Prove $ lim_z rightarrow z_o f(z) in mathbbC$ if $lim_z rightarrow z_o (z-z_0) f(z) = 0$

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Context: A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 9.1, 9.3 (asked about here)
Question
From (here): How do we have that $$lim_z rightarrow z_o f(z) in mathbbC textfor removable singularity z_0 tag1$$ under the characterisation that $f$ has a removable singularity if $$lim_z rightarrow z_o (z-z_0) f(z) = 0 tag2$$?
Update: Attempt is now moved to answer.
Some intuition:
$(2) Leftarrow (1)$
- If $L in mathbbC$, then $lim_z rightarrow z_o (z-z_0) f(z) = lim_z rightarrow z_o (z-z_0) lim_z rightarrow z_o f(z) = (0)(L) = 0$
$(2) Rightarrow (1)$
If $L = infty$, then $lim_z rightarrow z_o (z-z_0) f(z) = 0 cdot infty$, indeterminate.
If $L$ dne, then $lim_z rightarrow z_o (z-z_0) f(z)$ dne.
Upon reflection, this actually seems like a simple complex extension of a simple elementary real analysis proof.
real-analysis complex-analysis derivatives proof-verification residue-calculus
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up vote
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down vote
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Context: A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 9.1, 9.3 (asked about here)
Question
From (here): How do we have that $$lim_z rightarrow z_o f(z) in mathbbC textfor removable singularity z_0 tag1$$ under the characterisation that $f$ has a removable singularity if $$lim_z rightarrow z_o (z-z_0) f(z) = 0 tag2$$?
Update: Attempt is now moved to answer.
Some intuition:
$(2) Leftarrow (1)$
- If $L in mathbbC$, then $lim_z rightarrow z_o (z-z_0) f(z) = lim_z rightarrow z_o (z-z_0) lim_z rightarrow z_o f(z) = (0)(L) = 0$
$(2) Rightarrow (1)$
If $L = infty$, then $lim_z rightarrow z_o (z-z_0) f(z) = 0 cdot infty$, indeterminate.
If $L$ dne, then $lim_z rightarrow z_o (z-z_0) f(z)$ dne.
Upon reflection, this actually seems like a simple complex extension of a simple elementary real analysis proof.
real-analysis complex-analysis derivatives proof-verification residue-calculus
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Context: A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 9.1, 9.3 (asked about here)
Question
From (here): How do we have that $$lim_z rightarrow z_o f(z) in mathbbC textfor removable singularity z_0 tag1$$ under the characterisation that $f$ has a removable singularity if $$lim_z rightarrow z_o (z-z_0) f(z) = 0 tag2$$?
Update: Attempt is now moved to answer.
Some intuition:
$(2) Leftarrow (1)$
- If $L in mathbbC$, then $lim_z rightarrow z_o (z-z_0) f(z) = lim_z rightarrow z_o (z-z_0) lim_z rightarrow z_o f(z) = (0)(L) = 0$
$(2) Rightarrow (1)$
If $L = infty$, then $lim_z rightarrow z_o (z-z_0) f(z) = 0 cdot infty$, indeterminate.
If $L$ dne, then $lim_z rightarrow z_o (z-z_0) f(z)$ dne.
Upon reflection, this actually seems like a simple complex extension of a simple elementary real analysis proof.
real-analysis complex-analysis derivatives proof-verification residue-calculus
Context: A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 9.1, 9.3 (asked about here)
Question
From (here): How do we have that $$lim_z rightarrow z_o f(z) in mathbbC textfor removable singularity z_0 tag1$$ under the characterisation that $f$ has a removable singularity if $$lim_z rightarrow z_o (z-z_0) f(z) = 0 tag2$$?
Update: Attempt is now moved to answer.
Some intuition:
$(2) Leftarrow (1)$
- If $L in mathbbC$, then $lim_z rightarrow z_o (z-z_0) f(z) = lim_z rightarrow z_o (z-z_0) lim_z rightarrow z_o f(z) = (0)(L) = 0$
$(2) Rightarrow (1)$
If $L = infty$, then $lim_z rightarrow z_o (z-z_0) f(z) = 0 cdot infty$, indeterminate.
If $L$ dne, then $lim_z rightarrow z_o (z-z_0) f(z)$ dne.
Upon reflection, this actually seems like a simple complex extension of a simple elementary real analysis proof.
real-analysis complex-analysis derivatives proof-verification residue-calculus
edited Aug 15 at 8:58
asked Aug 8 at 7:00
BCLC
6,72322073
6,72322073
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1 Answer
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Pf $(2) iff (1)$:
Firstly, let $lim_z rightarrow z_o f(z) =: L$, a letter which could be a finite complex number, $infty$ or 'dne'.
Now for the forward implication:
$(2) Leftarrow (1)$:
Given: $forall varepsilon_1 > 0, exists delta_1 > 0: |f(z)-L|<varepsilon_1 textwhenever 0<|z-z_0|<delta_1$
TS: $forall varepsilon > 0, exists delta > 0: |(z-z_0)f(z)|<varepsilon textwhenever 0<|z-z_0|<delta$
Pf: Let $varepsilon > 0$. $$|(z-z_0)f(z)| = |(z-z_0)[f(z)-L+L]|$$
$$= |(z-z_0)| |f(z)-L+L| le |z-z_0| (|f(z)-L|+|L|)$$
$$stackrel(*)le |z-z_0| varepsilon_1+|L| < varepsilon textwhenever 0 < |z-z_0| < delta := minfracvarepsilonvarepsilon_1+,delta_1$$
(*) $textwhenever 0 < |z-z_0| < delta_1$
QED $(2) Leftarrow (1)$
Now for the backward implication:
$(2) Rightarrow (1)$
Given: $forall varepsilon > 0, exists delta > 0: |(z-z_0)f(z)|< varepsilon textwhenever 0<|z-z_0|<delta$
TS: $forall varepsilon_1 > 0, exists delta_1 > 0: |f(z)-L|< varepsilon_1 textwhenever 0<|z-z_0|<delta_1$
Pf: Let $varepsilon_1 > 0$. $$|f(z)-L| = |frac(z-z_0)[f(z)]z-z_0-L| le |frac(z-z_0)[f(z)]z-z_0| + |L|$$
$$stackrel(**)le |fracvarepsilonz-z_0| + |L| le varepsilon_1 textwhenever 0<|z-z_0|<delta_1 := mindelta,fracvarepsilonL$$
(**) $textwhenever 0 < |z-z_0| < delta$
QED $(2) Rightarrow (1)$
Now both the forward and backward implications have been proven.
QED $(2) iff (1)$
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Pf $(2) iff (1)$:
Firstly, let $lim_z rightarrow z_o f(z) =: L$, a letter which could be a finite complex number, $infty$ or 'dne'.
Now for the forward implication:
$(2) Leftarrow (1)$:
Given: $forall varepsilon_1 > 0, exists delta_1 > 0: |f(z)-L|<varepsilon_1 textwhenever 0<|z-z_0|<delta_1$
TS: $forall varepsilon > 0, exists delta > 0: |(z-z_0)f(z)|<varepsilon textwhenever 0<|z-z_0|<delta$
Pf: Let $varepsilon > 0$. $$|(z-z_0)f(z)| = |(z-z_0)[f(z)-L+L]|$$
$$= |(z-z_0)| |f(z)-L+L| le |z-z_0| (|f(z)-L|+|L|)$$
$$stackrel(*)le |z-z_0| varepsilon_1+|L| < varepsilon textwhenever 0 < |z-z_0| < delta := minfracvarepsilonvarepsilon_1+,delta_1$$
(*) $textwhenever 0 < |z-z_0| < delta_1$
QED $(2) Leftarrow (1)$
Now for the backward implication:
$(2) Rightarrow (1)$
Given: $forall varepsilon > 0, exists delta > 0: |(z-z_0)f(z)|< varepsilon textwhenever 0<|z-z_0|<delta$
TS: $forall varepsilon_1 > 0, exists delta_1 > 0: |f(z)-L|< varepsilon_1 textwhenever 0<|z-z_0|<delta_1$
Pf: Let $varepsilon_1 > 0$. $$|f(z)-L| = |frac(z-z_0)[f(z)]z-z_0-L| le |frac(z-z_0)[f(z)]z-z_0| + |L|$$
$$stackrel(**)le |fracvarepsilonz-z_0| + |L| le varepsilon_1 textwhenever 0<|z-z_0|<delta_1 := mindelta,fracvarepsilonL$$
(**) $textwhenever 0 < |z-z_0| < delta$
QED $(2) Rightarrow (1)$
Now both the forward and backward implications have been proven.
QED $(2) iff (1)$
add a comment |Â
up vote
1
down vote
accepted
Pf $(2) iff (1)$:
Firstly, let $lim_z rightarrow z_o f(z) =: L$, a letter which could be a finite complex number, $infty$ or 'dne'.
Now for the forward implication:
$(2) Leftarrow (1)$:
Given: $forall varepsilon_1 > 0, exists delta_1 > 0: |f(z)-L|<varepsilon_1 textwhenever 0<|z-z_0|<delta_1$
TS: $forall varepsilon > 0, exists delta > 0: |(z-z_0)f(z)|<varepsilon textwhenever 0<|z-z_0|<delta$
Pf: Let $varepsilon > 0$. $$|(z-z_0)f(z)| = |(z-z_0)[f(z)-L+L]|$$
$$= |(z-z_0)| |f(z)-L+L| le |z-z_0| (|f(z)-L|+|L|)$$
$$stackrel(*)le |z-z_0| varepsilon_1+|L| < varepsilon textwhenever 0 < |z-z_0| < delta := minfracvarepsilonvarepsilon_1+,delta_1$$
(*) $textwhenever 0 < |z-z_0| < delta_1$
QED $(2) Leftarrow (1)$
Now for the backward implication:
$(2) Rightarrow (1)$
Given: $forall varepsilon > 0, exists delta > 0: |(z-z_0)f(z)|< varepsilon textwhenever 0<|z-z_0|<delta$
TS: $forall varepsilon_1 > 0, exists delta_1 > 0: |f(z)-L|< varepsilon_1 textwhenever 0<|z-z_0|<delta_1$
Pf: Let $varepsilon_1 > 0$. $$|f(z)-L| = |frac(z-z_0)[f(z)]z-z_0-L| le |frac(z-z_0)[f(z)]z-z_0| + |L|$$
$$stackrel(**)le |fracvarepsilonz-z_0| + |L| le varepsilon_1 textwhenever 0<|z-z_0|<delta_1 := mindelta,fracvarepsilonL$$
(**) $textwhenever 0 < |z-z_0| < delta$
QED $(2) Rightarrow (1)$
Now both the forward and backward implications have been proven.
QED $(2) iff (1)$
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Pf $(2) iff (1)$:
Firstly, let $lim_z rightarrow z_o f(z) =: L$, a letter which could be a finite complex number, $infty$ or 'dne'.
Now for the forward implication:
$(2) Leftarrow (1)$:
Given: $forall varepsilon_1 > 0, exists delta_1 > 0: |f(z)-L|<varepsilon_1 textwhenever 0<|z-z_0|<delta_1$
TS: $forall varepsilon > 0, exists delta > 0: |(z-z_0)f(z)|<varepsilon textwhenever 0<|z-z_0|<delta$
Pf: Let $varepsilon > 0$. $$|(z-z_0)f(z)| = |(z-z_0)[f(z)-L+L]|$$
$$= |(z-z_0)| |f(z)-L+L| le |z-z_0| (|f(z)-L|+|L|)$$
$$stackrel(*)le |z-z_0| varepsilon_1+|L| < varepsilon textwhenever 0 < |z-z_0| < delta := minfracvarepsilonvarepsilon_1+,delta_1$$
(*) $textwhenever 0 < |z-z_0| < delta_1$
QED $(2) Leftarrow (1)$
Now for the backward implication:
$(2) Rightarrow (1)$
Given: $forall varepsilon > 0, exists delta > 0: |(z-z_0)f(z)|< varepsilon textwhenever 0<|z-z_0|<delta$
TS: $forall varepsilon_1 > 0, exists delta_1 > 0: |f(z)-L|< varepsilon_1 textwhenever 0<|z-z_0|<delta_1$
Pf: Let $varepsilon_1 > 0$. $$|f(z)-L| = |frac(z-z_0)[f(z)]z-z_0-L| le |frac(z-z_0)[f(z)]z-z_0| + |L|$$
$$stackrel(**)le |fracvarepsilonz-z_0| + |L| le varepsilon_1 textwhenever 0<|z-z_0|<delta_1 := mindelta,fracvarepsilonL$$
(**) $textwhenever 0 < |z-z_0| < delta$
QED $(2) Rightarrow (1)$
Now both the forward and backward implications have been proven.
QED $(2) iff (1)$
Pf $(2) iff (1)$:
Firstly, let $lim_z rightarrow z_o f(z) =: L$, a letter which could be a finite complex number, $infty$ or 'dne'.
Now for the forward implication:
$(2) Leftarrow (1)$:
Given: $forall varepsilon_1 > 0, exists delta_1 > 0: |f(z)-L|<varepsilon_1 textwhenever 0<|z-z_0|<delta_1$
TS: $forall varepsilon > 0, exists delta > 0: |(z-z_0)f(z)|<varepsilon textwhenever 0<|z-z_0|<delta$
Pf: Let $varepsilon > 0$. $$|(z-z_0)f(z)| = |(z-z_0)[f(z)-L+L]|$$
$$= |(z-z_0)| |f(z)-L+L| le |z-z_0| (|f(z)-L|+|L|)$$
$$stackrel(*)le |z-z_0| varepsilon_1+|L| < varepsilon textwhenever 0 < |z-z_0| < delta := minfracvarepsilonvarepsilon_1+,delta_1$$
(*) $textwhenever 0 < |z-z_0| < delta_1$
QED $(2) Leftarrow (1)$
Now for the backward implication:
$(2) Rightarrow (1)$
Given: $forall varepsilon > 0, exists delta > 0: |(z-z_0)f(z)|< varepsilon textwhenever 0<|z-z_0|<delta$
TS: $forall varepsilon_1 > 0, exists delta_1 > 0: |f(z)-L|< varepsilon_1 textwhenever 0<|z-z_0|<delta_1$
Pf: Let $varepsilon_1 > 0$. $$|f(z)-L| = |frac(z-z_0)[f(z)]z-z_0-L| le |frac(z-z_0)[f(z)]z-z_0| + |L|$$
$$stackrel(**)le |fracvarepsilonz-z_0| + |L| le varepsilon_1 textwhenever 0<|z-z_0|<delta_1 := mindelta,fracvarepsilonL$$
(**) $textwhenever 0 < |z-z_0| < delta$
QED $(2) Rightarrow (1)$
Now both the forward and backward implications have been proven.
QED $(2) iff (1)$
answered Aug 15 at 11:29
BCLC
6,72322073
6,72322073
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