$f(z) = 2z + 3overline z$, 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 $|f(z) â f(a)|$ and $|z â a|$, 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 $|f(z) â f(a)|$ and $|z â a|$, 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
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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 $|f(z) â f(a)|$ and $|z â a|$, 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
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 $|f(z) â f(a)|$ and $|z â a|$, 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
edited Aug 23 at 5:34
an4s
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2,0632417
asked Aug 23 at 4:39
Sophie Wines
213
213
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1 Answer
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Hint:
$$|2(z-a)+3(barz-bara)|leq|2(z-a)|+3|barz-bara|leq2|z-a|+3|z-a|$$
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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
Hint:
$$|2(z-a)+3(barz-bara)|leq|2(z-a)|+3|barz-bara|leq2|z-a|+3|z-a|$$
add a comment |Â
up vote
1
down vote
Hint:
$$|2(z-a)+3(barz-bara)|leq|2(z-a)|+3|barz-bara|leq2|z-a|+3|z-a|$$
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Hint:
$$|2(z-a)+3(barz-bara)|leq|2(z-a)|+3|barz-bara|leq2|z-a|+3|z-a|$$
Hint:
$$|2(z-a)+3(barz-bara)|leq|2(z-a)|+3|barz-bara|leq2|z-a|+3|z-a|$$
answered Aug 23 at 4:47
Nosrati
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