Use the definition of limit to prove that $lim_{zto z_0} (az + b) = az_0 + b$, a,b belongs to complex numbers

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$begingroup$


Here what i already done:



begin{align*} |az+b-(az_0+b)&+|az-az_0+b-b|\&=|a(z-z_0)|\&=|a||z-z_0|{le}|a|*{Delta}|z-z_0|
end{align*}



If ${Delta}=1$ and we know that $0{lt}|z-z_0|{lt}1$, then:



$||z|-|z_0||{le}|z-z_0|<1 \{to} |z|{lt}1+|z_o|$



Then:



$|z|<1+|z_0| {to} |z+z_0|{le}|z|+|z_0|{le}1+2|z_0|$



So:



${Delta}|a|*(1+2|z_0|)\ = {Delta}(|a|*(1+2|z_0|)={epsilon}\ {to} {Delta} = {epsilon}/(|a|(|1+2|z_0|)$



Then:



$|az+b - (az_0 + b)| {lt} |a|{Delta}|z-z_0|\ {le} {Delta}(|a|(1+2|(z_0)|\{lt}{epsilon}/(|a|(1+2|z_0|)*(|a|(1+2|z_0|) = {epsilon}$



That proves:



$lim_{zto z_0} (az + b) = az_0 + b$



Is that correct?






complex-analysis limits proof-verification complex-numbers







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  • $begingroup$
    For $aneq 0$, from the first three lines of your reasoning you could just select: $$0<|z-z_0|<delta=epsilon/|a| implies |az+b-(az_0+b)|<epsilon$$ And then consider the case where $a=0$.
    $endgroup$
    – projectilemotion
    Dec 12 '18 at 22:56












  • $begingroup$
    @projectilemotion You forget the case $a = 0$.
    $endgroup$
    – Rebellos
    Dec 12 '18 at 23:10










  • $begingroup$
    I didn't forget it, in that case you could just take any $delta>0$.
    $endgroup$
    – projectilemotion
    Dec 12 '18 at 23:15
















0












$begingroup$


Here what i already done:



begin{align*} |az+b-(az_0+b)&+|az-az_0+b-b|\&=|a(z-z_0)|\&=|a||z-z_0|{le}|a|*{Delta}|z-z_0|
end{align*}



If ${Delta}=1$ and we know that $0{lt}|z-z_0|{lt}1$, then:



$||z|-|z_0||{le}|z-z_0|<1 \{to} |z|{lt}1+|z_o|$



Then:



$|z|<1+|z_0| {to} |z+z_0|{le}|z|+|z_0|{le}1+2|z_0|$



So:



${Delta}|a|*(1+2|z_0|)\ = {Delta}(|a|*(1+2|z_0|)={epsilon}\ {to} {Delta} = {epsilon}/(|a|(|1+2|z_0|)$



Then:



$|az+b - (az_0 + b)| {lt} |a|{Delta}|z-z_0|\ {le} {Delta}(|a|(1+2|(z_0)|\{lt}{epsilon}/(|a|(1+2|z_0|)*(|a|(1+2|z_0|) = {epsilon}$



That proves:



$lim_{zto z_0} (az + b) = az_0 + b$



Is that correct?






complex-analysis limits proof-verification complex-numbers







share|cite|improve this question











$endgroup$












  • $begingroup$
    For $aneq 0$, from the first three lines of your reasoning you could just select: $$0<|z-z_0|<delta=epsilon/|a| implies |az+b-(az_0+b)|<epsilon$$ And then consider the case where $a=0$.
    $endgroup$
    – projectilemotion
    Dec 12 '18 at 22:56












  • $begingroup$
    @projectilemotion You forget the case $a = 0$.
    $endgroup$
    – Rebellos
    Dec 12 '18 at 23:10










  • $begingroup$
    I didn't forget it, in that case you could just take any $delta>0$.
    $endgroup$
    – projectilemotion
    Dec 12 '18 at 23:15














0












0








0





$begingroup$


Here what i already done:



begin{align*} |az+b-(az_0+b)&+|az-az_0+b-b|\&=|a(z-z_0)|\&=|a||z-z_0|{le}|a|*{Delta}|z-z_0|
end{align*}



If ${Delta}=1$ and we know that $0{lt}|z-z_0|{lt}1$, then:



$||z|-|z_0||{le}|z-z_0|<1 \{to} |z|{lt}1+|z_o|$



Then:



$|z|<1+|z_0| {to} |z+z_0|{le}|z|+|z_0|{le}1+2|z_0|$



So:



${Delta}|a|*(1+2|z_0|)\ = {Delta}(|a|*(1+2|z_0|)={epsilon}\ {to} {Delta} = {epsilon}/(|a|(|1+2|z_0|)$



Then:



$|az+b - (az_0 + b)| {lt} |a|{Delta}|z-z_0|\ {le} {Delta}(|a|(1+2|(z_0)|\{lt}{epsilon}/(|a|(1+2|z_0|)*(|a|(1+2|z_0|) = {epsilon}$



That proves:



$lim_{zto z_0} (az + b) = az_0 + b$



Is that correct?






complex-analysis limits proof-verification complex-numbers







share|cite|improve this question











$endgroup$




Here what i already done:



begin{align*} |az+b-(az_0+b)&+|az-az_0+b-b|\&=|a(z-z_0)|\&=|a||z-z_0|{le}|a|*{Delta}|z-z_0|
end{align*}



If ${Delta}=1$ and we know that $0{lt}|z-z_0|{lt}1$, then:



$||z|-|z_0||{le}|z-z_0|<1 \{to} |z|{lt}1+|z_o|$



Then:



$|z|<1+|z_0| {to} |z+z_0|{le}|z|+|z_0|{le}1+2|z_0|$



So:



${Delta}|a|*(1+2|z_0|)\ = {Delta}(|a|*(1+2|z_0|)={epsilon}\ {to} {Delta} = {epsilon}/(|a|(|1+2|z_0|)$



Then:



$|az+b - (az_0 + b)| {lt} |a|{Delta}|z-z_0|\ {le} {Delta}(|a|(1+2|(z_0)|\{lt}{epsilon}/(|a|(1+2|z_0|)*(|a|(1+2|z_0|) = {epsilon}$



That proves:



$lim_{zto z_0} (az + b) = az_0 + b$



Is that correct?






complex-analysis limits proof-verification complex-numbers




complex-analysis limits proof-verification complex-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 12 '18 at 22:48









platty

3,350320




3,350320










asked Dec 12 '18 at 22:44









RafaelRafael

12




12












  • $begingroup$
    For $aneq 0$, from the first three lines of your reasoning you could just select: $$0<|z-z_0|<delta=epsilon/|a| implies |az+b-(az_0+b)|<epsilon$$ And then consider the case where $a=0$.
    $endgroup$
    – projectilemotion
    Dec 12 '18 at 22:56












  • $begingroup$
    @projectilemotion You forget the case $a = 0$.
    $endgroup$
    – Rebellos
    Dec 12 '18 at 23:10










  • $begingroup$
    I didn't forget it, in that case you could just take any $delta>0$.
    $endgroup$
    – projectilemotion
    Dec 12 '18 at 23:15


















  • $begingroup$
    For $aneq 0$, from the first three lines of your reasoning you could just select: $$0<|z-z_0|<delta=epsilon/|a| implies |az+b-(az_0+b)|<epsilon$$ And then consider the case where $a=0$.
    $endgroup$
    – projectilemotion
    Dec 12 '18 at 22:56












  • $begingroup$
    @projectilemotion You forget the case $a = 0$.
    $endgroup$
    – Rebellos
    Dec 12 '18 at 23:10










  • $begingroup$
    I didn't forget it, in that case you could just take any $delta>0$.
    $endgroup$
    – projectilemotion
    Dec 12 '18 at 23:15
















$begingroup$
For $aneq 0$, from the first three lines of your reasoning you could just select: $$0<|z-z_0|<delta=epsilon/|a| implies |az+b-(az_0+b)|<epsilon$$ And then consider the case where $a=0$.
$endgroup$
– projectilemotion
Dec 12 '18 at 22:56






$begingroup$
For $aneq 0$, from the first three lines of your reasoning you could just select: $$0<|z-z_0|<delta=epsilon/|a| implies |az+b-(az_0+b)|<epsilon$$ And then consider the case where $a=0$.
$endgroup$
– projectilemotion
Dec 12 '18 at 22:56














$begingroup$
@projectilemotion You forget the case $a = 0$.
$endgroup$
– Rebellos
Dec 12 '18 at 23:10




$begingroup$
@projectilemotion You forget the case $a = 0$.
$endgroup$
– Rebellos
Dec 12 '18 at 23:10












$begingroup$
I didn't forget it, in that case you could just take any $delta>0$.
$endgroup$
– projectilemotion
Dec 12 '18 at 23:15




$begingroup$
I didn't forget it, in that case you could just take any $delta>0$.
$endgroup$
– projectilemotion
Dec 12 '18 at 23:15










1 Answer
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$begingroup$

Seems okay in general (but your approach is more complex than it should), just some fixes.



First of all, you need to consider the case $a=0 in mathbb C$. Then, if that's the case, you have nothing to prove, as it's straightforward. Then you can go on by considering the case $aneq 0 in mathbb C$.



But for a more fast and elegant approach, observe that



$$|az + b - (az_0 + b)| = |a||z-z_0| < varepsilon implies |z-z_0| < frac{varepsilon}{|a|+1}$$



considering $a=0$ and $aneq 0$ at the same time, thus you're finished.






share|cite|improve this answer









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    $begingroup$

    Seems okay in general (but your approach is more complex than it should), just some fixes.



    First of all, you need to consider the case $a=0 in mathbb C$. Then, if that's the case, you have nothing to prove, as it's straightforward. Then you can go on by considering the case $aneq 0 in mathbb C$.



    But for a more fast and elegant approach, observe that



    $$|az + b - (az_0 + b)| = |a||z-z_0| < varepsilon implies |z-z_0| < frac{varepsilon}{|a|+1}$$



    considering $a=0$ and $aneq 0$ at the same time, thus you're finished.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Seems okay in general (but your approach is more complex than it should), just some fixes.



      First of all, you need to consider the case $a=0 in mathbb C$. Then, if that's the case, you have nothing to prove, as it's straightforward. Then you can go on by considering the case $aneq 0 in mathbb C$.



      But for a more fast and elegant approach, observe that



      $$|az + b - (az_0 + b)| = |a||z-z_0| < varepsilon implies |z-z_0| < frac{varepsilon}{|a|+1}$$



      considering $a=0$ and $aneq 0$ at the same time, thus you're finished.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Seems okay in general (but your approach is more complex than it should), just some fixes.



        First of all, you need to consider the case $a=0 in mathbb C$. Then, if that's the case, you have nothing to prove, as it's straightforward. Then you can go on by considering the case $aneq 0 in mathbb C$.



        But for a more fast and elegant approach, observe that



        $$|az + b - (az_0 + b)| = |a||z-z_0| < varepsilon implies |z-z_0| < frac{varepsilon}{|a|+1}$$



        considering $a=0$ and $aneq 0$ at the same time, thus you're finished.






        share|cite|improve this answer









        $endgroup$



        Seems okay in general (but your approach is more complex than it should), just some fixes.



        First of all, you need to consider the case $a=0 in mathbb C$. Then, if that's the case, you have nothing to prove, as it's straightforward. Then you can go on by considering the case $aneq 0 in mathbb C$.



        But for a more fast and elegant approach, observe that



        $$|az + b - (az_0 + b)| = |a||z-z_0| < varepsilon implies |z-z_0| < frac{varepsilon}{|a|+1}$$



        considering $a=0$ and $aneq 0$ at the same time, thus you're finished.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 12 '18 at 22:59









        RebellosRebellos

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