$
$Argz_1+Argz_2=Arg(z_1z_2)$ is not true however $arg(z_1)+arg(z_2)=arg(z_1z_2 $
$endgroup$
– Heisenberg
Oct 20 '14 at 16:31
$begingroup$
$Argz_1+Argz_2=Arg(z_1z_2)$ is not true however $arg(z_1)+arg(z_2)=arg(z_1z_2 $
$endgroup$
– Heisenberg
Oct 20 '14 at 16:31
$begingroup$
I mean that is not true in general
$endgroup$
– Heisenberg
Oct 20 '14 at 16:31
$begingroup$
I mean that is not true in general
$endgroup$
– Heisenberg
Oct 20 '14 at 16:31
$begingroup$
Indeed, you are correct. But can you see how to justify it? We are talking about sets.
$endgroup$
– heropup
Oct 20 '14 at 16:33
$begingroup$
Indeed, you are correct. But can you see how to justify it? We are talking about sets.
$endgroup$
– heropup
Oct 20 '14 at 16:33
|
show 1 more comment
$begingroup$
It is true that
$$log(z_1z_2)=log(z_1)+log(z_2) tag1$$
when $(1)$ is interpreted as a set equivalence.
This means that any value of $log(z_1z_2)$ can be expressed as the sum of some value of $log(z_1)$ and some value of $log(z_2)$. And conversely, it means that the sum of any value of $log(z_1)$ and any value of $log(z_2)$ can be expressed as some value of $log(z_1z_2)$.
Note that $(1)$ is true since $log(|z_1z_2|)=log(|z_1|)+log(|z_2|)$ and $arg(z_1z_2)=arg(z_1)+arg(z_2)$.
To see that $arg(z_1z_2)=arg(z_1)+arg(z_2)$, we write $z_1 = |z_1| e^{iphi_1}$ and $z_2=|z_2|e^{iphi_2}$. Clearly, $z_1z_2=|z_1||z_2|e^{i(phi_1+phi_2)}$. So any argument of $z_1$ plus any argument of $z_2$ is an argument of $z_1z_2$. On the other hand consider any argument of $z_1z_2$, which is of the form $phi_1+phi_2+2npi$ for some integer $n$. We could simply take $arg(z_1)=phi_1$ and $arg(z_2)=phi_2+2npi$ and have the stated equality hold.
NOTE: It is important to understand that $(1)$ does not hold in general if $log(z)$ is taken on the Principal branch of the complex logarithm (or any other designated branch since then we lose a degree of freedom).
As an example, take $z_1=i$ and $z_2=-i$. Then, if we choose $log(i)=ipi/2$ and $log(-i)=-ipi/2$, then we must choose the branch of $log(z)$ for which $log(1)=0$. If on the other hand we choose $log(i)=-i3pi/2$ and $log(-i)=-ipi/2$, then we must choose the branch of $log(z)$ for which $log(1)=-i2pi$.
Conversely, if we choose the branch of $log(z)$ for which $log(1)=0$, then for $log(i)=-i3pi/2$, we must have $log(-i)=-ipi/2$.
answered Jan 28 at 20:02
Mark ViolaMark Viola
134k1278176
$endgroup$
$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
Jan 30 at 5:02
add a comment |
$begingroup$
It is true that
$$log(z_1z_2)=log(z_1)+log(z_2) tag1$$
when $(1)$ is interpreted as a set equivalence.
This means that any value of $log(z_1z_2)$ can be expressed as the sum of some value of $log(z_1)$ and some value of $log(z_2)$. And conversely, it means that the sum of any value of $log(z_1)$ and any value of $log(z_2)$ can be expressed as some value of $log(z_1z_2)$.
Note that $(1)$ is true since $log(|z_1z_2|)=log(|z_1|)+log(|z_2|)$ and $arg(z_1z_2)=arg(z_1)+arg(z_2)$.
To see that $arg(z_1z_2)=arg(z_1)+arg(z_2)$, we write $z_1 = |z_1| e^{iphi_1}$ and $z_2=|z_2|e^{iphi_2}$. Clearly, $z_1z_2=|z_1||z_2|e^{i(phi_1+phi_2)}$. So any argument of $z_1$ plus any argument of $z_2$ is an argument of $z_1z_2$. On the other hand consider any argument of $z_1z_2$, which is of the form $phi_1+phi_2+2npi$ for some integer $n$. We could simply take $arg(z_1)=phi_1$ and $arg(z_2)=phi_2+2npi$ and have the stated equality hold.
NOTE: It is important to understand that $(1)$ does not hold in general if $log(z)$ is taken on the Principal branch of the complex logarithm (or any other designated branch since then we lose a degree of freedom).
As an example, take $z_1=i$ and $z_2=-i$. Then, if we choose $log(i)=ipi/2$ and $log(-i)=-ipi/2$, then we must choose the branch of $log(z)$ for which $log(1)=0$. If on the other hand we choose $log(i)=-i3pi/2$ and $log(-i)=-ipi/2$, then we must choose the branch of $log(z)$ for which $log(1)=-i2pi$.
Conversely, if we choose the branch of $log(z)$ for which $log(1)=0$, then for $log(i)=-i3pi/2$, we must have $log(-i)=-ipi/2$.
answered Jan 28 at 20:02
Mark ViolaMark Viola
134k1278176
$endgroup$
$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
Jan 30 at 5:02
add a comment |
$begingroup$
It is true that
$$log(z_1z_2)=log(z_1)+log(z_2) tag1$$
when $(1)$ is interpreted as a set equivalence.
This means that any value of $log(z_1z_2)$ can be expressed as the sum of some value of $log(z_1)$ and some value of $log(z_2)$. And conversely, it means that the sum of any value of $log(z_1)$ and any value of $log(z_2)$ can be expressed as some value of $log(z_1z_2)$.
Note that $(1)$ is true since $log(|z_1z_2|)=log(|z_1|)+log(|z_2|)$ and $arg(z_1z_2)=arg(z_1)+arg(z_2)$.
To see that $arg(z_1z_2)=arg(z_1)+arg(z_2)$, we write $z_1 = |z_1| e^{iphi_1}$ and $z_2=|z_2|e^{iphi_2}$. Clearly, $z_1z_2=|z_1||z_2|e^{i(phi_1+phi_2)}$. So any argument of $z_1$ plus any argument of $z_2$ is an argument of $z_1z_2$. On the other hand consider any argument of $z_1z_2$, which is of the form $phi_1+phi_2+2npi$ for some integer $n$. We could simply take $arg(z_1)=phi_1$ and $arg(z_2)=phi_2+2npi$ and have the stated equality hold.
NOTE: It is important to understand that $(1)$ does not hold in general if $log(z)$ is taken on the Principal branch of the complex logarithm (or any other designated branch since then we lose a degree of freedom).
As an example, take $z_1=i$ and $z_2=-i$. Then, if we choose $log(i)=ipi/2$ and $log(-i)=-ipi/2$, then we must choose the branch of $log(z)$ for which $log(1)=0$. If on the other hand we choose $log(i)=-i3pi/2$ and $log(-i)=-ipi/2$, then we must choose the branch of $log(z)$ for which $log(1)=-i2pi$.
Conversely, if we choose the branch of $log(z)$ for which $log(1)=0$, then for $log(i)=-i3pi/2$, we must have $log(-i)=-ipi/2$.
answered Jan 28 at 20:02
Mark ViolaMark Viola
134k1278176
$endgroup$
It is true that
$$log(z_1z_2)=log(z_1)+log(z_2) tag1$$
when $(1)$ is interpreted as a set equivalence.
This means that any value of $log(z_1z_2)$ can be expressed as the sum of some value of $log(z_1)$ and some value of $log(z_2)$. And conversely, it means that the sum of any value of $log(z_1)$ and any value of $log(z_2)$ can be expressed as some value of $log(z_1z_2)$.
Note that $(1)$ is true since $log(|z_1z_2|)=log(|z_1|)+log(|z_2|)$ and $arg(z_1z_2)=arg(z_1)+arg(z_2)$.
To see that $arg(z_1z_2)=arg(z_1)+arg(z_2)$, we write $z_1 = |z_1| e^{iphi_1}$ and $z_2=|z_2|e^{iphi_2}$. Clearly, $z_1z_2=|z_1||z_2|e^{i(phi_1+phi_2)}$. So any argument of $z_1$ plus any argument of $z_2$ is an argument of $z_1z_2$. On the other hand consider any argument of $z_1z_2$, which is of the form $phi_1+phi_2+2npi$ for some integer $n$. We could simply take $arg(z_1)=phi_1$ and $arg(z_2)=phi_2+2npi$ and have the stated equality hold.
NOTE: It is important to understand that $(1)$ does not hold in general if $log(z)$ is taken on the Principal branch of the complex logarithm (or any other designated branch since then we lose a degree of freedom).
As an example, take $z_1=i$ and $z_2=-i$. Then, if we choose $log(i)=ipi/2$ and $log(-i)=-ipi/2$, then we must choose the branch of $log(z)$ for which $log(1)=0$. If on the other hand we choose $log(i)=-i3pi/2$ and $log(-i)=-ipi/2$, then we must choose the branch of $log(z)$ for which $log(1)=-i2pi$.
Conversely, if we choose the branch of $log(z)$ for which $log(1)=0$, then for $log(i)=-i3pi/2$, we must have $log(-i)=-ipi/2$.
answered Jan 28 at 20:02
Mark ViolaMark Viola
134k1278176
edited Jan 28 at 20:52
answered Jan 28 at 20:02
Mark ViolaMark Viola
134k1278176
answered Jan 28 at 20:02
Mark ViolaMark Viola
134k1278176
answered Jan 28 at 20:02
Mark ViolaMark Viola
134k1278176
134k1278176
$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
Jan 30 at 5:02
add a comment |
$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
Jan 30 at 5:02
$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
Jan 30 at 5:02
$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
Jan 30 at 5:02
add a comment |
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$begingroup$
this identity is simply not true for all complex values $z_1,z_2$ (and especially when one of them is $0$ ...)
$endgroup$
– mercio
Oct 20 '14 at 16:26
$begingroup$
@mercio Sorry about that edited the title thanks
$endgroup$
– Heisenberg
Oct 20 '14 at 16:27