Mixing
complex modulus and square root
$begingroup$
I am failing to understand something about complex square roots:
If we fix the argument $thetain(0,2pi],$ that is we take the positive real line as branch cut, than for $z=rmathrm{e}^{itheta}$, $sqrt{z}$ has argument in the interval $(0,pi].$ In other words, a positive real number will have a negative square root and thus
$$|sqrt{z}|neqsqrt{|z|}.$$
Is that true?
complex-analysis radicals
asked Jul 19 '16 at 19:34
themightymoosethemightymoose
38319
$endgroup$
add a comment |
$begingroup$
I am failing to understand something about complex square roots:
If we fix the argument $thetain(0,2pi],$ that is we take the positive real line as branch cut, than for $z=rmathrm{e}^{itheta}$, $sqrt{z}$ has argument in the interval $(0,pi].$ In other words, a positive real number will have a negative square root and thus
$$|sqrt{z}|neqsqrt{|z|}.$$
Is that true?
complex-analysis radicals
asked Jul 19 '16 at 19:34
themightymoosethemightymoose
38319
$endgroup$
add a comment |
$begingroup$
I am failing to understand something about complex square roots:
If we fix the argument $thetain(0,2pi],$ that is we take the positive real line as branch cut, than for $z=rmathrm{e}^{itheta}$, $sqrt{z}$ has argument in the interval $(0,pi].$ In other words, a positive real number will have a negative square root and thus
$$|sqrt{z}|neqsqrt{|z|}.$$
Is that true?
complex-analysis radicals
asked Jul 19 '16 at 19:34
themightymoosethemightymoose
38319
$endgroup$
I am failing to understand something about complex square roots:
If we fix the argument $thetain(0,2pi],$ that is we take the positive real line as branch cut, than for $z=rmathrm{e}^{itheta}$, $sqrt{z}$ has argument in the interval $(0,pi].$ In other words, a positive real number will have a negative square root and thus
$$|sqrt{z}|neqsqrt{|z|}.$$
Is that true?
complex-analysis radicals
complex-analysis radicals
asked Jul 19 '16 at 19:34
themightymoosethemightymoose
38319
asked Jul 19 '16 at 19:34
themightymoosethemightymoose
38319
asked Jul 19 '16 at 19:34
themightymoosethemightymoose
38319
asked Jul 19 '16 at 19:34
themightymoosethemightymoose
38319
asked Jul 19 '16 at 19:34
themightymoosethemightymoose
38319
38319
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
According to the definition, $sqrt{1}=-1$ and so
$$
|sqrt{1}|=1
$$
whereas
$$
sqrt{|1|}=sqrt{1}=-1
$$
For any positive real it's the same. If $a>0$, then
$$
|sqrt{a^2}|=lvert-arvert=a,
qquad
sqrt{|a^2|}=sqrt{a^2}=-a
$$
answered Jul 19 '16 at 19:58
egregegreg
181k1485203
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add a comment |
$begingroup$
To the OP : nice analysis, I upvoted. My understanding is that it is precisely because of the anomaly that you discovered that the principal Argument is restricted to $;(-pi, pi].;$ Part of the idea behind complex analysis is to consider real analysis as a special case and to preserve all of the pre-existing real analysis conventions.
By forcing Arg(z) into the range as I describe, zero radians is a permissible value for $;(1/2) times;$ Arg(z), while $pi$ radians is not a permissible value. This preserves the real analysis convention that the principal square root of a positive real number is positive.
answered Jan 10 at 8:26
user2661923user2661923
538112
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
According to the definition, $sqrt{1}=-1$ and so
$$
|sqrt{1}|=1
$$
whereas
$$
sqrt{|1|}=sqrt{1}=-1
$$
For any positive real it's the same. If $a>0$, then
$$
|sqrt{a^2}|=lvert-arvert=a,
qquad
sqrt{|a^2|}=sqrt{a^2}=-a
$$
answered Jul 19 '16 at 19:58
egregegreg
181k1485203
$endgroup$
add a comment |
$begingroup$
According to the definition, $sqrt{1}=-1$ and so
$$
|sqrt{1}|=1
$$
whereas
$$
sqrt{|1|}=sqrt{1}=-1
$$
For any positive real it's the same. If $a>0$, then
$$
|sqrt{a^2}|=lvert-arvert=a,
qquad
sqrt{|a^2|}=sqrt{a^2}=-a
$$
answered Jul 19 '16 at 19:58
egregegreg
181k1485203
$endgroup$
add a comment |
$begingroup$
According to the definition, $sqrt{1}=-1$ and so
$$
|sqrt{1}|=1
$$
whereas
$$
sqrt{|1|}=sqrt{1}=-1
$$
For any positive real it's the same. If $a>0$, then
$$
|sqrt{a^2}|=lvert-arvert=a,
qquad
sqrt{|a^2|}=sqrt{a^2}=-a
$$
answered Jul 19 '16 at 19:58
egregegreg
181k1485203
$endgroup$
According to the definition, $sqrt{1}=-1$ and so
$$
|sqrt{1}|=1
$$
whereas
$$
sqrt{|1|}=sqrt{1}=-1
$$
For any positive real it's the same. If $a>0$, then
$$
|sqrt{a^2}|=lvert-arvert=a,
qquad
sqrt{|a^2|}=sqrt{a^2}=-a
$$
answered Jul 19 '16 at 19:58
egregegreg
181k1485203
answered Jul 19 '16 at 19:58
egregegreg
181k1485203
answered Jul 19 '16 at 19:58
egregegreg
181k1485203
answered Jul 19 '16 at 19:58
egregegreg
181k1485203
181k1485203
add a comment |
add a comment |
$begingroup$
To the OP : nice analysis, I upvoted. My understanding is that it is precisely because of the anomaly that you discovered that the principal Argument is restricted to $;(-pi, pi].;$ Part of the idea behind complex analysis is to consider real analysis as a special case and to preserve all of the pre-existing real analysis conventions.
By forcing Arg(z) into the range as I describe, zero radians is a permissible value for $;(1/2) times;$ Arg(z), while $pi$ radians is not a permissible value. This preserves the real analysis convention that the principal square root of a positive real number is positive.
answered Jan 10 at 8:26
user2661923user2661923
538112
$endgroup$
add a comment |
$begingroup$
To the OP : nice analysis, I upvoted. My understanding is that it is precisely because of the anomaly that you discovered that the principal Argument is restricted to $;(-pi, pi].;$ Part of the idea behind complex analysis is to consider real analysis as a special case and to preserve all of the pre-existing real analysis conventions.
By forcing Arg(z) into the range as I describe, zero radians is a permissible value for $;(1/2) times;$ Arg(z), while $pi$ radians is not a permissible value. This preserves the real analysis convention that the principal square root of a positive real number is positive.
answered Jan 10 at 8:26
user2661923user2661923
538112
$endgroup$
add a comment |
$begingroup$
To the OP : nice analysis, I upvoted. My understanding is that it is precisely because of the anomaly that you discovered that the principal Argument is restricted to $;(-pi, pi].;$ Part of the idea behind complex analysis is to consider real analysis as a special case and to preserve all of the pre-existing real analysis conventions.
By forcing Arg(z) into the range as I describe, zero radians is a permissible value for $;(1/2) times;$ Arg(z), while $pi$ radians is not a permissible value. This preserves the real analysis convention that the principal square root of a positive real number is positive.
answered Jan 10 at 8:26
user2661923user2661923
538112
$endgroup$
To the OP : nice analysis, I upvoted. My understanding is that it is precisely because of the anomaly that you discovered that the principal Argument is restricted to $;(-pi, pi].;$ Part of the idea behind complex analysis is to consider real analysis as a special case and to preserve all of the pre-existing real analysis conventions.
By forcing Arg(z) into the range as I describe, zero radians is a permissible value for $;(1/2) times;$ Arg(z), while $pi$ radians is not a permissible value. This preserves the real analysis convention that the principal square root of a positive real number is positive.
answered Jan 10 at 8:26
user2661923user2661923
538112
answered Jan 10 at 8:26
user2661923user2661923
538112
answered Jan 10 at 8:26
user2661923user2661923
538112
answered Jan 10 at 8:26
user2661923user2661923
538112
538112
add a comment |
add a comment |
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