Mixing
Give principal branch of complex function
$begingroup$
Give the principal branch of complex-valued function $f(z)=sqrt{1-z}$.
My approach: We know that square root is multi-valued function.
Consider the set $D:=mathbb{C}-{z:Im(z)=0, Re(z)leq 1}$ which is region (open and connected subset) in $mathbb{C}$. Let's take the principal branch of logarithm, i.e. $$text{Log}(1-z)=log|1-z|+iarg(1-z).$$
Then the principal branch of $f(z)$ will be $$f_1(z)=expleft(frac{1}{2}(log|1-z|+iarg(1-z))right)$$ on the region $D$.
Also $(f_1(z))^2=expleft(log|1-z|+iarg(1-z)right)=exp(text{Log}(1-z))=1-z$.
We see that function $f_1(z)$ is single-valued on $D$, continuous on $D$ and also it's one of the values of function $sqrt{1-z}$.
Please, can anyone take a look at my solution and say is it ok? This the first time when I have solved problems on complex analysis. I've solved it after many hours of thinking and reading lecture materials. So please do not duplicate my question.
complex-analysis proof-verification
edited Feb 2 at 22:16
user
6,57011031
asked Feb 2 at 22:07
ZFRZFR
5,36131540
$endgroup$
add a comment |
$begingroup$
Give the principal branch of complex-valued function $f(z)=sqrt{1-z}$.
My approach: We know that square root is multi-valued function.
Consider the set $D:=mathbb{C}-{z:Im(z)=0, Re(z)leq 1}$ which is region (open and connected subset) in $mathbb{C}$. Let's take the principal branch of logarithm, i.e. $$text{Log}(1-z)=log|1-z|+iarg(1-z).$$
Then the principal branch of $f(z)$ will be $$f_1(z)=expleft(frac{1}{2}(log|1-z|+iarg(1-z))right)$$ on the region $D$.
Also $(f_1(z))^2=expleft(log|1-z|+iarg(1-z)right)=exp(text{Log}(1-z))=1-z$.
We see that function $f_1(z)$ is single-valued on $D$, continuous on $D$ and also it's one of the values of function $sqrt{1-z}$.
Please, can anyone take a look at my solution and say is it ok? This the first time when I have solved problems on complex analysis. I've solved it after many hours of thinking and reading lecture materials. So please do not duplicate my question.
complex-analysis proof-verification
edited Feb 2 at 22:16
user
6,57011031
asked Feb 2 at 22:07
ZFRZFR
5,36131540
$endgroup$
$begingroup$
The principal branch is $z mapsto e^{{1 over 2} Log (1-z)}$, defined on $mathbb{C} setminus [1,infty)$. Note the change in the real axis part of the definition.
$endgroup$
– copper.hat
Feb 3 at 4:01
add a comment |
$begingroup$
Give the principal branch of complex-valued function $f(z)=sqrt{1-z}$.
My approach: We know that square root is multi-valued function.
Consider the set $D:=mathbb{C}-{z:Im(z)=0, Re(z)leq 1}$ which is region (open and connected subset) in $mathbb{C}$. Let's take the principal branch of logarithm, i.e. $$text{Log}(1-z)=log|1-z|+iarg(1-z).$$
Then the principal branch of $f(z)$ will be $$f_1(z)=expleft(frac{1}{2}(log|1-z|+iarg(1-z))right)$$ on the region $D$.
Also $(f_1(z))^2=expleft(log|1-z|+iarg(1-z)right)=exp(text{Log}(1-z))=1-z$.
We see that function $f_1(z)$ is single-valued on $D$, continuous on $D$ and also it's one of the values of function $sqrt{1-z}$.
Please, can anyone take a look at my solution and say is it ok? This the first time when I have solved problems on complex analysis. I've solved it after many hours of thinking and reading lecture materials. So please do not duplicate my question.
complex-analysis proof-verification
edited Feb 2 at 22:16
user
6,57011031
asked Feb 2 at 22:07
ZFRZFR
5,36131540
$endgroup$
Give the principal branch of complex-valued function $f(z)=sqrt{1-z}$.
My approach: We know that square root is multi-valued function.
Consider the set $D:=mathbb{C}-{z:Im(z)=0, Re(z)leq 1}$ which is region (open and connected subset) in $mathbb{C}$. Let's take the principal branch of logarithm, i.e. $$text{Log}(1-z)=log|1-z|+iarg(1-z).$$
Then the principal branch of $f(z)$ will be $$f_1(z)=expleft(frac{1}{2}(log|1-z|+iarg(1-z))right)$$ on the region $D$.
Also $(f_1(z))^2=expleft(log|1-z|+iarg(1-z)right)=exp(text{Log}(1-z))=1-z$.
We see that function $f_1(z)$ is single-valued on $D$, continuous on $D$ and also it's one of the values of function $sqrt{1-z}$.
Please, can anyone take a look at my solution and say is it ok? This the first time when I have solved problems on complex analysis. I've solved it after many hours of thinking and reading lecture materials. So please do not duplicate my question.
complex-analysis proof-verification
complex-analysis proof-verification
edited Feb 2 at 22:16
user
6,57011031
asked Feb 2 at 22:07
ZFRZFR
5,36131540
edited Feb 2 at 22:16
user
6,57011031
asked Feb 2 at 22:07
ZFRZFR
5,36131540
edited Feb 2 at 22:16
user
6,57011031
edited Feb 2 at 22:16
user
6,57011031
edited Feb 2 at 22:16
user
6,57011031
6,57011031
asked Feb 2 at 22:07
ZFRZFR
5,36131540
asked Feb 2 at 22:07
ZFRZFR
5,36131540
asked Feb 2 at 22:07
ZFRZFR
5,36131540
5,36131540
$begingroup$
The principal branch is $z mapsto e^{{1 over 2} Log (1-z)}$, defined on $mathbb{C} setminus [1,infty)$. Note the change in the real axis part of the definition.
$endgroup$
– copper.hat
Feb 3 at 4:01
add a comment |
$begingroup$
The principal branch is $z mapsto e^{{1 over 2} Log (1-z)}$, defined on $mathbb{C} setminus [1,infty)$. Note the change in the real axis part of the definition.
$endgroup$
– copper.hat
Feb 3 at 4:01
$begingroup$
The principal branch is $z mapsto e^{{1 over 2} Log (1-z)}$, defined on $mathbb{C} setminus [1,infty)$. Note the change in the real axis part of the definition.
$endgroup$
– copper.hat
Feb 3 at 4:01
$begingroup$
The principal branch is $z mapsto e^{{1 over 2} Log (1-z)}$, defined on $mathbb{C} setminus [1,infty)$. Note the change in the real axis part of the definition.
$endgroup$
– copper.hat
Feb 3 at 4:01
add a comment |
1 Answer
1
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$begingroup$
The principle branch of logarithm is usually defined on $mathbb C setminus {z:Im(z)=0, Re (z) leq 0}$. If this is what you mean by Log then the correct answer is $mathbb C setminus {z:Im(z)=0, Re (z) geq 1}$. Note that $z=2$ is in the region you are considering but the principle branch of log is not defined at $1-2=-1$.
answered Feb 2 at 23:38
Kavi Rama MurthyKavi Rama Murthy
74.8k53270
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The principle branch of logarithm is usually defined on $mathbb C setminus {z:Im(z)=0, Re (z) leq 0}$. If this is what you mean by Log then the correct answer is $mathbb C setminus {z:Im(z)=0, Re (z) geq 1}$. Note that $z=2$ is in the region you are considering but the principle branch of log is not defined at $1-2=-1$.
answered Feb 2 at 23:38
Kavi Rama MurthyKavi Rama Murthy
74.8k53270
$endgroup$
add a comment |
$begingroup$
The principle branch of logarithm is usually defined on $mathbb C setminus {z:Im(z)=0, Re (z) leq 0}$. If this is what you mean by Log then the correct answer is $mathbb C setminus {z:Im(z)=0, Re (z) geq 1}$. Note that $z=2$ is in the region you are considering but the principle branch of log is not defined at $1-2=-1$.
answered Feb 2 at 23:38
Kavi Rama MurthyKavi Rama Murthy
74.8k53270
$endgroup$
add a comment |
$begingroup$
The principle branch of logarithm is usually defined on $mathbb C setminus {z:Im(z)=0, Re (z) leq 0}$. If this is what you mean by Log then the correct answer is $mathbb C setminus {z:Im(z)=0, Re (z) geq 1}$. Note that $z=2$ is in the region you are considering but the principle branch of log is not defined at $1-2=-1$.
answered Feb 2 at 23:38
Kavi Rama MurthyKavi Rama Murthy
74.8k53270
$endgroup$
The principle branch of logarithm is usually defined on $mathbb C setminus {z:Im(z)=0, Re (z) leq 0}$. If this is what you mean by Log then the correct answer is $mathbb C setminus {z:Im(z)=0, Re (z) geq 1}$. Note that $z=2$ is in the region you are considering but the principle branch of log is not defined at $1-2=-1$.
answered Feb 2 at 23:38
Kavi Rama MurthyKavi Rama Murthy
74.8k53270
answered Feb 2 at 23:38
Kavi Rama MurthyKavi Rama Murthy
74.8k53270
answered Feb 2 at 23:38
Kavi Rama MurthyKavi Rama Murthy
74.8k53270
answered Feb 2 at 23:38
Kavi Rama MurthyKavi Rama Murthy
74.8k53270
74.8k53270
add a comment |
add a comment |
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$begingroup$
The principal branch is $z mapsto e^{{1 over 2} Log (1-z)}$, defined on $mathbb{C} setminus [1,infty)$. Note the change in the real axis part of the definition.
$endgroup$
– copper.hat
Feb 3 at 4:01