Mixing
Clarification over Ahlfors page 116, 2.1 about winding numbers
$begingroup$
Everything on this question is in complex plane.
As the book describes a property of a winding number, it says that:
Outside of the [line segment from $a$ to $b$] the function $(z-a) / (z-b)$ is never real and $leq 0$.
Here, the above statement should be interpreted as "never (real and $leq 0$)".
If anyone could explain why this is true that would be great. I do get why any point on the line segment (other than $b$, in which case the denominator is $0$) has to satisfy the condition that $(z-a) / (z-b)$ is real and $leq 0$, but I am not sure how to prove why any point not on the line has to satisfy the condition also.
Here, $a$ and $b$ are arbitrary complex number in a region determined by a closed curve in the complex plane; both points lie on the same region.
complex-analysis winding-number
asked Oct 27 '18 at 3:12
Cute BrownieCute Brownie
990416
$endgroup$
add a comment |
$begingroup$
Everything on this question is in complex plane.
As the book describes a property of a winding number, it says that:
Outside of the [line segment from $a$ to $b$] the function $(z-a) / (z-b)$ is never real and $leq 0$.
Here, the above statement should be interpreted as "never (real and $leq 0$)".
If anyone could explain why this is true that would be great. I do get why any point on the line segment (other than $b$, in which case the denominator is $0$) has to satisfy the condition that $(z-a) / (z-b)$ is real and $leq 0$, but I am not sure how to prove why any point not on the line has to satisfy the condition also.
Here, $a$ and $b$ are arbitrary complex number in a region determined by a closed curve in the complex plane; both points lie on the same region.
complex-analysis winding-number
asked Oct 27 '18 at 3:12
Cute BrownieCute Brownie
990416
$endgroup$
add a comment |
$begingroup$
Everything on this question is in complex plane.
As the book describes a property of a winding number, it says that:
Outside of the [line segment from $a$ to $b$] the function $(z-a) / (z-b)$ is never real and $leq 0$.
Here, the above statement should be interpreted as "never (real and $leq 0$)".
If anyone could explain why this is true that would be great. I do get why any point on the line segment (other than $b$, in which case the denominator is $0$) has to satisfy the condition that $(z-a) / (z-b)$ is real and $leq 0$, but I am not sure how to prove why any point not on the line has to satisfy the condition also.
Here, $a$ and $b$ are arbitrary complex number in a region determined by a closed curve in the complex plane; both points lie on the same region.
complex-analysis winding-number
asked Oct 27 '18 at 3:12
Cute BrownieCute Brownie
990416
$endgroup$
Everything on this question is in complex plane.
As the book describes a property of a winding number, it says that:
Outside of the [line segment from $a$ to $b$] the function $(z-a) / (z-b)$ is never real and $leq 0$.
Here, the above statement should be interpreted as "never (real and $leq 0$)".
If anyone could explain why this is true that would be great. I do get why any point on the line segment (other than $b$, in which case the denominator is $0$) has to satisfy the condition that $(z-a) / (z-b)$ is real and $leq 0$, but I am not sure how to prove why any point not on the line has to satisfy the condition also.
Here, $a$ and $b$ are arbitrary complex number in a region determined by a closed curve in the complex plane; both points lie on the same region.
complex-analysis winding-number
complex-analysis winding-number
asked Oct 27 '18 at 3:12
Cute BrownieCute Brownie
990416
asked Oct 27 '18 at 3:12
Cute BrownieCute Brownie
990416
edited Oct 27 '18 at 3:35
Cute Brownie
asked Oct 27 '18 at 3:12
Cute BrownieCute Brownie
990416
asked Oct 27 '18 at 3:12
Cute BrownieCute Brownie
990416
asked Oct 27 '18 at 3:12
Cute BrownieCute Brownie
990416
990416
add a comment |
add a comment |
3 Answers
3
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oldest
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$begingroup$
Note that $frac{z-a}{z-b}$ is unchanged if we add the same number to each of $z$, $a$, and $b$. So, we may translate all of our points by $-a$ to assume that $a=0$. Now let $$t=frac{z}{z-b}.$$ Solving for $z$, we have $$z=frac{t}{t-1}b.$$ If $t$ is real, then we see that $z$ is a real multiple of $b$, so it is on the line between $a=0$ and $b$. More specifically, if $tleq 0$, then $frac{t}{t-1}in [0,1)$, so $z$ is in fact on the line segment between $a=0$ and $b$.
From a geometric perspective, $frac{z-a}{z-b}$ being negative means that the vector from $a$ to $z$ and the vector from $b$ to $z$ point in opposite directions. It should not be hard to convince yourself with a picture that this only happens when $z$ is on the line segment between $a$ and $b$.
answered Oct 27 '18 at 3:44
Eric WofseyEric Wofsey
184k13212338
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add a comment |
$begingroup$
Sorry, I misread the question at first; thanks to Eric for pointing it out. Also see his answer for the geometric perspective.
Anyways, if the statement were true, then there exists $c$ such that $k(c - b) = c - a$ for a real number $k le 0$. That is,
$$a - kb = (1 - k)c\
text{Let } l = (1 - k)^{-1}, text{where } 0 lt l le 1.\
c = {a - kbover 1 - k} = la + (1 - l)b = a + (1 - l)(-a + b)$$
Yet, $c$ cannot intersect the line segment from $a$ to $b$. That is, $c neq a + m(b - a)$ for any real number $0 le m le 1$. But we have just found $0 le 1 - l lt 1$ above. So the statement can only be false.
$$ $$
Original answer:
I'm not sure if that statement as displayed is actually true. Say f{z} = (z - a)/(z - b), and let a = -1 - i, b = 1 + i. f{c} = 3 for c = 2 + 2i.
answered Oct 27 '18 at 3:31
d0SO'Nd0SO'N
12
$endgroup$
$begingroup$
You seem to have missed the "$leq 0$" part of the statement.
$endgroup$
– Eric Wofsey
Oct 27 '18 at 3:32
add a comment |
$begingroup$
The imaginary part of $u=frac{z-a}{z-b}$ is $$frac{u-overline{u}}{2i}=frac{1}{2i}left(frac{z-a}{z-b}-overline{frac{z-a}{z-b}}right)=frac{1}{2i}left(frac{z-a}{z-b}-frac{overline{z}-overline{a}}{overline{z}-overline{b}}right)$$
So, if we want this to be real, we need
$$
begin{eqnarray*}
frac{1}{2i}left(frac{z-a}{z-b}-frac{overline{z}-overline{a}}{overline{z}-overline{b}}right)&=&0
\
(z-a)(overline{z}-overline{b})&=&(z-b)(overline{z}-overline{a}) \
zoverline{z}-zoverline{b}-aoverline{z}+aoverline{b}&=&zoverline{z}-zoverline{a}-boverline{z}+boverline{a}
\
0&=&z(overline{a}-overline{b})+b(overline{z}-overline{a})-a(overline{z}-overline{b})
\
0&=&-Re(b)Im(z)+Im(b)Re(z)+Re(a)(Im(z)-Im(b))+Im(a)(Re(b)-Re(z)) \
0&=&Re(b)Im{a}-Re(a)Im(b)-(Re(a)-Re(b))Im(z)+(Im(b)-Im(a))Re(z) \
Im(z)&=&frac{Im(b) Re(a) - Im(a) Re(b) + (Im(a) - Im(b)) Re(z)}{Re(a) - Re(b)}
end{eqnarray*}
$$
which is the complex point-slope form of the line connecting $a$ and $b$.
So, what about whether the real part is positive or negative? We'll use this formula for $z$ to compute $u$.
$$
begin{eqnarray*}
u&=&frac{Re(z)+iIm(z)-left(Re(a)-i Im(a)right)}{Re(z)+iIm(z)-left(Re(b)-i Im(b)right)}\
&=&
frac{Re(z)+ileft(frac{Im(b) Re(a) - Im(a) Re(b) + (Im(a) - Im(b)) Re(z)}{Re(a) - Re(b)}right)-(Re(a)+i Im(a))}{Re(z)+ileft(frac{Im(b) Re(a) - Im(a) Re(b) + (Im(a) - Im(b)) Re(z)}{Re(a) - Re(b)}right)-(Re(b)+i Im(b))}\
&=&
frac{left(Re(z)-Re(a)right)left(frac{Re(a)-Re(b)+i(-Im(a)+Im(b))}{Re(a)-Re(b)}right)}{left(Re(z)-Re(b)right)left(frac{Re(a)-Re(b)+i(-Im(a)+Im(b))}{Re(a)-Re(b)}right)}\
&=&
frac{Re(z)-Re(a)}{Re(z)-Re(b)}
end{eqnarray*}
$$
From there, it's simple algebra to see that $u=frac{Re(z)-Re(a)}{Re(z)-Re(b)}$ is negative only between $Re(a)$ and $Re(b)$, which corresponds to the line segment joining $a$ and $b$.
answered Oct 27 '18 at 4:43
Alexander Gruber♦Alexander Gruber
20k25102172
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that $frac{z-a}{z-b}$ is unchanged if we add the same number to each of $z$, $a$, and $b$. So, we may translate all of our points by $-a$ to assume that $a=0$. Now let $$t=frac{z}{z-b}.$$ Solving for $z$, we have $$z=frac{t}{t-1}b.$$ If $t$ is real, then we see that $z$ is a real multiple of $b$, so it is on the line between $a=0$ and $b$. More specifically, if $tleq 0$, then $frac{t}{t-1}in [0,1)$, so $z$ is in fact on the line segment between $a=0$ and $b$.
From a geometric perspective, $frac{z-a}{z-b}$ being negative means that the vector from $a$ to $z$ and the vector from $b$ to $z$ point in opposite directions. It should not be hard to convince yourself with a picture that this only happens when $z$ is on the line segment between $a$ and $b$.
answered Oct 27 '18 at 3:44
Eric WofseyEric Wofsey
184k13212338
$endgroup$
add a comment |
$begingroup$
Note that $frac{z-a}{z-b}$ is unchanged if we add the same number to each of $z$, $a$, and $b$. So, we may translate all of our points by $-a$ to assume that $a=0$. Now let $$t=frac{z}{z-b}.$$ Solving for $z$, we have $$z=frac{t}{t-1}b.$$ If $t$ is real, then we see that $z$ is a real multiple of $b$, so it is on the line between $a=0$ and $b$. More specifically, if $tleq 0$, then $frac{t}{t-1}in [0,1)$, so $z$ is in fact on the line segment between $a=0$ and $b$.
From a geometric perspective, $frac{z-a}{z-b}$ being negative means that the vector from $a$ to $z$ and the vector from $b$ to $z$ point in opposite directions. It should not be hard to convince yourself with a picture that this only happens when $z$ is on the line segment between $a$ and $b$.
answered Oct 27 '18 at 3:44
Eric WofseyEric Wofsey
184k13212338
$endgroup$
add a comment |
$begingroup$
Note that $frac{z-a}{z-b}$ is unchanged if we add the same number to each of $z$, $a$, and $b$. So, we may translate all of our points by $-a$ to assume that $a=0$. Now let $$t=frac{z}{z-b}.$$ Solving for $z$, we have $$z=frac{t}{t-1}b.$$ If $t$ is real, then we see that $z$ is a real multiple of $b$, so it is on the line between $a=0$ and $b$. More specifically, if $tleq 0$, then $frac{t}{t-1}in [0,1)$, so $z$ is in fact on the line segment between $a=0$ and $b$.
From a geometric perspective, $frac{z-a}{z-b}$ being negative means that the vector from $a$ to $z$ and the vector from $b$ to $z$ point in opposite directions. It should not be hard to convince yourself with a picture that this only happens when $z$ is on the line segment between $a$ and $b$.
answered Oct 27 '18 at 3:44
Eric WofseyEric Wofsey
184k13212338
$endgroup$
Note that $frac{z-a}{z-b}$ is unchanged if we add the same number to each of $z$, $a$, and $b$. So, we may translate all of our points by $-a$ to assume that $a=0$. Now let $$t=frac{z}{z-b}.$$ Solving for $z$, we have $$z=frac{t}{t-1}b.$$ If $t$ is real, then we see that $z$ is a real multiple of $b$, so it is on the line between $a=0$ and $b$. More specifically, if $tleq 0$, then $frac{t}{t-1}in [0,1)$, so $z$ is in fact on the line segment between $a=0$ and $b$.
From a geometric perspective, $frac{z-a}{z-b}$ being negative means that the vector from $a$ to $z$ and the vector from $b$ to $z$ point in opposite directions. It should not be hard to convince yourself with a picture that this only happens when $z$ is on the line segment between $a$ and $b$.
answered Oct 27 '18 at 3:44
Eric WofseyEric Wofsey
184k13212338
answered Oct 27 '18 at 3:44
Eric WofseyEric Wofsey
184k13212338
answered Oct 27 '18 at 3:44
Eric WofseyEric Wofsey
184k13212338
answered Oct 27 '18 at 3:44
Eric WofseyEric Wofsey
184k13212338
184k13212338
add a comment |
add a comment |
$begingroup$
Sorry, I misread the question at first; thanks to Eric for pointing it out. Also see his answer for the geometric perspective.
Anyways, if the statement were true, then there exists $c$ such that $k(c - b) = c - a$ for a real number $k le 0$. That is,
$$a - kb = (1 - k)c\
text{Let } l = (1 - k)^{-1}, text{where } 0 lt l le 1.\
c = {a - kbover 1 - k} = la + (1 - l)b = a + (1 - l)(-a + b)$$
Yet, $c$ cannot intersect the line segment from $a$ to $b$. That is, $c neq a + m(b - a)$ for any real number $0 le m le 1$. But we have just found $0 le 1 - l lt 1$ above. So the statement can only be false.
$$ $$
Original answer:
I'm not sure if that statement as displayed is actually true. Say f{z} = (z - a)/(z - b), and let a = -1 - i, b = 1 + i. f{c} = 3 for c = 2 + 2i.
answered Oct 27 '18 at 3:31
d0SO'Nd0SO'N
12
$endgroup$
$begingroup$
You seem to have missed the "$leq 0$" part of the statement.
$endgroup$
– Eric Wofsey
Oct 27 '18 at 3:32
add a comment |
$begingroup$
Sorry, I misread the question at first; thanks to Eric for pointing it out. Also see his answer for the geometric perspective.
Anyways, if the statement were true, then there exists $c$ such that $k(c - b) = c - a$ for a real number $k le 0$. That is,
$$a - kb = (1 - k)c\
text{Let } l = (1 - k)^{-1}, text{where } 0 lt l le 1.\
c = {a - kbover 1 - k} = la + (1 - l)b = a + (1 - l)(-a + b)$$
Yet, $c$ cannot intersect the line segment from $a$ to $b$. That is, $c neq a + m(b - a)$ for any real number $0 le m le 1$. But we have just found $0 le 1 - l lt 1$ above. So the statement can only be false.
$$ $$
Original answer:
I'm not sure if that statement as displayed is actually true. Say f{z} = (z - a)/(z - b), and let a = -1 - i, b = 1 + i. f{c} = 3 for c = 2 + 2i.
answered Oct 27 '18 at 3:31
d0SO'Nd0SO'N
12
$endgroup$
$begingroup$
You seem to have missed the "$leq 0$" part of the statement.
$endgroup$
– Eric Wofsey
Oct 27 '18 at 3:32
add a comment |
$begingroup$
Sorry, I misread the question at first; thanks to Eric for pointing it out. Also see his answer for the geometric perspective.
Anyways, if the statement were true, then there exists $c$ such that $k(c - b) = c - a$ for a real number $k le 0$. That is,
$$a - kb = (1 - k)c\
text{Let } l = (1 - k)^{-1}, text{where } 0 lt l le 1.\
c = {a - kbover 1 - k} = la + (1 - l)b = a + (1 - l)(-a + b)$$
Yet, $c$ cannot intersect the line segment from $a$ to $b$. That is, $c neq a + m(b - a)$ for any real number $0 le m le 1$. But we have just found $0 le 1 - l lt 1$ above. So the statement can only be false.
$$ $$
Original answer:
I'm not sure if that statement as displayed is actually true. Say f{z} = (z - a)/(z - b), and let a = -1 - i, b = 1 + i. f{c} = 3 for c = 2 + 2i.
answered Oct 27 '18 at 3:31
d0SO'Nd0SO'N
12
$endgroup$
Sorry, I misread the question at first; thanks to Eric for pointing it out. Also see his answer for the geometric perspective.
Anyways, if the statement were true, then there exists $c$ such that $k(c - b) = c - a$ for a real number $k le 0$. That is,
$$a - kb = (1 - k)c\
text{Let } l = (1 - k)^{-1}, text{where } 0 lt l le 1.\
c = {a - kbover 1 - k} = la + (1 - l)b = a + (1 - l)(-a + b)$$
Yet, $c$ cannot intersect the line segment from $a$ to $b$. That is, $c neq a + m(b - a)$ for any real number $0 le m le 1$. But we have just found $0 le 1 - l lt 1$ above. So the statement can only be false.
$$ $$
Original answer:
I'm not sure if that statement as displayed is actually true. Say f{z} = (z - a)/(z - b), and let a = -1 - i, b = 1 + i. f{c} = 3 for c = 2 + 2i.
answered Oct 27 '18 at 3:31
d0SO'Nd0SO'N
12
edited Oct 27 '18 at 4:37
answered Oct 27 '18 at 3:31
d0SO'Nd0SO'N
12
answered Oct 27 '18 at 3:31
d0SO'Nd0SO'N
12
answered Oct 27 '18 at 3:31
d0SO'Nd0SO'N
12
12
$begingroup$
You seem to have missed the "$leq 0$" part of the statement.
$endgroup$
– Eric Wofsey
Oct 27 '18 at 3:32
add a comment |
$begingroup$
You seem to have missed the "$leq 0$" part of the statement.
$endgroup$
– Eric Wofsey
Oct 27 '18 at 3:32
$begingroup$
You seem to have missed the "$leq 0$" part of the statement.
$endgroup$
– Eric Wofsey
Oct 27 '18 at 3:32
$begingroup$
You seem to have missed the "$leq 0$" part of the statement.
$endgroup$
– Eric Wofsey
Oct 27 '18 at 3:32
add a comment |
$begingroup$
The imaginary part of $u=frac{z-a}{z-b}$ is $$frac{u-overline{u}}{2i}=frac{1}{2i}left(frac{z-a}{z-b}-overline{frac{z-a}{z-b}}right)=frac{1}{2i}left(frac{z-a}{z-b}-frac{overline{z}-overline{a}}{overline{z}-overline{b}}right)$$
So, if we want this to be real, we need
$$
begin{eqnarray*}
frac{1}{2i}left(frac{z-a}{z-b}-frac{overline{z}-overline{a}}{overline{z}-overline{b}}right)&=&0
\
(z-a)(overline{z}-overline{b})&=&(z-b)(overline{z}-overline{a}) \
zoverline{z}-zoverline{b}-aoverline{z}+aoverline{b}&=&zoverline{z}-zoverline{a}-boverline{z}+boverline{a}
\
0&=&z(overline{a}-overline{b})+b(overline{z}-overline{a})-a(overline{z}-overline{b})
\
0&=&-Re(b)Im(z)+Im(b)Re(z)+Re(a)(Im(z)-Im(b))+Im(a)(Re(b)-Re(z)) \
0&=&Re(b)Im{a}-Re(a)Im(b)-(Re(a)-Re(b))Im(z)+(Im(b)-Im(a))Re(z) \
Im(z)&=&frac{Im(b) Re(a) - Im(a) Re(b) + (Im(a) - Im(b)) Re(z)}{Re(a) - Re(b)}
end{eqnarray*}
$$
which is the complex point-slope form of the line connecting $a$ and $b$.
So, what about whether the real part is positive or negative? We'll use this formula for $z$ to compute $u$.
$$
begin{eqnarray*}
u&=&frac{Re(z)+iIm(z)-left(Re(a)-i Im(a)right)}{Re(z)+iIm(z)-left(Re(b)-i Im(b)right)}\
&=&
frac{Re(z)+ileft(frac{Im(b) Re(a) - Im(a) Re(b) + (Im(a) - Im(b)) Re(z)}{Re(a) - Re(b)}right)-(Re(a)+i Im(a))}{Re(z)+ileft(frac{Im(b) Re(a) - Im(a) Re(b) + (Im(a) - Im(b)) Re(z)}{Re(a) - Re(b)}right)-(Re(b)+i Im(b))}\
&=&
frac{left(Re(z)-Re(a)right)left(frac{Re(a)-Re(b)+i(-Im(a)+Im(b))}{Re(a)-Re(b)}right)}{left(Re(z)-Re(b)right)left(frac{Re(a)-Re(b)+i(-Im(a)+Im(b))}{Re(a)-Re(b)}right)}\
&=&
frac{Re(z)-Re(a)}{Re(z)-Re(b)}
end{eqnarray*}
$$
From there, it's simple algebra to see that $u=frac{Re(z)-Re(a)}{Re(z)-Re(b)}$ is negative only between $Re(a)$ and $Re(b)$, which corresponds to the line segment joining $a$ and $b$.
answered Oct 27 '18 at 4:43
Alexander Gruber♦Alexander Gruber
20k25102172
$endgroup$
add a comment |
$begingroup$
The imaginary part of $u=frac{z-a}{z-b}$ is $$frac{u-overline{u}}{2i}=frac{1}{2i}left(frac{z-a}{z-b}-overline{frac{z-a}{z-b}}right)=frac{1}{2i}left(frac{z-a}{z-b}-frac{overline{z}-overline{a}}{overline{z}-overline{b}}right)$$
So, if we want this to be real, we need
$$
begin{eqnarray*}
frac{1}{2i}left(frac{z-a}{z-b}-frac{overline{z}-overline{a}}{overline{z}-overline{b}}right)&=&0
\
(z-a)(overline{z}-overline{b})&=&(z-b)(overline{z}-overline{a}) \
zoverline{z}-zoverline{b}-aoverline{z}+aoverline{b}&=&zoverline{z}-zoverline{a}-boverline{z}+boverline{a}
\
0&=&z(overline{a}-overline{b})+b(overline{z}-overline{a})-a(overline{z}-overline{b})
\
0&=&-Re(b)Im(z)+Im(b)Re(z)+Re(a)(Im(z)-Im(b))+Im(a)(Re(b)-Re(z)) \
0&=&Re(b)Im{a}-Re(a)Im(b)-(Re(a)-Re(b))Im(z)+(Im(b)-Im(a))Re(z) \
Im(z)&=&frac{Im(b) Re(a) - Im(a) Re(b) + (Im(a) - Im(b)) Re(z)}{Re(a) - Re(b)}
end{eqnarray*}
$$
which is the complex point-slope form of the line connecting $a$ and $b$.
So, what about whether the real part is positive or negative? We'll use this formula for $z$ to compute $u$.
$$
begin{eqnarray*}
u&=&frac{Re(z)+iIm(z)-left(Re(a)-i Im(a)right)}{Re(z)+iIm(z)-left(Re(b)-i Im(b)right)}\
&=&
frac{Re(z)+ileft(frac{Im(b) Re(a) - Im(a) Re(b) + (Im(a) - Im(b)) Re(z)}{Re(a) - Re(b)}right)-(Re(a)+i Im(a))}{Re(z)+ileft(frac{Im(b) Re(a) - Im(a) Re(b) + (Im(a) - Im(b)) Re(z)}{Re(a) - Re(b)}right)-(Re(b)+i Im(b))}\
&=&
frac{left(Re(z)-Re(a)right)left(frac{Re(a)-Re(b)+i(-Im(a)+Im(b))}{Re(a)-Re(b)}right)}{left(Re(z)-Re(b)right)left(frac{Re(a)-Re(b)+i(-Im(a)+Im(b))}{Re(a)-Re(b)}right)}\
&=&
frac{Re(z)-Re(a)}{Re(z)-Re(b)}
end{eqnarray*}
$$
From there, it's simple algebra to see that $u=frac{Re(z)-Re(a)}{Re(z)-Re(b)}$ is negative only between $Re(a)$ and $Re(b)$, which corresponds to the line segment joining $a$ and $b$.
answered Oct 27 '18 at 4:43
Alexander Gruber♦Alexander Gruber
20k25102172
$endgroup$
add a comment |
$begingroup$
The imaginary part of $u=frac{z-a}{z-b}$ is $$frac{u-overline{u}}{2i}=frac{1}{2i}left(frac{z-a}{z-b}-overline{frac{z-a}{z-b}}right)=frac{1}{2i}left(frac{z-a}{z-b}-frac{overline{z}-overline{a}}{overline{z}-overline{b}}right)$$
So, if we want this to be real, we need
$$
begin{eqnarray*}
frac{1}{2i}left(frac{z-a}{z-b}-frac{overline{z}-overline{a}}{overline{z}-overline{b}}right)&=&0
\
(z-a)(overline{z}-overline{b})&=&(z-b)(overline{z}-overline{a}) \
zoverline{z}-zoverline{b}-aoverline{z}+aoverline{b}&=&zoverline{z}-zoverline{a}-boverline{z}+boverline{a}
\
0&=&z(overline{a}-overline{b})+b(overline{z}-overline{a})-a(overline{z}-overline{b})
\
0&=&-Re(b)Im(z)+Im(b)Re(z)+Re(a)(Im(z)-Im(b))+Im(a)(Re(b)-Re(z)) \
0&=&Re(b)Im{a}-Re(a)Im(b)-(Re(a)-Re(b))Im(z)+(Im(b)-Im(a))Re(z) \
Im(z)&=&frac{Im(b) Re(a) - Im(a) Re(b) + (Im(a) - Im(b)) Re(z)}{Re(a) - Re(b)}
end{eqnarray*}
$$
which is the complex point-slope form of the line connecting $a$ and $b$.
So, what about whether the real part is positive or negative? We'll use this formula for $z$ to compute $u$.
$$
begin{eqnarray*}
u&=&frac{Re(z)+iIm(z)-left(Re(a)-i Im(a)right)}{Re(z)+iIm(z)-left(Re(b)-i Im(b)right)}\
&=&
frac{Re(z)+ileft(frac{Im(b) Re(a) - Im(a) Re(b) + (Im(a) - Im(b)) Re(z)}{Re(a) - Re(b)}right)-(Re(a)+i Im(a))}{Re(z)+ileft(frac{Im(b) Re(a) - Im(a) Re(b) + (Im(a) - Im(b)) Re(z)}{Re(a) - Re(b)}right)-(Re(b)+i Im(b))}\
&=&
frac{left(Re(z)-Re(a)right)left(frac{Re(a)-Re(b)+i(-Im(a)+Im(b))}{Re(a)-Re(b)}right)}{left(Re(z)-Re(b)right)left(frac{Re(a)-Re(b)+i(-Im(a)+Im(b))}{Re(a)-Re(b)}right)}\
&=&
frac{Re(z)-Re(a)}{Re(z)-Re(b)}
end{eqnarray*}
$$
From there, it's simple algebra to see that $u=frac{Re(z)-Re(a)}{Re(z)-Re(b)}$ is negative only between $Re(a)$ and $Re(b)$, which corresponds to the line segment joining $a$ and $b$.
answered Oct 27 '18 at 4:43
Alexander Gruber♦Alexander Gruber
20k25102172
$endgroup$
The imaginary part of $u=frac{z-a}{z-b}$ is $$frac{u-overline{u}}{2i}=frac{1}{2i}left(frac{z-a}{z-b}-overline{frac{z-a}{z-b}}right)=frac{1}{2i}left(frac{z-a}{z-b}-frac{overline{z}-overline{a}}{overline{z}-overline{b}}right)$$
So, if we want this to be real, we need
$$
begin{eqnarray*}
frac{1}{2i}left(frac{z-a}{z-b}-frac{overline{z}-overline{a}}{overline{z}-overline{b}}right)&=&0
\
(z-a)(overline{z}-overline{b})&=&(z-b)(overline{z}-overline{a}) \
zoverline{z}-zoverline{b}-aoverline{z}+aoverline{b}&=&zoverline{z}-zoverline{a}-boverline{z}+boverline{a}
\
0&=&z(overline{a}-overline{b})+b(overline{z}-overline{a})-a(overline{z}-overline{b})
\
0&=&-Re(b)Im(z)+Im(b)Re(z)+Re(a)(Im(z)-Im(b))+Im(a)(Re(b)-Re(z)) \
0&=&Re(b)Im{a}-Re(a)Im(b)-(Re(a)-Re(b))Im(z)+(Im(b)-Im(a))Re(z) \
Im(z)&=&frac{Im(b) Re(a) - Im(a) Re(b) + (Im(a) - Im(b)) Re(z)}{Re(a) - Re(b)}
end{eqnarray*}
$$
which is the complex point-slope form of the line connecting $a$ and $b$.
So, what about whether the real part is positive or negative? We'll use this formula for $z$ to compute $u$.
$$
begin{eqnarray*}
u&=&frac{Re(z)+iIm(z)-left(Re(a)-i Im(a)right)}{Re(z)+iIm(z)-left(Re(b)-i Im(b)right)}\
&=&
frac{Re(z)+ileft(frac{Im(b) Re(a) - Im(a) Re(b) + (Im(a) - Im(b)) Re(z)}{Re(a) - Re(b)}right)-(Re(a)+i Im(a))}{Re(z)+ileft(frac{Im(b) Re(a) - Im(a) Re(b) + (Im(a) - Im(b)) Re(z)}{Re(a) - Re(b)}right)-(Re(b)+i Im(b))}\
&=&
frac{left(Re(z)-Re(a)right)left(frac{Re(a)-Re(b)+i(-Im(a)+Im(b))}{Re(a)-Re(b)}right)}{left(Re(z)-Re(b)right)left(frac{Re(a)-Re(b)+i(-Im(a)+Im(b))}{Re(a)-Re(b)}right)}\
&=&
frac{Re(z)-Re(a)}{Re(z)-Re(b)}
end{eqnarray*}
$$
From there, it's simple algebra to see that $u=frac{Re(z)-Re(a)}{Re(z)-Re(b)}$ is negative only between $Re(a)$ and $Re(b)$, which corresponds to the line segment joining $a$ and $b$.
answered Oct 27 '18 at 4:43
Alexander Gruber♦Alexander Gruber
20k25102172
edited Jan 8 at 8:43
answered Oct 27 '18 at 4:43
Alexander Gruber♦Alexander Gruber
20k25102172
answered Oct 27 '18 at 4:43
Alexander Gruber♦Alexander Gruber
20k25102172
answered Oct 27 '18 at 4:43
Alexander Gruber♦Alexander Gruber
20k25102172
20k25102172
add a comment |
add a comment |
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