Mixing
Finding a continuous branch of $F(z)=sqrt{frac{(z^2-1)(z-2)}{z}}$
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I am having problems understanding branch cuts. For instance, I am given the following function, for $z in mathbb{C},$ let
$$F(z)=sqrt{frac{(z^2-1)(z-2)}{z}}=sqrt{frac{(z+1)(z-1)(z-2)}{z}}.$$
Determine if there is a continuous branch $f$ of $F,$ with $Re{(f(i))}>0$, that is defined on $mathbb{C} setminus { [-1,0] cup [1,2] }.$
I have seen examples where the function is written using the principal value of the square root function, and then the signs of the principal values are determined to fit the case at hand, but it seemed arbitrary.
I have no idea how to proceed. How does the process of finding a continuous branch works?
complex-analysis branch-cuts
asked Feb 1 at 5:00
Gaby AlfonsoGaby Alfonso
1,1951418
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add a comment |
$begingroup$
I am having problems understanding branch cuts. For instance, I am given the following function, for $z in mathbb{C},$ let
$$F(z)=sqrt{frac{(z^2-1)(z-2)}{z}}=sqrt{frac{(z+1)(z-1)(z-2)}{z}}.$$
Determine if there is a continuous branch $f$ of $F,$ with $Re{(f(i))}>0$, that is defined on $mathbb{C} setminus { [-1,0] cup [1,2] }.$
I have seen examples where the function is written using the principal value of the square root function, and then the signs of the principal values are determined to fit the case at hand, but it seemed arbitrary.
I have no idea how to proceed. How does the process of finding a continuous branch works?
complex-analysis branch-cuts
asked Feb 1 at 5:00
Gaby AlfonsoGaby Alfonso
1,1951418
$endgroup$
1
$begingroup$
For largish $z$ you can write $F(z)=zsqrt{(1-z^{-2})(1-2z^{-1})}$ which gives you one holomorphic function for $|z|>2$. Now you only need to explore how this extends inwards. Or perhaps the form $F(z)=(z-1)sqrt{1+z^{-1}}sqrt{1-(z-1)^{-1}}$ leads more directly to the desired result.
$endgroup$
– LutzL
Feb 1 at 9:46
add a comment |
$begingroup$
I am having problems understanding branch cuts. For instance, I am given the following function, for $z in mathbb{C},$ let
$$F(z)=sqrt{frac{(z^2-1)(z-2)}{z}}=sqrt{frac{(z+1)(z-1)(z-2)}{z}}.$$
Determine if there is a continuous branch $f$ of $F,$ with $Re{(f(i))}>0$, that is defined on $mathbb{C} setminus { [-1,0] cup [1,2] }.$
I have seen examples where the function is written using the principal value of the square root function, and then the signs of the principal values are determined to fit the case at hand, but it seemed arbitrary.
I have no idea how to proceed. How does the process of finding a continuous branch works?
complex-analysis branch-cuts
asked Feb 1 at 5:00
Gaby AlfonsoGaby Alfonso
1,1951418
$endgroup$
I am having problems understanding branch cuts. For instance, I am given the following function, for $z in mathbb{C},$ let
$$F(z)=sqrt{frac{(z^2-1)(z-2)}{z}}=sqrt{frac{(z+1)(z-1)(z-2)}{z}}.$$
Determine if there is a continuous branch $f$ of $F,$ with $Re{(f(i))}>0$, that is defined on $mathbb{C} setminus { [-1,0] cup [1,2] }.$
I have seen examples where the function is written using the principal value of the square root function, and then the signs of the principal values are determined to fit the case at hand, but it seemed arbitrary.
I have no idea how to proceed. How does the process of finding a continuous branch works?
complex-analysis branch-cuts
complex-analysis branch-cuts
asked Feb 1 at 5:00
Gaby AlfonsoGaby Alfonso
1,1951418
asked Feb 1 at 5:00
Gaby AlfonsoGaby Alfonso
1,1951418
edited Feb 1 at 6:34
Gaby Alfonso
asked Feb 1 at 5:00
Gaby AlfonsoGaby Alfonso
1,1951418
asked Feb 1 at 5:00
Gaby AlfonsoGaby Alfonso
1,1951418
asked Feb 1 at 5:00
Gaby AlfonsoGaby Alfonso
1,1951418
1,1951418
1
$begingroup$
For largish $z$ you can write $F(z)=zsqrt{(1-z^{-2})(1-2z^{-1})}$ which gives you one holomorphic function for $|z|>2$. Now you only need to explore how this extends inwards. Or perhaps the form $F(z)=(z-1)sqrt{1+z^{-1}}sqrt{1-(z-1)^{-1}}$ leads more directly to the desired result.
$endgroup$
– LutzL
Feb 1 at 9:46
add a comment |
1
$begingroup$
For largish $z$ you can write $F(z)=zsqrt{(1-z^{-2})(1-2z^{-1})}$ which gives you one holomorphic function for $|z|>2$. Now you only need to explore how this extends inwards. Or perhaps the form $F(z)=(z-1)sqrt{1+z^{-1}}sqrt{1-(z-1)^{-1}}$ leads more directly to the desired result.
$endgroup$
– LutzL
Feb 1 at 9:46
1
1
$begingroup$
For largish $z$ you can write $F(z)=zsqrt{(1-z^{-2})(1-2z^{-1})}$ which gives you one holomorphic function for $|z|>2$. Now you only need to explore how this extends inwards. Or perhaps the form $F(z)=(z-1)sqrt{1+z^{-1}}sqrt{1-(z-1)^{-1}}$ leads more directly to the desired result.
$endgroup$
– LutzL
Feb 1 at 9:46
$begingroup$
For largish $z$ you can write $F(z)=zsqrt{(1-z^{-2})(1-2z^{-1})}$ which gives you one holomorphic function for $|z|>2$. Now you only need to explore how this extends inwards. Or perhaps the form $F(z)=(z-1)sqrt{1+z^{-1}}sqrt{1-(z-1)^{-1}}$ leads more directly to the desired result.
$endgroup$
– LutzL
Feb 1 at 9:46
add a comment |
2 Answers
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Consider $g(z) = (z+1)(z-1)(z-2)/z$. Then
$$ frac{g'(z)}{g(z)} = frac{1}{z+1} - frac{1}{z} + frac{1}{z-1} + frac{1}{z-2}. $$
Moreover, if $gamma$ is any closed curve in $Omega = mathbb{C} setminus ([-1, 0] cup [1, 2])$, then it must wind $-1$ and $0$ the same time and $1$ and $2$ the same time. So if $W(gamma, z_0)$ denotes the winding number of $gamma$ at $z_0$, then
$$ int_{gamma} g(z) , mathrm{d}z = 2pi i left[ W(gamma, -1) - W(gamma, 0) + W(gamma,1) + W(gamma,2) right]. $$
By the previous comment, $W(gamma,-1) = W(gamma, 0)$ and $W(gamma, 1) = W(gamma, 2)$, and so, the above number is an even multiple of $2pi i$. So $frac{1}{2} int_{gamma} g(z) , mathrm{d}z$ is still an integer multiple of $2pi i$. This allows us to define $F(z)$ as
$$ F(z) = a expleft{ frac{1}{2} int_{i}^{z} frac{g'(w)}{g(w)} , mathrm{d}w right}, $$
where $a^2 = f(i)$ is chosen to satisfy $operatorname{Re}(a) > 0$ and the integral is taken over any path in $Omega$ joining from $i$ to $z$. This is well-defind since the difference of any two such integrals is an integer multiple of $2pi i$, which is cancelled out by the exponential function. Moreover, it is easy to check that $F(z)^2 = g(z)$. So $F$ is the square root of $g$ on $Omega$ satisfying the prescribed condition.
More generally, assume that $r(z)$ is a rational function. Then its logarithmic derivative takes the form
$$ frac{r'(z)}{r(z)} = sum_k frac{n_k}{z - z_k}, $$
for some non-zero integer $n_k$'s and $z_k in mathbb{C}$. Indeed, if $n_k geq 1$, then $z_k$ is a zero of order $n_k$. If $n_k leq -1$, then $z_k$ is a pole of order $-n_k$.
Then, on each domain $Omega subseteq mathbb{C}$, an $m$-th root of $r$ is well-defined if the following condition holds: For each bounded connected component $C$ of $mathbb{C}setminusOmega$, the sum of $n_k$'s for which $z_k in C$ is a multiple of $m$.
The reasoning is fairly the same as before: Given this condition, integral of $r'(z)/r(z)$ over any closed curve in $Omega$ is a multiple of $2pi i m$, and so, the logarithm can be defined in $mathbb{C} / 2pi i m mathbb{Z}$. So if we divide this logarithm by $m$ and composing with $exp$, such ambiguity disappears, yielding a well-defined function whose $m$-th power equals $r(z)$.
(I think this is an equivalent condition, but do not want to delve into technicality that I may encounter while attempting to prove the converse.)
answered Feb 2 at 1:42
Sangchul LeeSangchul Lee
96.5k12173283
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add a comment |
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We can investigate the behavior of $F$ by setting $0le a,b,c,d<2pi, $ putting
$z+1=|z+1|e^{ia}; z-1=|z-1|e^{ib}; z-2=|z-2|e^{ic}; z=|z|e^{id}, $ and checking that
$F(z)=sqrt{left |{frac{(z+1)(z-1)(z-2)}{z}} right |}e^{frac{1}{2}((a-d)+(b+c))i}$ is continuous on $mathbb{C} setminus { [-1,0] cup [1,2] }.$
But this is clear:
$1). $ For an arbitrary $z=|z|e^{itheta}in mathbb{C} setminus { [-1,0] cup [1,2] },$ as $theta$ makes its way from $0$ to $2pi, $ so do $a,b,c$ and $d$ and $F$ does not "jump" when this happens. More precisely, we have, by direct calculation,
$e^{frac{1}{2}(a+2pi-(d+2pi))}=e^{frac{1}{2}(a-d)}$ and $ e^{frac{1}{2}(b+2pi+c+2pi)i}=e^{frac{1}{2}(b+c)i}cdot e^{2pi i}=e^{frac{1}{2}(b+c)i}.$
$2).$ If $z=i, $ then $a=frac{pi}{4}; b=frac{7pi}{4}; c=pi -tan ^{-1}left(frac{1}{2}right)=pi(1-.147)=.852pi; d=frac{pi}{2}, $ so you can check that $cos left ( frac{1}{2}((a-d)+(b+c)) right )neq 0, $ so $frak R$ $f(i)neq 0$, and we can always make this positive by setting $G=-F$ if necessary.
answered Feb 1 at 23:44
MatematletaMatematleta
12.1k21020
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add a comment |
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2 Answers
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$begingroup$
Consider $g(z) = (z+1)(z-1)(z-2)/z$. Then
$$ frac{g'(z)}{g(z)} = frac{1}{z+1} - frac{1}{z} + frac{1}{z-1} + frac{1}{z-2}. $$
Moreover, if $gamma$ is any closed curve in $Omega = mathbb{C} setminus ([-1, 0] cup [1, 2])$, then it must wind $-1$ and $0$ the same time and $1$ and $2$ the same time. So if $W(gamma, z_0)$ denotes the winding number of $gamma$ at $z_0$, then
$$ int_{gamma} g(z) , mathrm{d}z = 2pi i left[ W(gamma, -1) - W(gamma, 0) + W(gamma,1) + W(gamma,2) right]. $$
By the previous comment, $W(gamma,-1) = W(gamma, 0)$ and $W(gamma, 1) = W(gamma, 2)$, and so, the above number is an even multiple of $2pi i$. So $frac{1}{2} int_{gamma} g(z) , mathrm{d}z$ is still an integer multiple of $2pi i$. This allows us to define $F(z)$ as
$$ F(z) = a expleft{ frac{1}{2} int_{i}^{z} frac{g'(w)}{g(w)} , mathrm{d}w right}, $$
where $a^2 = f(i)$ is chosen to satisfy $operatorname{Re}(a) > 0$ and the integral is taken over any path in $Omega$ joining from $i$ to $z$. This is well-defind since the difference of any two such integrals is an integer multiple of $2pi i$, which is cancelled out by the exponential function. Moreover, it is easy to check that $F(z)^2 = g(z)$. So $F$ is the square root of $g$ on $Omega$ satisfying the prescribed condition.
More generally, assume that $r(z)$ is a rational function. Then its logarithmic derivative takes the form
$$ frac{r'(z)}{r(z)} = sum_k frac{n_k}{z - z_k}, $$
for some non-zero integer $n_k$'s and $z_k in mathbb{C}$. Indeed, if $n_k geq 1$, then $z_k$ is a zero of order $n_k$. If $n_k leq -1$, then $z_k$ is a pole of order $-n_k$.
Then, on each domain $Omega subseteq mathbb{C}$, an $m$-th root of $r$ is well-defined if the following condition holds: For each bounded connected component $C$ of $mathbb{C}setminusOmega$, the sum of $n_k$'s for which $z_k in C$ is a multiple of $m$.
The reasoning is fairly the same as before: Given this condition, integral of $r'(z)/r(z)$ over any closed curve in $Omega$ is a multiple of $2pi i m$, and so, the logarithm can be defined in $mathbb{C} / 2pi i m mathbb{Z}$. So if we divide this logarithm by $m$ and composing with $exp$, such ambiguity disappears, yielding a well-defined function whose $m$-th power equals $r(z)$.
(I think this is an equivalent condition, but do not want to delve into technicality that I may encounter while attempting to prove the converse.)
answered Feb 2 at 1:42
Sangchul LeeSangchul Lee
96.5k12173283
$endgroup$
add a comment |
$begingroup$
Consider $g(z) = (z+1)(z-1)(z-2)/z$. Then
$$ frac{g'(z)}{g(z)} = frac{1}{z+1} - frac{1}{z} + frac{1}{z-1} + frac{1}{z-2}. $$
Moreover, if $gamma$ is any closed curve in $Omega = mathbb{C} setminus ([-1, 0] cup [1, 2])$, then it must wind $-1$ and $0$ the same time and $1$ and $2$ the same time. So if $W(gamma, z_0)$ denotes the winding number of $gamma$ at $z_0$, then
$$ int_{gamma} g(z) , mathrm{d}z = 2pi i left[ W(gamma, -1) - W(gamma, 0) + W(gamma,1) + W(gamma,2) right]. $$
By the previous comment, $W(gamma,-1) = W(gamma, 0)$ and $W(gamma, 1) = W(gamma, 2)$, and so, the above number is an even multiple of $2pi i$. So $frac{1}{2} int_{gamma} g(z) , mathrm{d}z$ is still an integer multiple of $2pi i$. This allows us to define $F(z)$ as
$$ F(z) = a expleft{ frac{1}{2} int_{i}^{z} frac{g'(w)}{g(w)} , mathrm{d}w right}, $$
where $a^2 = f(i)$ is chosen to satisfy $operatorname{Re}(a) > 0$ and the integral is taken over any path in $Omega$ joining from $i$ to $z$. This is well-defind since the difference of any two such integrals is an integer multiple of $2pi i$, which is cancelled out by the exponential function. Moreover, it is easy to check that $F(z)^2 = g(z)$. So $F$ is the square root of $g$ on $Omega$ satisfying the prescribed condition.
More generally, assume that $r(z)$ is a rational function. Then its logarithmic derivative takes the form
$$ frac{r'(z)}{r(z)} = sum_k frac{n_k}{z - z_k}, $$
for some non-zero integer $n_k$'s and $z_k in mathbb{C}$. Indeed, if $n_k geq 1$, then $z_k$ is a zero of order $n_k$. If $n_k leq -1$, then $z_k$ is a pole of order $-n_k$.
Then, on each domain $Omega subseteq mathbb{C}$, an $m$-th root of $r$ is well-defined if the following condition holds: For each bounded connected component $C$ of $mathbb{C}setminusOmega$, the sum of $n_k$'s for which $z_k in C$ is a multiple of $m$.
The reasoning is fairly the same as before: Given this condition, integral of $r'(z)/r(z)$ over any closed curve in $Omega$ is a multiple of $2pi i m$, and so, the logarithm can be defined in $mathbb{C} / 2pi i m mathbb{Z}$. So if we divide this logarithm by $m$ and composing with $exp$, such ambiguity disappears, yielding a well-defined function whose $m$-th power equals $r(z)$.
(I think this is an equivalent condition, but do not want to delve into technicality that I may encounter while attempting to prove the converse.)
answered Feb 2 at 1:42
Sangchul LeeSangchul Lee
96.5k12173283
$endgroup$
add a comment |
$begingroup$
Consider $g(z) = (z+1)(z-1)(z-2)/z$. Then
$$ frac{g'(z)}{g(z)} = frac{1}{z+1} - frac{1}{z} + frac{1}{z-1} + frac{1}{z-2}. $$
Moreover, if $gamma$ is any closed curve in $Omega = mathbb{C} setminus ([-1, 0] cup [1, 2])$, then it must wind $-1$ and $0$ the same time and $1$ and $2$ the same time. So if $W(gamma, z_0)$ denotes the winding number of $gamma$ at $z_0$, then
$$ int_{gamma} g(z) , mathrm{d}z = 2pi i left[ W(gamma, -1) - W(gamma, 0) + W(gamma,1) + W(gamma,2) right]. $$
By the previous comment, $W(gamma,-1) = W(gamma, 0)$ and $W(gamma, 1) = W(gamma, 2)$, and so, the above number is an even multiple of $2pi i$. So $frac{1}{2} int_{gamma} g(z) , mathrm{d}z$ is still an integer multiple of $2pi i$. This allows us to define $F(z)$ as
$$ F(z) = a expleft{ frac{1}{2} int_{i}^{z} frac{g'(w)}{g(w)} , mathrm{d}w right}, $$
where $a^2 = f(i)$ is chosen to satisfy $operatorname{Re}(a) > 0$ and the integral is taken over any path in $Omega$ joining from $i$ to $z$. This is well-defind since the difference of any two such integrals is an integer multiple of $2pi i$, which is cancelled out by the exponential function. Moreover, it is easy to check that $F(z)^2 = g(z)$. So $F$ is the square root of $g$ on $Omega$ satisfying the prescribed condition.
More generally, assume that $r(z)$ is a rational function. Then its logarithmic derivative takes the form
$$ frac{r'(z)}{r(z)} = sum_k frac{n_k}{z - z_k}, $$
for some non-zero integer $n_k$'s and $z_k in mathbb{C}$. Indeed, if $n_k geq 1$, then $z_k$ is a zero of order $n_k$. If $n_k leq -1$, then $z_k$ is a pole of order $-n_k$.
Then, on each domain $Omega subseteq mathbb{C}$, an $m$-th root of $r$ is well-defined if the following condition holds: For each bounded connected component $C$ of $mathbb{C}setminusOmega$, the sum of $n_k$'s for which $z_k in C$ is a multiple of $m$.
The reasoning is fairly the same as before: Given this condition, integral of $r'(z)/r(z)$ over any closed curve in $Omega$ is a multiple of $2pi i m$, and so, the logarithm can be defined in $mathbb{C} / 2pi i m mathbb{Z}$. So if we divide this logarithm by $m$ and composing with $exp$, such ambiguity disappears, yielding a well-defined function whose $m$-th power equals $r(z)$.
(I think this is an equivalent condition, but do not want to delve into technicality that I may encounter while attempting to prove the converse.)
answered Feb 2 at 1:42
Sangchul LeeSangchul Lee
96.5k12173283
$endgroup$
Consider $g(z) = (z+1)(z-1)(z-2)/z$. Then
$$ frac{g'(z)}{g(z)} = frac{1}{z+1} - frac{1}{z} + frac{1}{z-1} + frac{1}{z-2}. $$
Moreover, if $gamma$ is any closed curve in $Omega = mathbb{C} setminus ([-1, 0] cup [1, 2])$, then it must wind $-1$ and $0$ the same time and $1$ and $2$ the same time. So if $W(gamma, z_0)$ denotes the winding number of $gamma$ at $z_0$, then
$$ int_{gamma} g(z) , mathrm{d}z = 2pi i left[ W(gamma, -1) - W(gamma, 0) + W(gamma,1) + W(gamma,2) right]. $$
By the previous comment, $W(gamma,-1) = W(gamma, 0)$ and $W(gamma, 1) = W(gamma, 2)$, and so, the above number is an even multiple of $2pi i$. So $frac{1}{2} int_{gamma} g(z) , mathrm{d}z$ is still an integer multiple of $2pi i$. This allows us to define $F(z)$ as
$$ F(z) = a expleft{ frac{1}{2} int_{i}^{z} frac{g'(w)}{g(w)} , mathrm{d}w right}, $$
where $a^2 = f(i)$ is chosen to satisfy $operatorname{Re}(a) > 0$ and the integral is taken over any path in $Omega$ joining from $i$ to $z$. This is well-defind since the difference of any two such integrals is an integer multiple of $2pi i$, which is cancelled out by the exponential function. Moreover, it is easy to check that $F(z)^2 = g(z)$. So $F$ is the square root of $g$ on $Omega$ satisfying the prescribed condition.
More generally, assume that $r(z)$ is a rational function. Then its logarithmic derivative takes the form
$$ frac{r'(z)}{r(z)} = sum_k frac{n_k}{z - z_k}, $$
for some non-zero integer $n_k$'s and $z_k in mathbb{C}$. Indeed, if $n_k geq 1$, then $z_k$ is a zero of order $n_k$. If $n_k leq -1$, then $z_k$ is a pole of order $-n_k$.
Then, on each domain $Omega subseteq mathbb{C}$, an $m$-th root of $r$ is well-defined if the following condition holds: For each bounded connected component $C$ of $mathbb{C}setminusOmega$, the sum of $n_k$'s for which $z_k in C$ is a multiple of $m$.
The reasoning is fairly the same as before: Given this condition, integral of $r'(z)/r(z)$ over any closed curve in $Omega$ is a multiple of $2pi i m$, and so, the logarithm can be defined in $mathbb{C} / 2pi i m mathbb{Z}$. So if we divide this logarithm by $m$ and composing with $exp$, such ambiguity disappears, yielding a well-defined function whose $m$-th power equals $r(z)$.
(I think this is an equivalent condition, but do not want to delve into technicality that I may encounter while attempting to prove the converse.)
answered Feb 2 at 1:42
Sangchul LeeSangchul Lee
96.5k12173283
answered Feb 2 at 1:42
Sangchul LeeSangchul Lee
96.5k12173283
answered Feb 2 at 1:42
Sangchul LeeSangchul Lee
96.5k12173283
answered Feb 2 at 1:42
Sangchul LeeSangchul Lee
96.5k12173283
96.5k12173283
add a comment |
add a comment |
$begingroup$
We can investigate the behavior of $F$ by setting $0le a,b,c,d<2pi, $ putting
$z+1=|z+1|e^{ia}; z-1=|z-1|e^{ib}; z-2=|z-2|e^{ic}; z=|z|e^{id}, $ and checking that
$F(z)=sqrt{left |{frac{(z+1)(z-1)(z-2)}{z}} right |}e^{frac{1}{2}((a-d)+(b+c))i}$ is continuous on $mathbb{C} setminus { [-1,0] cup [1,2] }.$
But this is clear:
$1). $ For an arbitrary $z=|z|e^{itheta}in mathbb{C} setminus { [-1,0] cup [1,2] },$ as $theta$ makes its way from $0$ to $2pi, $ so do $a,b,c$ and $d$ and $F$ does not "jump" when this happens. More precisely, we have, by direct calculation,
$e^{frac{1}{2}(a+2pi-(d+2pi))}=e^{frac{1}{2}(a-d)}$ and $ e^{frac{1}{2}(b+2pi+c+2pi)i}=e^{frac{1}{2}(b+c)i}cdot e^{2pi i}=e^{frac{1}{2}(b+c)i}.$
$2).$ If $z=i, $ then $a=frac{pi}{4}; b=frac{7pi}{4}; c=pi -tan ^{-1}left(frac{1}{2}right)=pi(1-.147)=.852pi; d=frac{pi}{2}, $ so you can check that $cos left ( frac{1}{2}((a-d)+(b+c)) right )neq 0, $ so $frak R$ $f(i)neq 0$, and we can always make this positive by setting $G=-F$ if necessary.
answered Feb 1 at 23:44
MatematletaMatematleta
12.1k21020
$endgroup$
add a comment |
$begingroup$
We can investigate the behavior of $F$ by setting $0le a,b,c,d<2pi, $ putting
$z+1=|z+1|e^{ia}; z-1=|z-1|e^{ib}; z-2=|z-2|e^{ic}; z=|z|e^{id}, $ and checking that
$F(z)=sqrt{left |{frac{(z+1)(z-1)(z-2)}{z}} right |}e^{frac{1}{2}((a-d)+(b+c))i}$ is continuous on $mathbb{C} setminus { [-1,0] cup [1,2] }.$
But this is clear:
$1). $ For an arbitrary $z=|z|e^{itheta}in mathbb{C} setminus { [-1,0] cup [1,2] },$ as $theta$ makes its way from $0$ to $2pi, $ so do $a,b,c$ and $d$ and $F$ does not "jump" when this happens. More precisely, we have, by direct calculation,
$e^{frac{1}{2}(a+2pi-(d+2pi))}=e^{frac{1}{2}(a-d)}$ and $ e^{frac{1}{2}(b+2pi+c+2pi)i}=e^{frac{1}{2}(b+c)i}cdot e^{2pi i}=e^{frac{1}{2}(b+c)i}.$
$2).$ If $z=i, $ then $a=frac{pi}{4}; b=frac{7pi}{4}; c=pi -tan ^{-1}left(frac{1}{2}right)=pi(1-.147)=.852pi; d=frac{pi}{2}, $ so you can check that $cos left ( frac{1}{2}((a-d)+(b+c)) right )neq 0, $ so $frak R$ $f(i)neq 0$, and we can always make this positive by setting $G=-F$ if necessary.
answered Feb 1 at 23:44
MatematletaMatematleta
12.1k21020
$endgroup$
add a comment |
$begingroup$
We can investigate the behavior of $F$ by setting $0le a,b,c,d<2pi, $ putting
$z+1=|z+1|e^{ia}; z-1=|z-1|e^{ib}; z-2=|z-2|e^{ic}; z=|z|e^{id}, $ and checking that
$F(z)=sqrt{left |{frac{(z+1)(z-1)(z-2)}{z}} right |}e^{frac{1}{2}((a-d)+(b+c))i}$ is continuous on $mathbb{C} setminus { [-1,0] cup [1,2] }.$
But this is clear:
$1). $ For an arbitrary $z=|z|e^{itheta}in mathbb{C} setminus { [-1,0] cup [1,2] },$ as $theta$ makes its way from $0$ to $2pi, $ so do $a,b,c$ and $d$ and $F$ does not "jump" when this happens. More precisely, we have, by direct calculation,
$e^{frac{1}{2}(a+2pi-(d+2pi))}=e^{frac{1}{2}(a-d)}$ and $ e^{frac{1}{2}(b+2pi+c+2pi)i}=e^{frac{1}{2}(b+c)i}cdot e^{2pi i}=e^{frac{1}{2}(b+c)i}.$
$2).$ If $z=i, $ then $a=frac{pi}{4}; b=frac{7pi}{4}; c=pi -tan ^{-1}left(frac{1}{2}right)=pi(1-.147)=.852pi; d=frac{pi}{2}, $ so you can check that $cos left ( frac{1}{2}((a-d)+(b+c)) right )neq 0, $ so $frak R$ $f(i)neq 0$, and we can always make this positive by setting $G=-F$ if necessary.
answered Feb 1 at 23:44
MatematletaMatematleta
12.1k21020
$endgroup$
We can investigate the behavior of $F$ by setting $0le a,b,c,d<2pi, $ putting
$z+1=|z+1|e^{ia}; z-1=|z-1|e^{ib}; z-2=|z-2|e^{ic}; z=|z|e^{id}, $ and checking that
$F(z)=sqrt{left |{frac{(z+1)(z-1)(z-2)}{z}} right |}e^{frac{1}{2}((a-d)+(b+c))i}$ is continuous on $mathbb{C} setminus { [-1,0] cup [1,2] }.$
But this is clear:
$1). $ For an arbitrary $z=|z|e^{itheta}in mathbb{C} setminus { [-1,0] cup [1,2] },$ as $theta$ makes its way from $0$ to $2pi, $ so do $a,b,c$ and $d$ and $F$ does not "jump" when this happens. More precisely, we have, by direct calculation,
$e^{frac{1}{2}(a+2pi-(d+2pi))}=e^{frac{1}{2}(a-d)}$ and $ e^{frac{1}{2}(b+2pi+c+2pi)i}=e^{frac{1}{2}(b+c)i}cdot e^{2pi i}=e^{frac{1}{2}(b+c)i}.$
$2).$ If $z=i, $ then $a=frac{pi}{4}; b=frac{7pi}{4}; c=pi -tan ^{-1}left(frac{1}{2}right)=pi(1-.147)=.852pi; d=frac{pi}{2}, $ so you can check that $cos left ( frac{1}{2}((a-d)+(b+c)) right )neq 0, $ so $frak R$ $f(i)neq 0$, and we can always make this positive by setting $G=-F$ if necessary.
answered Feb 1 at 23:44
MatematletaMatematleta
12.1k21020
edited Feb 2 at 0:04
answered Feb 1 at 23:44
MatematletaMatematleta
12.1k21020
answered Feb 1 at 23:44
MatematletaMatematleta
12.1k21020
answered Feb 1 at 23:44
MatematletaMatematleta
12.1k21020
12.1k21020
add a comment |
add a comment |
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$begingroup$
For largish $z$ you can write $F(z)=zsqrt{(1-z^{-2})(1-2z^{-1})}$ which gives you one holomorphic function for $|z|>2$. Now you only need to explore how this extends inwards. Or perhaps the form $F(z)=(z-1)sqrt{1+z^{-1}}sqrt{1-(z-1)^{-1}}$ leads more directly to the desired result.
$endgroup$
– LutzL
Feb 1 at 9:46