is the given function differentiable at origin
$begingroup$
it is given that f(z) = Im(z^2)/(z bar) when zis not equal to 0 and f(z) =0 when z=0
so when i find the limit of f(z) it depends on a parameter m
so limit doesnot exist uniquely and hence the given function is not differentiable at origin
but when i tried to find its limit by polar coordinates then its limit came out to b 0
i.e x=r cos theta and y = r sin theta
then
f(r, thetha) = (sin2 theta) r exp(i theta)
on putting limit r tends to 0 we get
zero
my query is that if the function is not differentiable at origin then whatever approach we apply its limit should not exist uniquely or finitely
but here wen i try to solve it by polar approach i m getting 0 ie unqiue and finite limit
& on the other hand by previous method i found that it depend on parameter m
is there any mistake i am making while solving it by latter method ? please solve
complex-analysis
asked Jan 9 at 11:26
HenryHenry
327
$endgroup$
add a comment |
$begingroup$
it is given that f(z) = Im(z^2)/(z bar) when zis not equal to 0 and f(z) =0 when z=0
so when i find the limit of f(z) it depends on a parameter m
so limit doesnot exist uniquely and hence the given function is not differentiable at origin
but when i tried to find its limit by polar coordinates then its limit came out to b 0
i.e x=r cos theta and y = r sin theta
then
f(r, thetha) = (sin2 theta) r exp(i theta)
on putting limit r tends to 0 we get
zero
my query is that if the function is not differentiable at origin then whatever approach we apply its limit should not exist uniquely or finitely
but here wen i try to solve it by polar approach i m getting 0 ie unqiue and finite limit
& on the other hand by previous method i found that it depend on parameter m
is there any mistake i am making while solving it by latter method ? please solve
complex-analysis
asked Jan 9 at 11:26
HenryHenry
327
$endgroup$
add a comment |
$begingroup$
it is given that f(z) = Im(z^2)/(z bar) when zis not equal to 0 and f(z) =0 when z=0
so when i find the limit of f(z) it depends on a parameter m
so limit doesnot exist uniquely and hence the given function is not differentiable at origin
but when i tried to find its limit by polar coordinates then its limit came out to b 0
i.e x=r cos theta and y = r sin theta
then
f(r, thetha) = (sin2 theta) r exp(i theta)
on putting limit r tends to 0 we get
zero
my query is that if the function is not differentiable at origin then whatever approach we apply its limit should not exist uniquely or finitely
but here wen i try to solve it by polar approach i m getting 0 ie unqiue and finite limit
& on the other hand by previous method i found that it depend on parameter m
is there any mistake i am making while solving it by latter method ? please solve
complex-analysis
asked Jan 9 at 11:26
HenryHenry
327
$endgroup$
it is given that f(z) = Im(z^2)/(z bar) when zis not equal to 0 and f(z) =0 when z=0
so when i find the limit of f(z) it depends on a parameter m
so limit doesnot exist uniquely and hence the given function is not differentiable at origin
but when i tried to find its limit by polar coordinates then its limit came out to b 0
i.e x=r cos theta and y = r sin theta
then
f(r, thetha) = (sin2 theta) r exp(i theta)
on putting limit r tends to 0 we get
zero
my query is that if the function is not differentiable at origin then whatever approach we apply its limit should not exist uniquely or finitely
but here wen i try to solve it by polar approach i m getting 0 ie unqiue and finite limit
& on the other hand by previous method i found that it depend on parameter m
is there any mistake i am making while solving it by latter method ? please solve
complex-analysis
complex-analysis
asked Jan 9 at 11:26
HenryHenry
327
asked Jan 9 at 11:26
HenryHenry
327
edited Jan 9 at 11:43
Henry
asked Jan 9 at 11:26
HenryHenry
327
asked Jan 9 at 11:26
HenryHenry
327
asked Jan 9 at 11:26
HenryHenry
327
327
add a comment |
add a comment |
2 Answers
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$begingroup$
I'm a little confused about what you're trying to prove. Your calculation with polar coordinates shows correctly that $lim_{zrightarrow 0} fleft(zright) = 0$ . But this merely establishes that $f$ is continuous at $z = 0$, not that it's differentiable there. To determine whether it's differentiable you need to establish what happens to $frac{fleft(zright) - fleft(0right)}{z - 0} = frac{fleft(zright)}{z}$ as $zrightarrow 0$ .
If you divide your expression for $fleft(zright)$ in terms of polar coordinates by $z = r e^{itheta}$ you're left with $sinleft(2thetaright)$ , independent of $r .$ This is equal to $1$ for $theta = frac{pi}{4}$ and $0$ for $theta = 0$, so there are values of $z$ arbitrarily close to $0$ for which $frac{fleft(zright)}{z} = 1$ and others for which $frac{fleft(zright)}{z} = 0$. It therefore can't have a limit as $zrightarrow 0$ , and consequently $f$ is not differentiable at $z= 0$ .
answered Jan 9 at 13:22
lonza leggieralonza leggiera
89117
$endgroup$
$begingroup$
no no that i understood very well ..m query is about the limit ...wen i try to solve it with the help of polar coordinates ...limit exists and equal to zero ...but wen i try to solve it with an alternative approach like along different path ...it depends on parameter m n hence the limit doesnt exist in this case ....so why is this difference ? if the limit doesnt exist then no matter whatever approach u go with it will never exist
$endgroup$
– Henry
Jan 9 at 14:43
$begingroup$
ooops! my bad ...i just skipped a step in calculation ....how could i do such silly mistakes .....totally agree with u ....thankyou so much :)
$endgroup$
– Henry
Jan 9 at 14:49
add a comment |
$begingroup$
The limit does not depend on a parameter !
Since $|Im(z^2)| le |z|^2=|z|^2$, we get
$|f(z)| le |z|$ for all $z $.
answered Jan 9 at 12:19
FredFred
46.9k1848
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
I'm a little confused about what you're trying to prove. Your calculation with polar coordinates shows correctly that $lim_{zrightarrow 0} fleft(zright) = 0$ . But this merely establishes that $f$ is continuous at $z = 0$, not that it's differentiable there. To determine whether it's differentiable you need to establish what happens to $frac{fleft(zright) - fleft(0right)}{z - 0} = frac{fleft(zright)}{z}$ as $zrightarrow 0$ .
If you divide your expression for $fleft(zright)$ in terms of polar coordinates by $z = r e^{itheta}$ you're left with $sinleft(2thetaright)$ , independent of $r .$ This is equal to $1$ for $theta = frac{pi}{4}$ and $0$ for $theta = 0$, so there are values of $z$ arbitrarily close to $0$ for which $frac{fleft(zright)}{z} = 1$ and others for which $frac{fleft(zright)}{z} = 0$. It therefore can't have a limit as $zrightarrow 0$ , and consequently $f$ is not differentiable at $z= 0$ .
answered Jan 9 at 13:22
lonza leggieralonza leggiera
89117
$endgroup$
$begingroup$
no no that i understood very well ..m query is about the limit ...wen i try to solve it with the help of polar coordinates ...limit exists and equal to zero ...but wen i try to solve it with an alternative approach like along different path ...it depends on parameter m n hence the limit doesnt exist in this case ....so why is this difference ? if the limit doesnt exist then no matter whatever approach u go with it will never exist
$endgroup$
– Henry
Jan 9 at 14:43
$begingroup$
ooops! my bad ...i just skipped a step in calculation ....how could i do such silly mistakes .....totally agree with u ....thankyou so much :)
$endgroup$
– Henry
Jan 9 at 14:49
add a comment |
$begingroup$
I'm a little confused about what you're trying to prove. Your calculation with polar coordinates shows correctly that $lim_{zrightarrow 0} fleft(zright) = 0$ . But this merely establishes that $f$ is continuous at $z = 0$, not that it's differentiable there. To determine whether it's differentiable you need to establish what happens to $frac{fleft(zright) - fleft(0right)}{z - 0} = frac{fleft(zright)}{z}$ as $zrightarrow 0$ .
If you divide your expression for $fleft(zright)$ in terms of polar coordinates by $z = r e^{itheta}$ you're left with $sinleft(2thetaright)$ , independent of $r .$ This is equal to $1$ for $theta = frac{pi}{4}$ and $0$ for $theta = 0$, so there are values of $z$ arbitrarily close to $0$ for which $frac{fleft(zright)}{z} = 1$ and others for which $frac{fleft(zright)}{z} = 0$. It therefore can't have a limit as $zrightarrow 0$ , and consequently $f$ is not differentiable at $z= 0$ .
answered Jan 9 at 13:22
lonza leggieralonza leggiera
89117
$endgroup$
$begingroup$
no no that i understood very well ..m query is about the limit ...wen i try to solve it with the help of polar coordinates ...limit exists and equal to zero ...but wen i try to solve it with an alternative approach like along different path ...it depends on parameter m n hence the limit doesnt exist in this case ....so why is this difference ? if the limit doesnt exist then no matter whatever approach u go with it will never exist
$endgroup$
– Henry
Jan 9 at 14:43
$begingroup$
ooops! my bad ...i just skipped a step in calculation ....how could i do such silly mistakes .....totally agree with u ....thankyou so much :)
$endgroup$
– Henry
Jan 9 at 14:49
add a comment |
$begingroup$
I'm a little confused about what you're trying to prove. Your calculation with polar coordinates shows correctly that $lim_{zrightarrow 0} fleft(zright) = 0$ . But this merely establishes that $f$ is continuous at $z = 0$, not that it's differentiable there. To determine whether it's differentiable you need to establish what happens to $frac{fleft(zright) - fleft(0right)}{z - 0} = frac{fleft(zright)}{z}$ as $zrightarrow 0$ .
If you divide your expression for $fleft(zright)$ in terms of polar coordinates by $z = r e^{itheta}$ you're left with $sinleft(2thetaright)$ , independent of $r .$ This is equal to $1$ for $theta = frac{pi}{4}$ and $0$ for $theta = 0$, so there are values of $z$ arbitrarily close to $0$ for which $frac{fleft(zright)}{z} = 1$ and others for which $frac{fleft(zright)}{z} = 0$. It therefore can't have a limit as $zrightarrow 0$ , and consequently $f$ is not differentiable at $z= 0$ .
answered Jan 9 at 13:22
lonza leggieralonza leggiera
89117
$endgroup$
I'm a little confused about what you're trying to prove. Your calculation with polar coordinates shows correctly that $lim_{zrightarrow 0} fleft(zright) = 0$ . But this merely establishes that $f$ is continuous at $z = 0$, not that it's differentiable there. To determine whether it's differentiable you need to establish what happens to $frac{fleft(zright) - fleft(0right)}{z - 0} = frac{fleft(zright)}{z}$ as $zrightarrow 0$ .
If you divide your expression for $fleft(zright)$ in terms of polar coordinates by $z = r e^{itheta}$ you're left with $sinleft(2thetaright)$ , independent of $r .$ This is equal to $1$ for $theta = frac{pi}{4}$ and $0$ for $theta = 0$, so there are values of $z$ arbitrarily close to $0$ for which $frac{fleft(zright)}{z} = 1$ and others for which $frac{fleft(zright)}{z} = 0$. It therefore can't have a limit as $zrightarrow 0$ , and consequently $f$ is not differentiable at $z= 0$ .
answered Jan 9 at 13:22
lonza leggieralonza leggiera
89117
edited Jan 9 at 14:06
answered Jan 9 at 13:22
lonza leggieralonza leggiera
89117
answered Jan 9 at 13:22
lonza leggieralonza leggiera
89117
answered Jan 9 at 13:22
lonza leggieralonza leggiera
89117
89117
$begingroup$
no no that i understood very well ..m query is about the limit ...wen i try to solve it with the help of polar coordinates ...limit exists and equal to zero ...but wen i try to solve it with an alternative approach like along different path ...it depends on parameter m n hence the limit doesnt exist in this case ....so why is this difference ? if the limit doesnt exist then no matter whatever approach u go with it will never exist
$endgroup$
– Henry
Jan 9 at 14:43
$begingroup$
ooops! my bad ...i just skipped a step in calculation ....how could i do such silly mistakes .....totally agree with u ....thankyou so much :)
$endgroup$
– Henry
Jan 9 at 14:49
add a comment |
$begingroup$
no no that i understood very well ..m query is about the limit ...wen i try to solve it with the help of polar coordinates ...limit exists and equal to zero ...but wen i try to solve it with an alternative approach like along different path ...it depends on parameter m n hence the limit doesnt exist in this case ....so why is this difference ? if the limit doesnt exist then no matter whatever approach u go with it will never exist
$endgroup$
– Henry
Jan 9 at 14:43
$begingroup$
ooops! my bad ...i just skipped a step in calculation ....how could i do such silly mistakes .....totally agree with u ....thankyou so much :)
$endgroup$
– Henry
Jan 9 at 14:49
$begingroup$
no no that i understood very well ..m query is about the limit ...wen i try to solve it with the help of polar coordinates ...limit exists and equal to zero ...but wen i try to solve it with an alternative approach like along different path ...it depends on parameter m n hence the limit doesnt exist in this case ....so why is this difference ? if the limit doesnt exist then no matter whatever approach u go with it will never exist
$endgroup$
– Henry
Jan 9 at 14:43
$begingroup$
no no that i understood very well ..m query is about the limit ...wen i try to solve it with the help of polar coordinates ...limit exists and equal to zero ...but wen i try to solve it with an alternative approach like along different path ...it depends on parameter m n hence the limit doesnt exist in this case ....so why is this difference ? if the limit doesnt exist then no matter whatever approach u go with it will never exist
$endgroup$
– Henry
Jan 9 at 14:43
$begingroup$
ooops! my bad ...i just skipped a step in calculation ....how could i do such silly mistakes .....totally agree with u ....thankyou so much :)
$endgroup$
– Henry
Jan 9 at 14:49
$begingroup$
ooops! my bad ...i just skipped a step in calculation ....how could i do such silly mistakes .....totally agree with u ....thankyou so much :)
$endgroup$
– Henry
Jan 9 at 14:49
add a comment |
$begingroup$
The limit does not depend on a parameter !
Since $|Im(z^2)| le |z|^2=|z|^2$, we get
$|f(z)| le |z|$ for all $z $.
answered Jan 9 at 12:19
FredFred
46.9k1848
$endgroup$
add a comment |
$begingroup$
The limit does not depend on a parameter !
Since $|Im(z^2)| le |z|^2=|z|^2$, we get
$|f(z)| le |z|$ for all $z $.
answered Jan 9 at 12:19
FredFred
46.9k1848
$endgroup$
add a comment |
$begingroup$
The limit does not depend on a parameter !
Since $|Im(z^2)| le |z|^2=|z|^2$, we get
$|f(z)| le |z|$ for all $z $.
answered Jan 9 at 12:19
FredFred
46.9k1848
$endgroup$
The limit does not depend on a parameter !
Since $|Im(z^2)| le |z|^2=|z|^2$, we get
$|f(z)| le |z|$ for all $z $.
answered Jan 9 at 12:19
FredFred
46.9k1848
answered Jan 9 at 12:19
FredFred
46.9k1848
answered Jan 9 at 12:19
FredFred
46.9k1848
answered Jan 9 at 12:19
FredFred
46.9k1848
46.9k1848
add a comment |
add a comment |
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