Why is arg $0$ not defined for any branch?
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Of course, any branch of arg $z$ must have a jump of $2pi$ somewhere. The branch depicted in Fig. 1.10(b) is discontinuous on the positive real axis, taking values from the interval $(0,2pi]$. The branch in Fig. 1.10(c) has the same branch cut but selects values from the interval $(2pi,4pi]$
The notation arg$_tauz$ is used for the branch of arg $z$ taking values from the interval $(tau, tau + 2pi]$. Thus arg$_-piz$ is the principal value Arg $z$, and the branches depicted in Fig. 1.10(b) and 1.10(c), respectively, are arg$_0z$ and arg$_2piz$. Note that arg $0$ cannot be sensibly defined for any branch.
I 'm having trouble understanding why arg $0$ can't be defined for any branch. What if we chose the interval $(-2pi,0]$, for example?
I've also included the images mentioned in the above text.
complex-analysis
asked Aug 16 at 3:23
K.M
487312
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Of course, any branch of arg $z$ must have a jump of $2pi$ somewhere. The branch depicted in Fig. 1.10(b) is discontinuous on the positive real axis, taking values from the interval $(0,2pi]$. The branch in Fig. 1.10(c) has the same branch cut but selects values from the interval $(2pi,4pi]$
The notation arg$_tauz$ is used for the branch of arg $z$ taking values from the interval $(tau, tau + 2pi]$. Thus arg$_-piz$ is the principal value Arg $z$, and the branches depicted in Fig. 1.10(b) and 1.10(c), respectively, are arg$_0z$ and arg$_2piz$. Note that arg $0$ cannot be sensibly defined for any branch.
I 'm having trouble understanding why arg $0$ can't be defined for any branch. What if we chose the interval $(-2pi,0]$, for example?
I've also included the images mentioned in the above text.
complex-analysis
asked Aug 16 at 3:23
K.M
487312
1
$0=0e^it$ for any $t$. So you could say that any real number is an "argument" for $0$. Why bother singling any of them out/
– Lord Shark the Unknown
Aug 16 at 3:52
1
$0=0+i.0$ so $arg(0)=tan^-1frac00$ is not defined.
– Mathlover
Aug 16 at 3:56
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Of course, any branch of arg $z$ must have a jump of $2pi$ somewhere. The branch depicted in Fig. 1.10(b) is discontinuous on the positive real axis, taking values from the interval $(0,2pi]$. The branch in Fig. 1.10(c) has the same branch cut but selects values from the interval $(2pi,4pi]$
The notation arg$_tauz$ is used for the branch of arg $z$ taking values from the interval $(tau, tau + 2pi]$. Thus arg$_-piz$ is the principal value Arg $z$, and the branches depicted in Fig. 1.10(b) and 1.10(c), respectively, are arg$_0z$ and arg$_2piz$. Note that arg $0$ cannot be sensibly defined for any branch.
I 'm having trouble understanding why arg $0$ can't be defined for any branch. What if we chose the interval $(-2pi,0]$, for example?
I've also included the images mentioned in the above text.
complex-analysis
asked Aug 16 at 3:23
K.M
487312
Of course, any branch of arg $z$ must have a jump of $2pi$ somewhere. The branch depicted in Fig. 1.10(b) is discontinuous on the positive real axis, taking values from the interval $(0,2pi]$. The branch in Fig. 1.10(c) has the same branch cut but selects values from the interval $(2pi,4pi]$
The notation arg$_tauz$ is used for the branch of arg $z$ taking values from the interval $(tau, tau + 2pi]$. Thus arg$_-piz$ is the principal value Arg $z$, and the branches depicted in Fig. 1.10(b) and 1.10(c), respectively, are arg$_0z$ and arg$_2piz$. Note that arg $0$ cannot be sensibly defined for any branch.
I 'm having trouble understanding why arg $0$ can't be defined for any branch. What if we chose the interval $(-2pi,0]$, for example?
I've also included the images mentioned in the above text.
complex-analysis
asked Aug 16 at 3:23
K.M
487312
asked Aug 16 at 3:23
K.M
487312
asked Aug 16 at 3:23
K.M
487312
asked Aug 16 at 3:23
K.M
487312
487312
1
$0=0e^it$ for any $t$. So you could say that any real number is an "argument" for $0$. Why bother singling any of them out/
– Lord Shark the Unknown
Aug 16 at 3:52
1
$0=0+i.0$ so $arg(0)=tan^-1frac00$ is not defined.
– Mathlover
Aug 16 at 3:56
add a comment |Â
1
$0=0e^it$ for any $t$. So you could say that any real number is an "argument" for $0$. Why bother singling any of them out/
– Lord Shark the Unknown
Aug 16 at 3:52
1
$0=0+i.0$ so $arg(0)=tan^-1frac00$ is not defined.
– Mathlover
Aug 16 at 3:56
1
1
$0=0e^it$ for any $t$. So you could say that any real number is an "argument" for $0$. Why bother singling any of them out/
– Lord Shark the Unknown
Aug 16 at 3:52
$0=0e^it$ for any $t$. So you could say that any real number is an "argument" for $0$. Why bother singling any of them out/
– Lord Shark the Unknown
Aug 16 at 3:52
1
1
$0=0+i.0$ so $arg(0)=tan^-1frac00$ is not defined.
– Mathlover
Aug 16 at 3:56
$0=0+i.0$ so $arg(0)=tan^-1frac00$ is not defined.
– Mathlover
Aug 16 at 3:56
add a comment |Â
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1
$0=0e^it$ for any $t$. So you could say that any real number is an "argument" for $0$. Why bother singling any of them out/
– Lord Shark the Unknown
Aug 16 at 3:52
1
$0=0+i.0$ so $arg(0)=tan^-1frac00$ is not defined.
– Mathlover
Aug 16 at 3:56