How to show the function $fcolon zmapstolvert zrvert^2$ is continuous on the entire complex plane?












1














I am required to show this using the epsilon-delta definition of continuity and I'm having trouble applying what I know about real functions to complex ones.
I've let $z_0$ be a complex number in $mathbb{C}$ and then considered all points in the open-disc $D(z_0,1):=lbrace z:lvert z-z_0rvert<1rbrace$.
I know that $lvert f(z)-f(z_0)rvert={largelvert}lvert zrvert^2-lvert z_0rvert^2{largelvert}$ but I'm not sure how I can relate this to $lvert z-z_0rvert<1$.
Is difference of two squares the method required?






complex-analysis continuity







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  • Just think of it as a real valued function of two variables: $(x,y) to x^2 + y^2$. And please edit the question to use mathjax formatting: math.meta.stackexchange.com/questions/5020/…
    – Ethan Bolker
    Nov 19 at 1:05










  • I don't understand how that helps solve the problem.
    – Anteater23
    Nov 19 at 1:13










  • Why is $|f(z)-f(z_0)|=|z|^2-|z_0|^2$?
    – Michael Burr
    Nov 19 at 1:16










  • @Anteater23 That function of two variables is precisely the function whose continuity you want to prove. You could also carefully read a proof of the continuity of $x to x^2$ in the one variable case and see how to apply it almost verbatim to the complex case.
    – Ethan Bolker
    Nov 19 at 1:18












  • Thank you michael, I fixed it.
    – Anteater23
    Nov 19 at 1:21
















1














I am required to show this using the epsilon-delta definition of continuity and I'm having trouble applying what I know about real functions to complex ones.
I've let $z_0$ be a complex number in $mathbb{C}$ and then considered all points in the open-disc $D(z_0,1):=lbrace z:lvert z-z_0rvert<1rbrace$.
I know that $lvert f(z)-f(z_0)rvert={largelvert}lvert zrvert^2-lvert z_0rvert^2{largelvert}$ but I'm not sure how I can relate this to $lvert z-z_0rvert<1$.
Is difference of two squares the method required?






complex-analysis continuity







share|cite|improve this question
























  • Just think of it as a real valued function of two variables: $(x,y) to x^2 + y^2$. And please edit the question to use mathjax formatting: math.meta.stackexchange.com/questions/5020/…
    – Ethan Bolker
    Nov 19 at 1:05










  • I don't understand how that helps solve the problem.
    – Anteater23
    Nov 19 at 1:13










  • Why is $|f(z)-f(z_0)|=|z|^2-|z_0|^2$?
    – Michael Burr
    Nov 19 at 1:16










  • @Anteater23 That function of two variables is precisely the function whose continuity you want to prove. You could also carefully read a proof of the continuity of $x to x^2$ in the one variable case and see how to apply it almost verbatim to the complex case.
    – Ethan Bolker
    Nov 19 at 1:18












  • Thank you michael, I fixed it.
    – Anteater23
    Nov 19 at 1:21














1












1








1







I am required to show this using the epsilon-delta definition of continuity and I'm having trouble applying what I know about real functions to complex ones.
I've let $z_0$ be a complex number in $mathbb{C}$ and then considered all points in the open-disc $D(z_0,1):=lbrace z:lvert z-z_0rvert<1rbrace$.
I know that $lvert f(z)-f(z_0)rvert={largelvert}lvert zrvert^2-lvert z_0rvert^2{largelvert}$ but I'm not sure how I can relate this to $lvert z-z_0rvert<1$.
Is difference of two squares the method required?






complex-analysis continuity







share|cite|improve this question















I am required to show this using the epsilon-delta definition of continuity and I'm having trouble applying what I know about real functions to complex ones.
I've let $z_0$ be a complex number in $mathbb{C}$ and then considered all points in the open-disc $D(z_0,1):=lbrace z:lvert z-z_0rvert<1rbrace$.
I know that $lvert f(z)-f(z_0)rvert={largelvert}lvert zrvert^2-lvert z_0rvert^2{largelvert}$ but I'm not sure how I can relate this to $lvert z-z_0rvert<1$.
Is difference of two squares the method required?






complex-analysis continuity




complex-analysis continuity






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share|cite|improve this question













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edited Nov 19 at 1:21









user10354138

7,4042824




7,4042824










asked Nov 19 at 1:00









Anteater23

62




62












  • Just think of it as a real valued function of two variables: $(x,y) to x^2 + y^2$. And please edit the question to use mathjax formatting: math.meta.stackexchange.com/questions/5020/…
    – Ethan Bolker
    Nov 19 at 1:05










  • I don't understand how that helps solve the problem.
    – Anteater23
    Nov 19 at 1:13










  • Why is $|f(z)-f(z_0)|=|z|^2-|z_0|^2$?
    – Michael Burr
    Nov 19 at 1:16










  • @Anteater23 That function of two variables is precisely the function whose continuity you want to prove. You could also carefully read a proof of the continuity of $x to x^2$ in the one variable case and see how to apply it almost verbatim to the complex case.
    – Ethan Bolker
    Nov 19 at 1:18












  • Thank you michael, I fixed it.
    – Anteater23
    Nov 19 at 1:21


















  • Just think of it as a real valued function of two variables: $(x,y) to x^2 + y^2$. And please edit the question to use mathjax formatting: math.meta.stackexchange.com/questions/5020/…
    – Ethan Bolker
    Nov 19 at 1:05










  • I don't understand how that helps solve the problem.
    – Anteater23
    Nov 19 at 1:13










  • Why is $|f(z)-f(z_0)|=|z|^2-|z_0|^2$?
    – Michael Burr
    Nov 19 at 1:16










  • @Anteater23 That function of two variables is precisely the function whose continuity you want to prove. You could also carefully read a proof of the continuity of $x to x^2$ in the one variable case and see how to apply it almost verbatim to the complex case.
    – Ethan Bolker
    Nov 19 at 1:18












  • Thank you michael, I fixed it.
    – Anteater23
    Nov 19 at 1:21
















Just think of it as a real valued function of two variables: $(x,y) to x^2 + y^2$. And please edit the question to use mathjax formatting: math.meta.stackexchange.com/questions/5020/…
– Ethan Bolker
Nov 19 at 1:05




Just think of it as a real valued function of two variables: $(x,y) to x^2 + y^2$. And please edit the question to use mathjax formatting: math.meta.stackexchange.com/questions/5020/…
– Ethan Bolker
Nov 19 at 1:05












I don't understand how that helps solve the problem.
– Anteater23
Nov 19 at 1:13




I don't understand how that helps solve the problem.
– Anteater23
Nov 19 at 1:13












Why is $|f(z)-f(z_0)|=|z|^2-|z_0|^2$?
– Michael Burr
Nov 19 at 1:16




Why is $|f(z)-f(z_0)|=|z|^2-|z_0|^2$?
– Michael Burr
Nov 19 at 1:16












@Anteater23 That function of two variables is precisely the function whose continuity you want to prove. You could also carefully read a proof of the continuity of $x to x^2$ in the one variable case and see how to apply it almost verbatim to the complex case.
– Ethan Bolker
Nov 19 at 1:18






@Anteater23 That function of two variables is precisely the function whose continuity you want to prove. You could also carefully read a proof of the continuity of $x to x^2$ in the one variable case and see how to apply it almost verbatim to the complex case.
– Ethan Bolker
Nov 19 at 1:18














Thank you michael, I fixed it.
– Anteater23
Nov 19 at 1:21




Thank you michael, I fixed it.
– Anteater23
Nov 19 at 1:21










3 Answers
3






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I think you can try to rewrite the original function $f(z) = |z|^{2}$ as $f(z) = z z^{*}$ where $z^{*}$ is the complex conjugacy.






share|cite|improve this answer





























    0














    Let $z = x+mathrm{i}y$ and $delta = delta_x + mathrm{i} delta_y$. Then begin{align*}
    |z+delta| &= |(x+delta_x) + mathrm{i}(y +delta_y)| \
    &leq |x+delta_x| + |y +delta_y| text{.}
    end{align*}



    With a little triangle inequality manipulation, you can get the inequality you need.






    share|cite|improve this answer





























      0














      For any sequence $(z_n)_{n in mathbb{N}}$ with $z_n stackrel{n to infty}{longrightarrow} z_0$ you have
      $$begin{eqnarray*}
      left| |z_n|^2 - |z_0|^2right|& = & left| (|z_n| - |z_0|)(|z_n| + |z_0|)right| \
      & = & left| |z_n| - |z_0|right|cdotleft(|z_n| + |z_0|right) \
      & leq & left| z_n - z_0right|cdot left(|z_n| + |z_0|right) \
      & stackrel{n to infty}{longrightarrow} & 0cdot 2|z_0| = 0
      end{eqnarray*}$$



      So $f$ is continuous on $mathbb{C}$.






      share|cite|improve this answer





















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        3 Answers
        3






        active

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        3 Answers
        3






        active

        oldest

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        active

        oldest

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        active

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        0














        I think you can try to rewrite the original function $f(z) = |z|^{2}$ as $f(z) = z z^{*}$ where $z^{*}$ is the complex conjugacy.






        share|cite|improve this answer


























          0














          I think you can try to rewrite the original function $f(z) = |z|^{2}$ as $f(z) = z z^{*}$ where $z^{*}$ is the complex conjugacy.






          share|cite|improve this answer
























            0












            0








            0






            I think you can try to rewrite the original function $f(z) = |z|^{2}$ as $f(z) = z z^{*}$ where $z^{*}$ is the complex conjugacy.






            share|cite|improve this answer












            I think you can try to rewrite the original function $f(z) = |z|^{2}$ as $f(z) = z z^{*}$ where $z^{*}$ is the complex conjugacy.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 19 at 1:51









            Dong Le

            517




            517























                0














                Let $z = x+mathrm{i}y$ and $delta = delta_x + mathrm{i} delta_y$. Then begin{align*}
                |z+delta| &= |(x+delta_x) + mathrm{i}(y +delta_y)| \
                &leq |x+delta_x| + |y +delta_y| text{.}
                end{align*}



                With a little triangle inequality manipulation, you can get the inequality you need.






                share|cite|improve this answer


























                  0














                  Let $z = x+mathrm{i}y$ and $delta = delta_x + mathrm{i} delta_y$. Then begin{align*}
                  |z+delta| &= |(x+delta_x) + mathrm{i}(y +delta_y)| \
                  &leq |x+delta_x| + |y +delta_y| text{.}
                  end{align*}



                  With a little triangle inequality manipulation, you can get the inequality you need.






                  share|cite|improve this answer
























                    0












                    0








                    0






                    Let $z = x+mathrm{i}y$ and $delta = delta_x + mathrm{i} delta_y$. Then begin{align*}
                    |z+delta| &= |(x+delta_x) + mathrm{i}(y +delta_y)| \
                    &leq |x+delta_x| + |y +delta_y| text{.}
                    end{align*}



                    With a little triangle inequality manipulation, you can get the inequality you need.






                    share|cite|improve this answer












                    Let $z = x+mathrm{i}y$ and $delta = delta_x + mathrm{i} delta_y$. Then begin{align*}
                    |z+delta| &= |(x+delta_x) + mathrm{i}(y +delta_y)| \
                    &leq |x+delta_x| + |y +delta_y| text{.}
                    end{align*}



                    With a little triangle inequality manipulation, you can get the inequality you need.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 19 at 3:37









                    Eric Towers

                    31.8k22265




                    31.8k22265























                        0














                        For any sequence $(z_n)_{n in mathbb{N}}$ with $z_n stackrel{n to infty}{longrightarrow} z_0$ you have
                        $$begin{eqnarray*}
                        left| |z_n|^2 - |z_0|^2right|& = & left| (|z_n| - |z_0|)(|z_n| + |z_0|)right| \
                        & = & left| |z_n| - |z_0|right|cdotleft(|z_n| + |z_0|right) \
                        & leq & left| z_n - z_0right|cdot left(|z_n| + |z_0|right) \
                        & stackrel{n to infty}{longrightarrow} & 0cdot 2|z_0| = 0
                        end{eqnarray*}$$



                        So $f$ is continuous on $mathbb{C}$.






                        share|cite|improve this answer


























                          0














                          For any sequence $(z_n)_{n in mathbb{N}}$ with $z_n stackrel{n to infty}{longrightarrow} z_0$ you have
                          $$begin{eqnarray*}
                          left| |z_n|^2 - |z_0|^2right|& = & left| (|z_n| - |z_0|)(|z_n| + |z_0|)right| \
                          & = & left| |z_n| - |z_0|right|cdotleft(|z_n| + |z_0|right) \
                          & leq & left| z_n - z_0right|cdot left(|z_n| + |z_0|right) \
                          & stackrel{n to infty}{longrightarrow} & 0cdot 2|z_0| = 0
                          end{eqnarray*}$$



                          So $f$ is continuous on $mathbb{C}$.






                          share|cite|improve this answer
























                            0












                            0








                            0






                            For any sequence $(z_n)_{n in mathbb{N}}$ with $z_n stackrel{n to infty}{longrightarrow} z_0$ you have
                            $$begin{eqnarray*}
                            left| |z_n|^2 - |z_0|^2right|& = & left| (|z_n| - |z_0|)(|z_n| + |z_0|)right| \
                            & = & left| |z_n| - |z_0|right|cdotleft(|z_n| + |z_0|right) \
                            & leq & left| z_n - z_0right|cdot left(|z_n| + |z_0|right) \
                            & stackrel{n to infty}{longrightarrow} & 0cdot 2|z_0| = 0
                            end{eqnarray*}$$



                            So $f$ is continuous on $mathbb{C}$.






                            share|cite|improve this answer












                            For any sequence $(z_n)_{n in mathbb{N}}$ with $z_n stackrel{n to infty}{longrightarrow} z_0$ you have
                            $$begin{eqnarray*}
                            left| |z_n|^2 - |z_0|^2right|& = & left| (|z_n| - |z_0|)(|z_n| + |z_0|)right| \
                            & = & left| |z_n| - |z_0|right|cdotleft(|z_n| + |z_0|right) \
                            & leq & left| z_n - z_0right|cdot left(|z_n| + |z_0|right) \
                            & stackrel{n to infty}{longrightarrow} & 0cdot 2|z_0| = 0
                            end{eqnarray*}$$



                            So $f$ is continuous on $mathbb{C}$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 19 at 7:14









                            trancelocation

                            9,1051521




                            9,1051521






























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