Prove the limit below by using epsilon-delta definition

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Prove $$lim_zrightarrow4+3ioverlinez^2 =(4-3i)^2$$ by using epsilon-delta definition.



My working:



Assume $|z-(4-3i)|<1 implies |z|<6$



beginalign
|overlinez^2-(4-3i)^2| &=|[overlinez-(4-3i)][overlinez+(4-3i)]|\
&=|overlinez-(4-3i)||overlinez+(4-3i)|\
&=|overlinez-(4+3i)+6i||overlinez-(4+3i)+8|\
&=|overlinez-(4+3i)|left[left|1+frac6ioverlinez-(4+3i)right|left|1+frac8overlinez-(4+3i)right|right]\
&le delta left[left(1+frac6ioverlinez-(4+3i)right)left(1+frac8overlinez-(4+3i)right)right]\
&< 63delta
endalign



Is this correct?




complex-analysis




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  • 1




    You should show some work first, and let us know where you got stuck. This is not a "do my homework" site.
    – Dzoooks
    Jul 24 at 15:48














up vote
1
down vote

favorite
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Prove $$lim_zrightarrow4+3ioverlinez^2 =(4-3i)^2$$ by using epsilon-delta definition.



My working:



Assume $|z-(4-3i)|<1 implies |z|<6$



beginalign
|overlinez^2-(4-3i)^2| &=|[overlinez-(4-3i)][overlinez+(4-3i)]|\
&=|overlinez-(4-3i)||overlinez+(4-3i)|\
&=|overlinez-(4+3i)+6i||overlinez-(4+3i)+8|\
&=|overlinez-(4+3i)|left[left|1+frac6ioverlinez-(4+3i)right|left|1+frac8overlinez-(4+3i)right|right]\
&le delta left[left(1+frac6ioverlinez-(4+3i)right)left(1+frac8overlinez-(4+3i)right)right]\
&< 63delta
endalign



Is this correct?




complex-analysis




share|cite|improve this question

















  • 1




    You should show some work first, and let us know where you got stuck. This is not a "do my homework" site.
    – Dzoooks
    Jul 24 at 15:48












up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1





Prove $$lim_zrightarrow4+3ioverlinez^2 =(4-3i)^2$$ by using epsilon-delta definition.



My working:



Assume $|z-(4-3i)|<1 implies |z|<6$



beginalign
|overlinez^2-(4-3i)^2| &=|[overlinez-(4-3i)][overlinez+(4-3i)]|\
&=|overlinez-(4-3i)||overlinez+(4-3i)|\
&=|overlinez-(4+3i)+6i||overlinez-(4+3i)+8|\
&=|overlinez-(4+3i)|left[left|1+frac6ioverlinez-(4+3i)right|left|1+frac8overlinez-(4+3i)right|right]\
&le delta left[left(1+frac6ioverlinez-(4+3i)right)left(1+frac8overlinez-(4+3i)right)right]\
&< 63delta
endalign



Is this correct?




complex-analysis




share|cite|improve this question













Prove $$lim_zrightarrow4+3ioverlinez^2 =(4-3i)^2$$ by using epsilon-delta definition.



My working:



Assume $|z-(4-3i)|<1 implies |z|<6$



beginalign
|overlinez^2-(4-3i)^2| &=|[overlinez-(4-3i)][overlinez+(4-3i)]|\
&=|overlinez-(4-3i)||overlinez+(4-3i)|\
&=|overlinez-(4+3i)+6i||overlinez-(4+3i)+8|\
&=|overlinez-(4+3i)|left[left|1+frac6ioverlinez-(4+3i)right|left|1+frac8overlinez-(4+3i)right|right]\
&le delta left[left(1+frac6ioverlinez-(4+3i)right)left(1+frac8overlinez-(4+3i)right)right]\
&< 63delta
endalign



Is this correct?





complex-analysis





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edited Jul 24 at 17:11









mechanodroid

22.2k52041




22.2k52041









asked Jul 24 at 15:26









Steven Lau Zhou Sheng

483




483







  • 1




    You should show some work first, and let us know where you got stuck. This is not a "do my homework" site.
    – Dzoooks
    Jul 24 at 15:48












  • 1




    You should show some work first, and let us know where you got stuck. This is not a "do my homework" site.
    – Dzoooks
    Jul 24 at 15:48







1




1




You should show some work first, and let us know where you got stuck. This is not a "do my homework" site.
– Dzoooks
Jul 24 at 15:48




You should show some work first, and let us know where you got stuck. This is not a "do my homework" site.
– Dzoooks
Jul 24 at 15:48










1 Answer
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Careful, you can assume $|z - (4+3i)| < delta$, not $|overlinez - (4+3i)| < delta$.



If we pick $0 < delta < -5+sqrt25+varepsilon$ we have



beginalign
left|overlinez^2 - (4-3i)^2right| &= |overlinez - (4-3i)|cdot |overlinez + (4-3i)|\
&= left|overlinez - (4+3i)right|cdot left|overlinez + (4+3i)right|\
&= |z - (4+3i)|cdot |z + (4+3i)|\
&= |z - (4+3i)|cdot |z - (4+3i) + (8+6i)|\
&le |z - (4+3i)|cdot big(|z - (4+3i)| + |8+6i|big)\
&< delta(delta + 10)\
&< varepsilon
endalign






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    Careful, you can assume $|z - (4+3i)| < delta$, not $|overlinez - (4+3i)| < delta$.



    If we pick $0 < delta < -5+sqrt25+varepsilon$ we have



    beginalign
    left|overlinez^2 - (4-3i)^2right| &= |overlinez - (4-3i)|cdot |overlinez + (4-3i)|\
    &= left|overlinez - (4+3i)right|cdot left|overlinez + (4+3i)right|\
    &= |z - (4+3i)|cdot |z + (4+3i)|\
    &= |z - (4+3i)|cdot |z - (4+3i) + (8+6i)|\
    &le |z - (4+3i)|cdot big(|z - (4+3i)| + |8+6i|big)\
    &< delta(delta + 10)\
    &< varepsilon
    endalign






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      up vote
      0
      down vote













      Careful, you can assume $|z - (4+3i)| < delta$, not $|overlinez - (4+3i)| < delta$.



      If we pick $0 < delta < -5+sqrt25+varepsilon$ we have



      beginalign
      left|overlinez^2 - (4-3i)^2right| &= |overlinez - (4-3i)|cdot |overlinez + (4-3i)|\
      &= left|overlinez - (4+3i)right|cdot left|overlinez + (4+3i)right|\
      &= |z - (4+3i)|cdot |z + (4+3i)|\
      &= |z - (4+3i)|cdot |z - (4+3i) + (8+6i)|\
      &le |z - (4+3i)|cdot big(|z - (4+3i)| + |8+6i|big)\
      &< delta(delta + 10)\
      &< varepsilon
      endalign






      share|cite|improve this answer























        up vote
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        up vote
        0
        down vote









        Careful, you can assume $|z - (4+3i)| < delta$, not $|overlinez - (4+3i)| < delta$.



        If we pick $0 < delta < -5+sqrt25+varepsilon$ we have



        beginalign
        left|overlinez^2 - (4-3i)^2right| &= |overlinez - (4-3i)|cdot |overlinez + (4-3i)|\
        &= left|overlinez - (4+3i)right|cdot left|overlinez + (4+3i)right|\
        &= |z - (4+3i)|cdot |z + (4+3i)|\
        &= |z - (4+3i)|cdot |z - (4+3i) + (8+6i)|\
        &le |z - (4+3i)|cdot big(|z - (4+3i)| + |8+6i|big)\
        &< delta(delta + 10)\
        &< varepsilon
        endalign






        share|cite|improve this answer













        Careful, you can assume $|z - (4+3i)| < delta$, not $|overlinez - (4+3i)| < delta$.



        If we pick $0 < delta < -5+sqrt25+varepsilon$ we have



        beginalign
        left|overlinez^2 - (4-3i)^2right| &= |overlinez - (4-3i)|cdot |overlinez + (4-3i)|\
        &= left|overlinez - (4+3i)right|cdot left|overlinez + (4+3i)right|\
        &= |z - (4+3i)|cdot |z + (4+3i)|\
        &= |z - (4+3i)|cdot |z - (4+3i) + (8+6i)|\
        &le |z - (4+3i)|cdot big(|z - (4+3i)| + |8+6i|big)\
        &< delta(delta + 10)\
        &< varepsilon
        endalign







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Jul 24 at 17:08









        mechanodroid

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