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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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
edited Dec 12 '18 at 22:48
platty
3,350320
asked Dec 12 '18 at 22:44
RafaelRafael
12
$endgroup$
add a comment |
$begingroup$
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
edited Dec 12 '18 at 22:48
platty
3,350320
asked Dec 12 '18 at 22:44
RafaelRafael
12
$endgroup$
$begingroup$
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
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@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
add a comment |
$begingroup$
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
edited Dec 12 '18 at 22:48
platty
3,350320
asked Dec 12 '18 at 22:44
RafaelRafael
12
$endgroup$
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
edited Dec 12 '18 at 22:48
platty
3,350320
asked Dec 12 '18 at 22:44
RafaelRafael
12
edited Dec 12 '18 at 22:48
platty
3,350320
asked Dec 12 '18 at 22:44
RafaelRafael
12
edited Dec 12 '18 at 22:48
platty
3,350320
edited Dec 12 '18 at 22:48
platty
3,350320
edited Dec 12 '18 at 22:48
platty
3,350320
3,350320
asked Dec 12 '18 at 22:44
RafaelRafael
12
asked Dec 12 '18 at 22:44
RafaelRafael
12
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
add a comment |
$begingroup$
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
add a comment |
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.
answered Dec 12 '18 at 22:59
RebellosRebellos
15.7k31250
$endgroup$
add a comment |
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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.
answered Dec 12 '18 at 22:59
RebellosRebellos
15.7k31250
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 12 '18 at 22:59
RebellosRebellos
15.7k31250
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 12 '18 at 22:59
RebellosRebellos
15.7k31250
$endgroup$
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.
answered Dec 12 '18 at 22:59
RebellosRebellos
15.7k31250
answered Dec 12 '18 at 22:59
RebellosRebellos
15.7k31250
answered Dec 12 '18 at 22:59
RebellosRebellos
15.7k31250
answered Dec 12 '18 at 22:59
RebellosRebellos
15.7k31250
15.7k31250
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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