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.










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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

















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.










share|cite|improve this question
























  • 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















up vote
1
down vote

favorite









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.










share|cite|improve this question















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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edited Jan 30 '16 at 15:37









Martin Sleziak

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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




















  • 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












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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, $
    $$|overline{z}-overline{z_0}| = |overline{z-z_0}| = |z-z_0|<epsilon.$$



    $$text{Hence, } lim_{zto z_0} overline{z} = overline{z_0}.$$






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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, $
      $$|overline{z}-overline{z_0}| = |overline{z-z_0}| = |z-z_0|<epsilon.$$



      $$text{Hence, } lim_{zto z_0} overline{z} = overline{z_0}.$$






      share|cite|improve this 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, $
        $$|overline{z}-overline{z_0}| = |overline{z-z_0}| = |z-z_0|<epsilon.$$



        $$text{Hence, } lim_{zto z_0} overline{z} = overline{z_0}.$$






        share|cite|improve this answer












        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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        answered Jan 30 '16 at 14:51









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