$f(z) = 2z + 3overline z$, show using the limit definition that $f$ is continuous at every point $ainmathbb C$.

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For the function $f : mathbb Ctomathbb C$ defined by $f(z) = 2z + 3overline z,forall zinmathbb C$, show using the limit definition, by finding an inequality between $|f(z) − f(a)|$ and $|z − a|$, that $f$ is continuous at every point $ainmathbb C$.




When I expand $|f(z) - f(a)|$ I get $2(z-a) + 3(overline z - overline a)$. I am unsure how to complete the proof after this?




complex-numbers




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    For the function $f : mathbb Ctomathbb C$ defined by $f(z) = 2z + 3overline z,forall zinmathbb C$, show using the limit definition, by finding an inequality between $|f(z) − f(a)|$ and $|z − a|$, that $f$ is continuous at every point $ainmathbb C$.




    When I expand $|f(z) - f(a)|$ I get $2(z-a) + 3(overline z - overline a)$. I am unsure how to complete the proof after this?




    complex-numbers




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      up vote
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      For the function $f : mathbb Ctomathbb C$ defined by $f(z) = 2z + 3overline z,forall zinmathbb C$, show using the limit definition, by finding an inequality between $|f(z) − f(a)|$ and $|z − a|$, that $f$ is continuous at every point $ainmathbb C$.




      When I expand $|f(z) - f(a)|$ I get $2(z-a) + 3(overline z - overline a)$. I am unsure how to complete the proof after this?




      complex-numbers




      share|cite|improve this question















      For the function $f : mathbb Ctomathbb C$ defined by $f(z) = 2z + 3overline z,forall zinmathbb C$, show using the limit definition, by finding an inequality between $|f(z) − f(a)|$ and $|z − a|$, that $f$ is continuous at every point $ainmathbb C$.




      When I expand $|f(z) - f(a)|$ I get $2(z-a) + 3(overline z - overline a)$. I am unsure how to complete the proof after this?





      complex-numbers





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      edited Aug 23 at 5:34









      an4s

      2,0632417




      2,0632417










      asked Aug 23 at 4:39









      Sophie Wines

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          Hint:
          $$|2(z-a)+3(barz-bara)|leq|2(z-a)|+3|barz-bara|leq2|z-a|+3|z-a|$$






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            Hint:
            $$|2(z-a)+3(barz-bara)|leq|2(z-a)|+3|barz-bara|leq2|z-a|+3|z-a|$$






            share|cite|improve this answer
























              up vote
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              Hint:
              $$|2(z-a)+3(barz-bara)|leq|2(z-a)|+3|barz-bara|leq2|z-a|+3|z-a|$$






              share|cite|improve this answer






















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                Hint:
                $$|2(z-a)+3(barz-bara)|leq|2(z-a)|+3|barz-bara|leq2|z-a|+3|z-a|$$






                share|cite|improve this answer












                Hint:
                $$|2(z-a)+3(barz-bara)|leq|2(z-a)|+3|barz-bara|leq2|z-a|+3|z-a|$$







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                answered Aug 23 at 4:47









                Nosrati

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