To prove how limit exists using definition given.
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The definition of a limit of a complex function goes as follows:- Let a function f be defined at all points z in some deleted neighbourhood of z°. Then the limit of f(z) as z approaches z° is a number w° . That means for each positive number ε , there is a positive number δ such that |f(z)- w°| is less than ε whenever| z-z° | lies between 0 & δ.
Now I am asked to prove
i). limiting value of ( Re z) as z tends to z° equals (Re z°).
ii). limiting value of conjugate of z equals conjugate of z° as z tends to z°.
using definition given above.
I have difficulty as I don't know how to start.
complex-analysis limits
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The definition of a limit of a complex function goes as follows:- Let a function f be defined at all points z in some deleted neighbourhood of z°. Then the limit of f(z) as z approaches z° is a number w° . That means for each positive number ε , there is a positive number δ such that |f(z)- w°| is less than ε whenever| z-z° | lies between 0 & δ.
Now I am asked to prove
i). limiting value of ( Re z) as z tends to z° equals (Re z°).
ii). limiting value of conjugate of z equals conjugate of z° as z tends to z°.
using definition given above.
I have difficulty as I don't know how to start.
complex-analysis limits
Start by noting that $mathrm{Re}(z)-mathrm{Re}(z_0) = mathrm{Re}(z-z_0)$ and $overline{z}-overline{z_0}=overline{z-z_0}$. Now try to show that $|mathrm{Re}(z-z_0)|$, $|overline{z-z_0}|$ cannot be very large if $|z-z_0|$ is small.
– Jonathan Y.
Jan 30 '16 at 14:46
Start with $|Re(z)-Re(z^o)|=|Re(z-z^o)|leq |z-z^o|$ and $|Im(z)-Im(z^o)|=|Im(z-z^o)|le |z-z^o|.$
– DanielWainfleet
Jul 12 at 3:36
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up vote
1
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up vote
1
down vote
favorite
The definition of a limit of a complex function goes as follows:- Let a function f be defined at all points z in some deleted neighbourhood of z°. Then the limit of f(z) as z approaches z° is a number w° . That means for each positive number ε , there is a positive number δ such that |f(z)- w°| is less than ε whenever| z-z° | lies between 0 & δ.
Now I am asked to prove
i). limiting value of ( Re z) as z tends to z° equals (Re z°).
ii). limiting value of conjugate of z equals conjugate of z° as z tends to z°.
using definition given above.
I have difficulty as I don't know how to start.
complex-analysis limits
The definition of a limit of a complex function goes as follows:- Let a function f be defined at all points z in some deleted neighbourhood of z°. Then the limit of f(z) as z approaches z° is a number w° . That means for each positive number ε , there is a positive number δ such that |f(z)- w°| is less than ε whenever| z-z° | lies between 0 & δ.
Now I am asked to prove
i). limiting value of ( Re z) as z tends to z° equals (Re z°).
ii). limiting value of conjugate of z equals conjugate of z° as z tends to z°.
using definition given above.
I have difficulty as I don't know how to start.
complex-analysis limits
complex-analysis limits
edited Jan 30 '16 at 15:37
Martin Sleziak
44.5k7115268
44.5k7115268
asked Jan 30 '16 at 14:17
Kavita Sahu
13011
13011
Start by noting that $mathrm{Re}(z)-mathrm{Re}(z_0) = mathrm{Re}(z-z_0)$ and $overline{z}-overline{z_0}=overline{z-z_0}$. Now try to show that $|mathrm{Re}(z-z_0)|$, $|overline{z-z_0}|$ cannot be very large if $|z-z_0|$ is small.
– Jonathan Y.
Jan 30 '16 at 14:46
Start with $|Re(z)-Re(z^o)|=|Re(z-z^o)|leq |z-z^o|$ and $|Im(z)-Im(z^o)|=|Im(z-z^o)|le |z-z^o|.$
– DanielWainfleet
Jul 12 at 3:36
add a comment |
Start by noting that $mathrm{Re}(z)-mathrm{Re}(z_0) = mathrm{Re}(z-z_0)$ and $overline{z}-overline{z_0}=overline{z-z_0}$. Now try to show that $|mathrm{Re}(z-z_0)|$, $|overline{z-z_0}|$ cannot be very large if $|z-z_0|$ is small.
– Jonathan Y.
Jan 30 '16 at 14:46
Start with $|Re(z)-Re(z^o)|=|Re(z-z^o)|leq |z-z^o|$ and $|Im(z)-Im(z^o)|=|Im(z-z^o)|le |z-z^o|.$
– DanielWainfleet
Jul 12 at 3:36
Start by noting that $mathrm{Re}(z)-mathrm{Re}(z_0) = mathrm{Re}(z-z_0)$ and $overline{z}-overline{z_0}=overline{z-z_0}$. Now try to show that $|mathrm{Re}(z-z_0)|$, $|overline{z-z_0}|$ cannot be very large if $|z-z_0|$ is small.
– Jonathan Y.
Jan 30 '16 at 14:46
Start by noting that $mathrm{Re}(z)-mathrm{Re}(z_0) = mathrm{Re}(z-z_0)$ and $overline{z}-overline{z_0}=overline{z-z_0}$. Now try to show that $|mathrm{Re}(z-z_0)|$, $|overline{z-z_0}|$ cannot be very large if $|z-z_0|$ is small.
– Jonathan Y.
Jan 30 '16 at 14:46
Start with $|Re(z)-Re(z^o)|=|Re(z-z^o)|leq |z-z^o|$ and $|Im(z)-Im(z^o)|=|Im(z-z^o)|le |z-z^o|.$
– DanielWainfleet
Jul 12 at 3:36
Start with $|Re(z)-Re(z^o)|=|Re(z-z^o)|leq |z-z^o|$ and $|Im(z)-Im(z^o)|=|Im(z-z^o)|le |z-z^o|.$
– DanielWainfleet
Jul 12 at 3:36
add a comment |
1 Answer
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i) Note that as $z to z_0$, $forall epsilon > 0,$ take $delta := epsilon$ such that $forall z: 0 < |z-z_0| < delta, $
$$|mathrm{Re}(z) - mathrm{Re}(z_0)| le sqrt{(mathrm{Re}(z) - mathrm{Re}(z_0))^2 + (mathrm{Im}(z) - mathrm{Im}(z_0))^2} = |z-z_0|<epsilon.$$
$$text{Hence, } lim_{zto z_0} mathrm{Re}(z) = mathrm{Re}(z_0).$$
ii) As $z to z_0$, $forall epsilon > 0,$ take $delta := epsilon$ such that $forall z: 0 < |z-z_0| < delta, $
$$|overline{z}-overline{z_0}| = |overline{z-z_0}| = |z-z_0|<epsilon.$$
$$text{Hence, } lim_{zto z_0} overline{z} = overline{z_0}.$$
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1 Answer
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1 Answer
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active
oldest
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active
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active
oldest
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up vote
0
down vote
i) Note that as $z to z_0$, $forall epsilon > 0,$ take $delta := epsilon$ such that $forall z: 0 < |z-z_0| < delta, $
$$|mathrm{Re}(z) - mathrm{Re}(z_0)| le sqrt{(mathrm{Re}(z) - mathrm{Re}(z_0))^2 + (mathrm{Im}(z) - mathrm{Im}(z_0))^2} = |z-z_0|<epsilon.$$
$$text{Hence, } lim_{zto z_0} mathrm{Re}(z) = mathrm{Re}(z_0).$$
ii) As $z to z_0$, $forall epsilon > 0,$ take $delta := epsilon$ such that $forall z: 0 < |z-z_0| < delta, $
$$|overline{z}-overline{z_0}| = |overline{z-z_0}| = |z-z_0|<epsilon.$$
$$text{Hence, } lim_{zto z_0} overline{z} = overline{z_0}.$$
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i) Note that as $z to z_0$, $forall epsilon > 0,$ take $delta := epsilon$ such that $forall z: 0 < |z-z_0| < delta, $
$$|mathrm{Re}(z) - mathrm{Re}(z_0)| le sqrt{(mathrm{Re}(z) - mathrm{Re}(z_0))^2 + (mathrm{Im}(z) - mathrm{Im}(z_0))^2} = |z-z_0|<epsilon.$$
$$text{Hence, } lim_{zto z_0} mathrm{Re}(z) = mathrm{Re}(z_0).$$
ii) As $z to z_0$, $forall epsilon > 0,$ take $delta := epsilon$ such that $forall z: 0 < |z-z_0| < delta, $
$$|overline{z}-overline{z_0}| = |overline{z-z_0}| = |z-z_0|<epsilon.$$
$$text{Hence, } lim_{zto z_0} overline{z} = overline{z_0}.$$
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i) Note that as $z to z_0$, $forall epsilon > 0,$ take $delta := epsilon$ such that $forall z: 0 < |z-z_0| < delta, $
$$|mathrm{Re}(z) - mathrm{Re}(z_0)| le sqrt{(mathrm{Re}(z) - mathrm{Re}(z_0))^2 + (mathrm{Im}(z) - mathrm{Im}(z_0))^2} = |z-z_0|<epsilon.$$
$$text{Hence, } lim_{zto z_0} mathrm{Re}(z) = mathrm{Re}(z_0).$$
ii) As $z to z_0$, $forall epsilon > 0,$ take $delta := epsilon$ such that $forall z: 0 < |z-z_0| < delta, $
$$|overline{z}-overline{z_0}| = |overline{z-z_0}| = |z-z_0|<epsilon.$$
$$text{Hence, } lim_{zto z_0} overline{z} = overline{z_0}.$$
i) Note that as $z to z_0$, $forall epsilon > 0,$ take $delta := epsilon$ such that $forall z: 0 < |z-z_0| < delta, $
$$|mathrm{Re}(z) - mathrm{Re}(z_0)| le sqrt{(mathrm{Re}(z) - mathrm{Re}(z_0))^2 + (mathrm{Im}(z) - mathrm{Im}(z_0))^2} = |z-z_0|<epsilon.$$
$$text{Hence, } lim_{zto z_0} mathrm{Re}(z) = mathrm{Re}(z_0).$$
ii) As $z to z_0$, $forall epsilon > 0,$ take $delta := epsilon$ such that $forall z: 0 < |z-z_0| < delta, $
$$|overline{z}-overline{z_0}| = |overline{z-z_0}| = |z-z_0|<epsilon.$$
$$text{Hence, } lim_{zto z_0} overline{z} = overline{z_0}.$$
answered Jan 30 '16 at 14:51
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Start by noting that $mathrm{Re}(z)-mathrm{Re}(z_0) = mathrm{Re}(z-z_0)$ and $overline{z}-overline{z_0}=overline{z-z_0}$. Now try to show that $|mathrm{Re}(z-z_0)|$, $|overline{z-z_0}|$ cannot be very large if $|z-z_0|$ is small.
– Jonathan Y.
Jan 30 '16 at 14:46
Start with $|Re(z)-Re(z^o)|=|Re(z-z^o)|leq |z-z^o|$ and $|Im(z)-Im(z^o)|=|Im(z-z^o)|le |z-z^o|.$
– DanielWainfleet
Jul 12 at 3:36