In $mathbb C$ does $sqrt z=z^{1/2}$?
$begingroup$
First of all, is $zmapsto sqrt{z}$ really a function? Indeed, each complex number has two square roots, so I have the impression that $sqrt z$ is not well-defined as a function, is it? Then, I was wondering if $sqrt z=z^{frac{1}{2}}$ was still true. Indeed, $z^{1/2}:=e^{frac{1}{2}log(z)}$, and this is holomorphic in some domain. Whereas, I even have doubt if $zmapsto sqrt z$ is really a function. What do you think ?
complex-analysis functions
edited Dec 10 '18 at 12:00
Gaby Alfonso
1,1811318
asked Dec 10 '18 at 11:40
user623855user623855
1457
$endgroup$
add a comment |
$begingroup$
First of all, is $zmapsto sqrt{z}$ really a function? Indeed, each complex number has two square roots, so I have the impression that $sqrt z$ is not well-defined as a function, is it? Then, I was wondering if $sqrt z=z^{frac{1}{2}}$ was still true. Indeed, $z^{1/2}:=e^{frac{1}{2}log(z)}$, and this is holomorphic in some domain. Whereas, I even have doubt if $zmapsto sqrt z$ is really a function. What do you think ?
complex-analysis functions
edited Dec 10 '18 at 12:00
Gaby Alfonso
1,1811318
asked Dec 10 '18 at 11:40
user623855user623855
1457
$endgroup$
add a comment |
$begingroup$
First of all, is $zmapsto sqrt{z}$ really a function? Indeed, each complex number has two square roots, so I have the impression that $sqrt z$ is not well-defined as a function, is it? Then, I was wondering if $sqrt z=z^{frac{1}{2}}$ was still true. Indeed, $z^{1/2}:=e^{frac{1}{2}log(z)}$, and this is holomorphic in some domain. Whereas, I even have doubt if $zmapsto sqrt z$ is really a function. What do you think ?
complex-analysis functions
edited Dec 10 '18 at 12:00
Gaby Alfonso
1,1811318
asked Dec 10 '18 at 11:40
user623855user623855
1457
$endgroup$
First of all, is $zmapsto sqrt{z}$ really a function? Indeed, each complex number has two square roots, so I have the impression that $sqrt z$ is not well-defined as a function, is it? Then, I was wondering if $sqrt z=z^{frac{1}{2}}$ was still true. Indeed, $z^{1/2}:=e^{frac{1}{2}log(z)}$, and this is holomorphic in some domain. Whereas, I even have doubt if $zmapsto sqrt z$ is really a function. What do you think ?
complex-analysis functions
complex-analysis functions
edited Dec 10 '18 at 12:00
Gaby Alfonso
1,1811318
asked Dec 10 '18 at 11:40
user623855user623855
1457
edited Dec 10 '18 at 12:00
Gaby Alfonso
1,1811318
asked Dec 10 '18 at 11:40
user623855user623855
1457
edited Dec 10 '18 at 12:00
Gaby Alfonso
1,1811318
edited Dec 10 '18 at 12:00
Gaby Alfonso
1,1811318
edited Dec 10 '18 at 12:00
Gaby Alfonso
1,1811318
1,1811318
asked Dec 10 '18 at 11:40
user623855user623855
1457
asked Dec 10 '18 at 11:40
user623855user623855
1457
asked Dec 10 '18 at 11:40
user623855user623855
1457
1457
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
The notations $sqrt{z}$ and $z^{1/2}$ both refer to the principal square root of $z$, which is defined as the solution of $w^2=z$ with argument in $(-pi/2,pi/2]$. Using $log(z)$ you run into the same problem because $exp(w)=z$ even has infinitely many solutions $winBbb C$. This is why for complex numbers you would write $operatorname{Log}(z)$ to refer to the principal value of the logarithm, which is the solution of $exp(w)=z$ with imaginary part in $(-pi,pi]$.
Note that the two choices of principal branches agree so that indeed
$$
sqrt{z} = expleft(frac12 operatorname{Log}(z)right) = z^{1/2}.
$$
answered Dec 10 '18 at 11:45
ChristophChristoph
12.5k1642
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1 Answer
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1 Answer
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$begingroup$
The notations $sqrt{z}$ and $z^{1/2}$ both refer to the principal square root of $z$, which is defined as the solution of $w^2=z$ with argument in $(-pi/2,pi/2]$. Using $log(z)$ you run into the same problem because $exp(w)=z$ even has infinitely many solutions $winBbb C$. This is why for complex numbers you would write $operatorname{Log}(z)$ to refer to the principal value of the logarithm, which is the solution of $exp(w)=z$ with imaginary part in $(-pi,pi]$.
Note that the two choices of principal branches agree so that indeed
$$
sqrt{z} = expleft(frac12 operatorname{Log}(z)right) = z^{1/2}.
$$
answered Dec 10 '18 at 11:45
ChristophChristoph
12.5k1642
$endgroup$
add a comment |
$begingroup$
The notations $sqrt{z}$ and $z^{1/2}$ both refer to the principal square root of $z$, which is defined as the solution of $w^2=z$ with argument in $(-pi/2,pi/2]$. Using $log(z)$ you run into the same problem because $exp(w)=z$ even has infinitely many solutions $winBbb C$. This is why for complex numbers you would write $operatorname{Log}(z)$ to refer to the principal value of the logarithm, which is the solution of $exp(w)=z$ with imaginary part in $(-pi,pi]$.
Note that the two choices of principal branches agree so that indeed
$$
sqrt{z} = expleft(frac12 operatorname{Log}(z)right) = z^{1/2}.
$$
answered Dec 10 '18 at 11:45
ChristophChristoph
12.5k1642
$endgroup$
add a comment |
$begingroup$
The notations $sqrt{z}$ and $z^{1/2}$ both refer to the principal square root of $z$, which is defined as the solution of $w^2=z$ with argument in $(-pi/2,pi/2]$. Using $log(z)$ you run into the same problem because $exp(w)=z$ even has infinitely many solutions $winBbb C$. This is why for complex numbers you would write $operatorname{Log}(z)$ to refer to the principal value of the logarithm, which is the solution of $exp(w)=z$ with imaginary part in $(-pi,pi]$.
Note that the two choices of principal branches agree so that indeed
$$
sqrt{z} = expleft(frac12 operatorname{Log}(z)right) = z^{1/2}.
$$
answered Dec 10 '18 at 11:45
ChristophChristoph
12.5k1642
$endgroup$
The notations $sqrt{z}$ and $z^{1/2}$ both refer to the principal square root of $z$, which is defined as the solution of $w^2=z$ with argument in $(-pi/2,pi/2]$. Using $log(z)$ you run into the same problem because $exp(w)=z$ even has infinitely many solutions $winBbb C$. This is why for complex numbers you would write $operatorname{Log}(z)$ to refer to the principal value of the logarithm, which is the solution of $exp(w)=z$ with imaginary part in $(-pi,pi]$.
Note that the two choices of principal branches agree so that indeed
$$
sqrt{z} = expleft(frac12 operatorname{Log}(z)right) = z^{1/2}.
$$
answered Dec 10 '18 at 11:45
ChristophChristoph
12.5k1642
edited Dec 10 '18 at 12:03
answered Dec 10 '18 at 11:45
ChristophChristoph
12.5k1642
answered Dec 10 '18 at 11:45
ChristophChristoph
12.5k1642
answered Dec 10 '18 at 11:45
ChristophChristoph
12.5k1642
12.5k1642
add a comment |
add a comment |
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