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Proving $arg(zw)=arg(z)+arg(w)$
The 2019 Stack Overflow Developer Survey Results Are InProving $arg(𝑧𝑤)=arg(𝑧)+arg(𝑤)$?Geometric meaning of $Arg(fracz_1z_2)=2Arg(fracz_3-z_1z_3-z_2)+2pi k$ for $z_i$ on the unit circleLimits involving logarithm and argument in the complex planeThe annulus $A_r,s(z_0)=< s$ is openIs $+$ open or closed?Finding the set of analytic functions whose image is a subset of a given setPrimitive of an analytic function - Proof verificationProve there is no branch of arg $z$ on $0 < z < 1$.How to row reduce a matrix with complex entries?Complex Contour Integral Involving Arg(z)Confusion in complex analysis
$begingroup$
This is my attempt I know this is incomplete or may even be wrong.
Let $θ_1 in arg(z)$ and $θ_2 in arg(w)$. Then, $θ_1+θ_2 in arg(z)+arg(w)$.
Also, $θ_1+θ_2 in arg(zw)$. Is this sufficient for the proof or correct at all? Hope someone could help me out. Thanks.
EDIT: Just to avoid any confusions I will add how $arg z$ is defined.
$arg z=textArg, z+2pi kmid ktext is an integer$
complex-analysis complex-numbers
$endgroup$
add a comment |
$begingroup$
This is my attempt I know this is incomplete or may even be wrong.
Let $θ_1 in arg(z)$ and $θ_2 in arg(w)$. Then, $θ_1+θ_2 in arg(z)+arg(w)$.
Also, $θ_1+θ_2 in arg(zw)$. Is this sufficient for the proof or correct at all? Hope someone could help me out. Thanks.
EDIT: Just to avoid any confusions I will add how $arg z$ is defined.
$arg z=textArg, z+2pi kmid ktext is an integer$
complex-analysis complex-numbers
$endgroup$
5
$begingroup$
Why use $in$ instead of $=$ ?
$endgroup$
– Winther
Sep 9 '14 at 19:00
5
$begingroup$
@Winther: Presumably because it makes the statement true? (If you define $operatornamearg(z)$ to be the set of all possible arguments of $z$.)
$endgroup$
– Micah
Sep 9 '14 at 21:45
add a comment |
$begingroup$
This is my attempt I know this is incomplete or may even be wrong.
Let $θ_1 in arg(z)$ and $θ_2 in arg(w)$. Then, $θ_1+θ_2 in arg(z)+arg(w)$.
Also, $θ_1+θ_2 in arg(zw)$. Is this sufficient for the proof or correct at all? Hope someone could help me out. Thanks.
EDIT: Just to avoid any confusions I will add how $arg z$ is defined.
$arg z=textArg, z+2pi kmid ktext is an integer$
complex-analysis complex-numbers
$endgroup$
This is my attempt I know this is incomplete or may even be wrong.
Let $θ_1 in arg(z)$ and $θ_2 in arg(w)$. Then, $θ_1+θ_2 in arg(z)+arg(w)$.
Also, $θ_1+θ_2 in arg(zw)$. Is this sufficient for the proof or correct at all? Hope someone could help me out. Thanks.
EDIT: Just to avoid any confusions I will add how $arg z$ is defined.
$arg z=textArg, z+2pi kmid ktext is an integer$
complex-analysis complex-numbers
complex-analysis complex-numbers
edited Mar 23 at 15:59
Thomas Shelby
4,7362727
4,7362727
asked Sep 9 '14 at 18:36
HeisenbergHeisenberg
1,3281742
1,3281742
5
$begingroup$
Why use $in$ instead of $=$ ?
$endgroup$
– Winther
Sep 9 '14 at 19:00
5
$begingroup$
@Winther: Presumably because it makes the statement true? (If you define $operatornamearg(z)$ to be the set of all possible arguments of $z$.)
$endgroup$
– Micah
Sep 9 '14 at 21:45
add a comment |
5
$begingroup$
Why use $in$ instead of $=$ ?
$endgroup$
– Winther
Sep 9 '14 at 19:00
5
$begingroup$
@Winther: Presumably because it makes the statement true? (If you define $operatornamearg(z)$ to be the set of all possible arguments of $z$.)
$endgroup$
– Micah
Sep 9 '14 at 21:45
5
5
$begingroup$
Why use $in$ instead of $=$ ?
$endgroup$
– Winther
Sep 9 '14 at 19:00
$begingroup$
Why use $in$ instead of $=$ ?
$endgroup$
– Winther
Sep 9 '14 at 19:00
5
5
$begingroup$
@Winther: Presumably because it makes the statement true? (If you define $operatornamearg(z)$ to be the set of all possible arguments of $z$.)
$endgroup$
– Micah
Sep 9 '14 at 21:45
$begingroup$
@Winther: Presumably because it makes the statement true? (If you define $operatornamearg(z)$ to be the set of all possible arguments of $z$.)
$endgroup$
– Micah
Sep 9 '14 at 21:45
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Let $z=r_1 e^i theta_1$ and $w=r_2 e^i theta_2$, then
$
zw=r_1 r_2 e^i (theta_1+theta_2).
$
Hence
$$
arg(z)=theta_1 + 2kpi, kin mathbfZ\
arg(w)=theta_2+2kpi, kin mathbfZ\
arg(zw)=theta_1+theta_2+2kpi, kin mathbfZ.
$$
There it is: $arg(zw)=arg(z)+arg(w)$.
$endgroup$
add a comment |
$begingroup$
Here's a proof: Let $z=r e^itheta$ and $w=s e^iphi$. Then $$textArg(zw)=textArgleft(rse^ithetae^iphiright)=textArgleft(rse^i(theta+phi)right)=textArg(z)+textArg(w)quad(textmod 2pi),$$
where $textArg(z)in[0,2pi)$ is the principal argument of $z$.
However, as Micah commented above, it looks like you're considering sets of all arguments, i.e. to obtain all representations, $$z=re^i(theta + 2pi n)$$ and $$w=se^i(phi+2pi m),$$ where $n,minmathbbZ$.
$endgroup$
$begingroup$
Since $arg(-1+i)=3pi/4$ and $(-1+i)^2=-2i$, your answer means that $arg(-2i)=3pi/4+3pi/4=3pi/2$? I thought that $arg(-i)$ was equal to $-pi/2$. Please explain.
$endgroup$
– Did
Sep 9 '14 at 21:03
$begingroup$
I think that fixes it, but I'm sure you will throw another spanner my way @ Did.
$endgroup$
– Antinous
Sep 9 '14 at 21:32
$begingroup$
If arguments are in $[0,2pi)$, try $z=w=-i$. (But I am surprised, do you still think your computations are rigorous?)
$endgroup$
– Did
Sep 9 '14 at 21:34
$begingroup$
In that case, $textArg(-i)=3pi/2$, and so $textArg(-i)+textArg(-i)=3piequiv pi$. But $textArg(-itimes-i)=textArg(-1)=pi$. This is correct given arguments in $[0,2pi)$.
$endgroup$
– Antinous
Sep 9 '14 at 21:39
$begingroup$
Sorry but this won't do: your answer does state that the usual sum of the arguments is the argument of the sum.
$endgroup$
– Did
Sep 9 '14 at 21:41
|
show 5 more comments
$begingroup$
I'm not sure we can use the $w = e^itheta_w$ here because this is a notation that is partially induced from what he is trying to prove. Here's a proof that doesn't need this.
So we have $textarg(w) in theta_w$ and $textarg(z) in theta_z$.
This means $w = |w|(costheta_w+sintheta_w)$ and $z = |z|(costheta_z+sintheta_z)$.
By multiplication : $wz = |w| |z| ((cos theta_w cos theta_z - sin theta_w sin theta_z) + i (cos theta_w sin theta_z + sin theta_w cos theta_z))$ which happens to be equal to $|w| |z| (cos (theta_w + theta_z) + i sin (theta_w + theta_z)).$
Hence, noting $wz = x + iy$ and identifying the real and the imaginary part, we have $dfrac x wz = cos (theta_w + theta_z)$ and $dfrac y wz = sin (theta_w + theta_z)$.
This is the definition of $(theta_w + theta_z)$ belonging to $textarg(wz)$.
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$begingroup$
Since $arg(-1+i)=3pi/4$ and $(-1+i)^2=-2i$, your answer means that $arg(-2i)=3pi/4+3pi/4=3pi/2$? I thought that $arg(-i)$ was equal to $-pi/2$. Please explain.
$endgroup$
– Did
Sep 9 '14 at 21:02
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@Did Yes, $-i$ and $-2i$ have the same argument. Draw them on the complex plane. E: Wait, there's another discussion on a comment with same content (by Did) in a different answer. :confused:
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– kekkonen
Sep 9 '14 at 21:58
1
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@Did I added the definition of argz to the question hoping it would resolve any confusions
$endgroup$
– Heisenberg
Sep 11 '14 at 3:35
1
$begingroup$
@did Yes it works for sets but not always for principal arguments
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– Heisenberg
Sep 11 '14 at 3:36
1
$begingroup$
@Traklonq the sum need not be in [0,2π[ right?
$endgroup$
– Heisenberg
Sep 11 '14 at 3:39
|
show 10 more comments
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $z=r_1 e^i theta_1$ and $w=r_2 e^i theta_2$, then
$
zw=r_1 r_2 e^i (theta_1+theta_2).
$
Hence
$$
arg(z)=theta_1 + 2kpi, kin mathbfZ\
arg(w)=theta_2+2kpi, kin mathbfZ\
arg(zw)=theta_1+theta_2+2kpi, kin mathbfZ.
$$
There it is: $arg(zw)=arg(z)+arg(w)$.
$endgroup$
add a comment |
$begingroup$
Let $z=r_1 e^i theta_1$ and $w=r_2 e^i theta_2$, then
$
zw=r_1 r_2 e^i (theta_1+theta_2).
$
Hence
$$
arg(z)=theta_1 + 2kpi, kin mathbfZ\
arg(w)=theta_2+2kpi, kin mathbfZ\
arg(zw)=theta_1+theta_2+2kpi, kin mathbfZ.
$$
There it is: $arg(zw)=arg(z)+arg(w)$.
$endgroup$
add a comment |
$begingroup$
Let $z=r_1 e^i theta_1$ and $w=r_2 e^i theta_2$, then
$
zw=r_1 r_2 e^i (theta_1+theta_2).
$
Hence
$$
arg(z)=theta_1 + 2kpi, kin mathbfZ\
arg(w)=theta_2+2kpi, kin mathbfZ\
arg(zw)=theta_1+theta_2+2kpi, kin mathbfZ.
$$
There it is: $arg(zw)=arg(z)+arg(w)$.
$endgroup$
Let $z=r_1 e^i theta_1$ and $w=r_2 e^i theta_2$, then
$
zw=r_1 r_2 e^i (theta_1+theta_2).
$
Hence
$$
arg(z)=theta_1 + 2kpi, kin mathbfZ\
arg(w)=theta_2+2kpi, kin mathbfZ\
arg(zw)=theta_1+theta_2+2kpi, kin mathbfZ.
$$
There it is: $arg(zw)=arg(z)+arg(w)$.
edited Aug 16 '17 at 21:00
ness_boy
1032
1032
answered Sep 11 '14 at 3:54
Pan YanPan Yan
472515
472515
add a comment |
add a comment |
$begingroup$
Here's a proof: Let $z=r e^itheta$ and $w=s e^iphi$. Then $$textArg(zw)=textArgleft(rse^ithetae^iphiright)=textArgleft(rse^i(theta+phi)right)=textArg(z)+textArg(w)quad(textmod 2pi),$$
where $textArg(z)in[0,2pi)$ is the principal argument of $z$.
However, as Micah commented above, it looks like you're considering sets of all arguments, i.e. to obtain all representations, $$z=re^i(theta + 2pi n)$$ and $$w=se^i(phi+2pi m),$$ where $n,minmathbbZ$.
$endgroup$
$begingroup$
Since $arg(-1+i)=3pi/4$ and $(-1+i)^2=-2i$, your answer means that $arg(-2i)=3pi/4+3pi/4=3pi/2$? I thought that $arg(-i)$ was equal to $-pi/2$. Please explain.
$endgroup$
– Did
Sep 9 '14 at 21:03
$begingroup$
I think that fixes it, but I'm sure you will throw another spanner my way @ Did.
$endgroup$
– Antinous
Sep 9 '14 at 21:32
$begingroup$
If arguments are in $[0,2pi)$, try $z=w=-i$. (But I am surprised, do you still think your computations are rigorous?)
$endgroup$
– Did
Sep 9 '14 at 21:34
$begingroup$
In that case, $textArg(-i)=3pi/2$, and so $textArg(-i)+textArg(-i)=3piequiv pi$. But $textArg(-itimes-i)=textArg(-1)=pi$. This is correct given arguments in $[0,2pi)$.
$endgroup$
– Antinous
Sep 9 '14 at 21:39
$begingroup$
Sorry but this won't do: your answer does state that the usual sum of the arguments is the argument of the sum.
$endgroup$
– Did
Sep 9 '14 at 21:41
|
show 5 more comments
$begingroup$
Here's a proof: Let $z=r e^itheta$ and $w=s e^iphi$. Then $$textArg(zw)=textArgleft(rse^ithetae^iphiright)=textArgleft(rse^i(theta+phi)right)=textArg(z)+textArg(w)quad(textmod 2pi),$$
where $textArg(z)in[0,2pi)$ is the principal argument of $z$.
However, as Micah commented above, it looks like you're considering sets of all arguments, i.e. to obtain all representations, $$z=re^i(theta + 2pi n)$$ and $$w=se^i(phi+2pi m),$$ where $n,minmathbbZ$.
$endgroup$
$begingroup$
Since $arg(-1+i)=3pi/4$ and $(-1+i)^2=-2i$, your answer means that $arg(-2i)=3pi/4+3pi/4=3pi/2$? I thought that $arg(-i)$ was equal to $-pi/2$. Please explain.
$endgroup$
– Did
Sep 9 '14 at 21:03
$begingroup$
I think that fixes it, but I'm sure you will throw another spanner my way @ Did.
$endgroup$
– Antinous
Sep 9 '14 at 21:32
$begingroup$
If arguments are in $[0,2pi)$, try $z=w=-i$. (But I am surprised, do you still think your computations are rigorous?)
$endgroup$
– Did
Sep 9 '14 at 21:34
$begingroup$
In that case, $textArg(-i)=3pi/2$, and so $textArg(-i)+textArg(-i)=3piequiv pi$. But $textArg(-itimes-i)=textArg(-1)=pi$. This is correct given arguments in $[0,2pi)$.
$endgroup$
– Antinous
Sep 9 '14 at 21:39
$begingroup$
Sorry but this won't do: your answer does state that the usual sum of the arguments is the argument of the sum.
$endgroup$
– Did
Sep 9 '14 at 21:41
|
show 5 more comments
$begingroup$
Here's a proof: Let $z=r e^itheta$ and $w=s e^iphi$. Then $$textArg(zw)=textArgleft(rse^ithetae^iphiright)=textArgleft(rse^i(theta+phi)right)=textArg(z)+textArg(w)quad(textmod 2pi),$$
where $textArg(z)in[0,2pi)$ is the principal argument of $z$.
However, as Micah commented above, it looks like you're considering sets of all arguments, i.e. to obtain all representations, $$z=re^i(theta + 2pi n)$$ and $$w=se^i(phi+2pi m),$$ where $n,minmathbbZ$.
$endgroup$
Here's a proof: Let $z=r e^itheta$ and $w=s e^iphi$. Then $$textArg(zw)=textArgleft(rse^ithetae^iphiright)=textArgleft(rse^i(theta+phi)right)=textArg(z)+textArg(w)quad(textmod 2pi),$$
where $textArg(z)in[0,2pi)$ is the principal argument of $z$.
However, as Micah commented above, it looks like you're considering sets of all arguments, i.e. to obtain all representations, $$z=re^i(theta + 2pi n)$$ and $$w=se^i(phi+2pi m),$$ where $n,minmathbbZ$.
edited Sep 10 '14 at 3:31
answered Sep 9 '14 at 19:05
AntinousAntinous
5,84842453
5,84842453
$begingroup$
Since $arg(-1+i)=3pi/4$ and $(-1+i)^2=-2i$, your answer means that $arg(-2i)=3pi/4+3pi/4=3pi/2$? I thought that $arg(-i)$ was equal to $-pi/2$. Please explain.
$endgroup$
– Did
Sep 9 '14 at 21:03
$begingroup$
I think that fixes it, but I'm sure you will throw another spanner my way @ Did.
$endgroup$
– Antinous
Sep 9 '14 at 21:32
$begingroup$
If arguments are in $[0,2pi)$, try $z=w=-i$. (But I am surprised, do you still think your computations are rigorous?)
$endgroup$
– Did
Sep 9 '14 at 21:34
$begingroup$
In that case, $textArg(-i)=3pi/2$, and so $textArg(-i)+textArg(-i)=3piequiv pi$. But $textArg(-itimes-i)=textArg(-1)=pi$. This is correct given arguments in $[0,2pi)$.
$endgroup$
– Antinous
Sep 9 '14 at 21:39
$begingroup$
Sorry but this won't do: your answer does state that the usual sum of the arguments is the argument of the sum.
$endgroup$
– Did
Sep 9 '14 at 21:41
|
show 5 more comments
$begingroup$
Since $arg(-1+i)=3pi/4$ and $(-1+i)^2=-2i$, your answer means that $arg(-2i)=3pi/4+3pi/4=3pi/2$? I thought that $arg(-i)$ was equal to $-pi/2$. Please explain.
$endgroup$
– Did
Sep 9 '14 at 21:03
$begingroup$
I think that fixes it, but I'm sure you will throw another spanner my way @ Did.
$endgroup$
– Antinous
Sep 9 '14 at 21:32
$begingroup$
If arguments are in $[0,2pi)$, try $z=w=-i$. (But I am surprised, do you still think your computations are rigorous?)
$endgroup$
– Did
Sep 9 '14 at 21:34
$begingroup$
In that case, $textArg(-i)=3pi/2$, and so $textArg(-i)+textArg(-i)=3piequiv pi$. But $textArg(-itimes-i)=textArg(-1)=pi$. This is correct given arguments in $[0,2pi)$.
$endgroup$
– Antinous
Sep 9 '14 at 21:39
$begingroup$
Sorry but this won't do: your answer does state that the usual sum of the arguments is the argument of the sum.
$endgroup$
– Did
Sep 9 '14 at 21:41
$begingroup$
Since $arg(-1+i)=3pi/4$ and $(-1+i)^2=-2i$, your answer means that $arg(-2i)=3pi/4+3pi/4=3pi/2$? I thought that $arg(-i)$ was equal to $-pi/2$. Please explain.
$endgroup$
– Did
Sep 9 '14 at 21:03
$begingroup$
Since $arg(-1+i)=3pi/4$ and $(-1+i)^2=-2i$, your answer means that $arg(-2i)=3pi/4+3pi/4=3pi/2$? I thought that $arg(-i)$ was equal to $-pi/2$. Please explain.
$endgroup$
– Did
Sep 9 '14 at 21:03
$begingroup$
I think that fixes it, but I'm sure you will throw another spanner my way @ Did.
$endgroup$
– Antinous
Sep 9 '14 at 21:32
$begingroup$
I think that fixes it, but I'm sure you will throw another spanner my way @ Did.
$endgroup$
– Antinous
Sep 9 '14 at 21:32
$begingroup$
If arguments are in $[0,2pi)$, try $z=w=-i$. (But I am surprised, do you still think your computations are rigorous?)
$endgroup$
– Did
Sep 9 '14 at 21:34
$begingroup$
If arguments are in $[0,2pi)$, try $z=w=-i$. (But I am surprised, do you still think your computations are rigorous?)
$endgroup$
– Did
Sep 9 '14 at 21:34
$begingroup$
In that case, $textArg(-i)=3pi/2$, and so $textArg(-i)+textArg(-i)=3piequiv pi$. But $textArg(-itimes-i)=textArg(-1)=pi$. This is correct given arguments in $[0,2pi)$.
$endgroup$
– Antinous
Sep 9 '14 at 21:39
$begingroup$
In that case, $textArg(-i)=3pi/2$, and so $textArg(-i)+textArg(-i)=3piequiv pi$. But $textArg(-itimes-i)=textArg(-1)=pi$. This is correct given arguments in $[0,2pi)$.
$endgroup$
– Antinous
Sep 9 '14 at 21:39
$begingroup$
Sorry but this won't do: your answer does state that the usual sum of the arguments is the argument of the sum.
$endgroup$
– Did
Sep 9 '14 at 21:41
$begingroup$
Sorry but this won't do: your answer does state that the usual sum of the arguments is the argument of the sum.
$endgroup$
– Did
Sep 9 '14 at 21:41
|
show 5 more comments
$begingroup$
I'm not sure we can use the $w = e^itheta_w$ here because this is a notation that is partially induced from what he is trying to prove. Here's a proof that doesn't need this.
So we have $textarg(w) in theta_w$ and $textarg(z) in theta_z$.
This means $w = |w|(costheta_w+sintheta_w)$ and $z = |z|(costheta_z+sintheta_z)$.
By multiplication : $wz = |w| |z| ((cos theta_w cos theta_z - sin theta_w sin theta_z) + i (cos theta_w sin theta_z + sin theta_w cos theta_z))$ which happens to be equal to $|w| |z| (cos (theta_w + theta_z) + i sin (theta_w + theta_z)).$
Hence, noting $wz = x + iy$ and identifying the real and the imaginary part, we have $dfrac x wz = cos (theta_w + theta_z)$ and $dfrac y wz = sin (theta_w + theta_z)$.
This is the definition of $(theta_w + theta_z)$ belonging to $textarg(wz)$.
$endgroup$
$begingroup$
Since $arg(-1+i)=3pi/4$ and $(-1+i)^2=-2i$, your answer means that $arg(-2i)=3pi/4+3pi/4=3pi/2$? I thought that $arg(-i)$ was equal to $-pi/2$. Please explain.
$endgroup$
– Did
Sep 9 '14 at 21:02
$begingroup$
@Did Yes, $-i$ and $-2i$ have the same argument. Draw them on the complex plane. E: Wait, there's another discussion on a comment with same content (by Did) in a different answer. :confused:
$endgroup$
– kekkonen
Sep 9 '14 at 21:58
1
$begingroup$
@Did I added the definition of argz to the question hoping it would resolve any confusions
$endgroup$
– Heisenberg
Sep 11 '14 at 3:35
1
$begingroup$
@did Yes it works for sets but not always for principal arguments
$endgroup$
– Heisenberg
Sep 11 '14 at 3:36
1
$begingroup$
@Traklonq the sum need not be in [0,2π[ right?
$endgroup$
– Heisenberg
Sep 11 '14 at 3:39
|
show 10 more comments
$begingroup$
I'm not sure we can use the $w = e^itheta_w$ here because this is a notation that is partially induced from what he is trying to prove. Here's a proof that doesn't need this.
So we have $textarg(w) in theta_w$ and $textarg(z) in theta_z$.
This means $w = |w|(costheta_w+sintheta_w)$ and $z = |z|(costheta_z+sintheta_z)$.
By multiplication : $wz = |w| |z| ((cos theta_w cos theta_z - sin theta_w sin theta_z) + i (cos theta_w sin theta_z + sin theta_w cos theta_z))$ which happens to be equal to $|w| |z| (cos (theta_w + theta_z) + i sin (theta_w + theta_z)).$
Hence, noting $wz = x + iy$ and identifying the real and the imaginary part, we have $dfrac x wz = cos (theta_w + theta_z)$ and $dfrac y wz = sin (theta_w + theta_z)$.
This is the definition of $(theta_w + theta_z)$ belonging to $textarg(wz)$.
$endgroup$
$begingroup$
Since $arg(-1+i)=3pi/4$ and $(-1+i)^2=-2i$, your answer means that $arg(-2i)=3pi/4+3pi/4=3pi/2$? I thought that $arg(-i)$ was equal to $-pi/2$. Please explain.
$endgroup$
– Did
Sep 9 '14 at 21:02
$begingroup$
@Did Yes, $-i$ and $-2i$ have the same argument. Draw them on the complex plane. E: Wait, there's another discussion on a comment with same content (by Did) in a different answer. :confused:
$endgroup$
– kekkonen
Sep 9 '14 at 21:58
1
$begingroup$
@Did I added the definition of argz to the question hoping it would resolve any confusions
$endgroup$
– Heisenberg
Sep 11 '14 at 3:35
1
$begingroup$
@did Yes it works for sets but not always for principal arguments
$endgroup$
– Heisenberg
Sep 11 '14 at 3:36
1
$begingroup$
@Traklonq the sum need not be in [0,2π[ right?
$endgroup$
– Heisenberg
Sep 11 '14 at 3:39
|
show 10 more comments
$begingroup$
I'm not sure we can use the $w = e^itheta_w$ here because this is a notation that is partially induced from what he is trying to prove. Here's a proof that doesn't need this.
So we have $textarg(w) in theta_w$ and $textarg(z) in theta_z$.
This means $w = |w|(costheta_w+sintheta_w)$ and $z = |z|(costheta_z+sintheta_z)$.
By multiplication : $wz = |w| |z| ((cos theta_w cos theta_z - sin theta_w sin theta_z) + i (cos theta_w sin theta_z + sin theta_w cos theta_z))$ which happens to be equal to $|w| |z| (cos (theta_w + theta_z) + i sin (theta_w + theta_z)).$
Hence, noting $wz = x + iy$ and identifying the real and the imaginary part, we have $dfrac x wz = cos (theta_w + theta_z)$ and $dfrac y wz = sin (theta_w + theta_z)$.
This is the definition of $(theta_w + theta_z)$ belonging to $textarg(wz)$.
$endgroup$
I'm not sure we can use the $w = e^itheta_w$ here because this is a notation that is partially induced from what he is trying to prove. Here's a proof that doesn't need this.
So we have $textarg(w) in theta_w$ and $textarg(z) in theta_z$.
This means $w = |w|(costheta_w+sintheta_w)$ and $z = |z|(costheta_z+sintheta_z)$.
By multiplication : $wz = |w| |z| ((cos theta_w cos theta_z - sin theta_w sin theta_z) + i (cos theta_w sin theta_z + sin theta_w cos theta_z))$ which happens to be equal to $|w| |z| (cos (theta_w + theta_z) + i sin (theta_w + theta_z)).$
Hence, noting $wz = x + iy$ and identifying the real and the imaginary part, we have $dfrac x wz = cos (theta_w + theta_z)$ and $dfrac y wz = sin (theta_w + theta_z)$.
This is the definition of $(theta_w + theta_z)$ belonging to $textarg(wz)$.
edited Sep 11 '14 at 6:50
answered Sep 9 '14 at 19:50
TraklonTraklon
2,71711127
2,71711127
$begingroup$
Since $arg(-1+i)=3pi/4$ and $(-1+i)^2=-2i$, your answer means that $arg(-2i)=3pi/4+3pi/4=3pi/2$? I thought that $arg(-i)$ was equal to $-pi/2$. Please explain.
$endgroup$
– Did
Sep 9 '14 at 21:02
$begingroup$
@Did Yes, $-i$ and $-2i$ have the same argument. Draw them on the complex plane. E: Wait, there's another discussion on a comment with same content (by Did) in a different answer. :confused:
$endgroup$
– kekkonen
Sep 9 '14 at 21:58
1
$begingroup$
@Did I added the definition of argz to the question hoping it would resolve any confusions
$endgroup$
– Heisenberg
Sep 11 '14 at 3:35
1
$begingroup$
@did Yes it works for sets but not always for principal arguments
$endgroup$
– Heisenberg
Sep 11 '14 at 3:36
1
$begingroup$
@Traklonq the sum need not be in [0,2π[ right?
$endgroup$
– Heisenberg
Sep 11 '14 at 3:39
|
show 10 more comments
$begingroup$
Since $arg(-1+i)=3pi/4$ and $(-1+i)^2=-2i$, your answer means that $arg(-2i)=3pi/4+3pi/4=3pi/2$? I thought that $arg(-i)$ was equal to $-pi/2$. Please explain.
$endgroup$
– Did
Sep 9 '14 at 21:02
$begingroup$
@Did Yes, $-i$ and $-2i$ have the same argument. Draw them on the complex plane. E: Wait, there's another discussion on a comment with same content (by Did) in a different answer. :confused:
$endgroup$
– kekkonen
Sep 9 '14 at 21:58
1
$begingroup$
@Did I added the definition of argz to the question hoping it would resolve any confusions
$endgroup$
– Heisenberg
Sep 11 '14 at 3:35
1
$begingroup$
@did Yes it works for sets but not always for principal arguments
$endgroup$
– Heisenberg
Sep 11 '14 at 3:36
1
$begingroup$
@Traklonq the sum need not be in [0,2π[ right?
$endgroup$
– Heisenberg
Sep 11 '14 at 3:39
$begingroup$
Since $arg(-1+i)=3pi/4$ and $(-1+i)^2=-2i$, your answer means that $arg(-2i)=3pi/4+3pi/4=3pi/2$? I thought that $arg(-i)$ was equal to $-pi/2$. Please explain.
$endgroup$
– Did
Sep 9 '14 at 21:02
$begingroup$
Since $arg(-1+i)=3pi/4$ and $(-1+i)^2=-2i$, your answer means that $arg(-2i)=3pi/4+3pi/4=3pi/2$? I thought that $arg(-i)$ was equal to $-pi/2$. Please explain.
$endgroup$
– Did
Sep 9 '14 at 21:02
$begingroup$
@Did Yes, $-i$ and $-2i$ have the same argument. Draw them on the complex plane. E: Wait, there's another discussion on a comment with same content (by Did) in a different answer. :confused:
$endgroup$
– kekkonen
Sep 9 '14 at 21:58
$begingroup$
@Did Yes, $-i$ and $-2i$ have the same argument. Draw them on the complex plane. E: Wait, there's another discussion on a comment with same content (by Did) in a different answer. :confused:
$endgroup$
– kekkonen
Sep 9 '14 at 21:58
1
1
$begingroup$
@Did I added the definition of argz to the question hoping it would resolve any confusions
$endgroup$
– Heisenberg
Sep 11 '14 at 3:35
$begingroup$
@Did I added the definition of argz to the question hoping it would resolve any confusions
$endgroup$
– Heisenberg
Sep 11 '14 at 3:35
1
1
$begingroup$
@did Yes it works for sets but not always for principal arguments
$endgroup$
– Heisenberg
Sep 11 '14 at 3:36
$begingroup$
@did Yes it works for sets but not always for principal arguments
$endgroup$
– Heisenberg
Sep 11 '14 at 3:36
1
1
$begingroup$
@Traklonq the sum need not be in [0,2π[ right?
$endgroup$
– Heisenberg
Sep 11 '14 at 3:39
$begingroup$
@Traklonq the sum need not be in [0,2π[ right?
$endgroup$
– Heisenberg
Sep 11 '14 at 3:39
|
show 10 more comments
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$begingroup$
Why use $in$ instead of $=$ ?
$endgroup$
– Winther
Sep 9 '14 at 19:00
5
$begingroup$
@Winther: Presumably because it makes the statement true? (If you define $operatornamearg(z)$ to be the set of all possible arguments of $z$.)
$endgroup$
– Micah
Sep 9 '14 at 21:45