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Image of a unit disk under a complex function
Clash Royale CLAN TAG#URR8PPP
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My question is the following:
Prove that the image of the disk $D=z$ under the function $f(z)=cos z$ is contained in the disk $D'=w in mathbbC : $. (Hint: $e^leq|e^z|leq e^$, for all $z in mathbbC$)
By the definition, we have
$$
|w|=|cos z|=frac12left|e^zi+e^-ziright|leqfrac12(|e^zi|+|e^-zi|).
$$
Now, using the hint,
beginequation
|w|leqfrac12(e^+e^).
endequation
And then, as $zin D$, we have $|zi|=|z||i|=|z|leq1$. Therefore
$$
|w|leqfrac12(e^+e^)leqfrac12(e+e^-1)=frace^2+12e.
$$
But I am not sure that I may use the hint to conclude that $|e^-zi|leq e^$. Can anyone have another solution to give me or even how to prove this inequality?
complex-analysis
asked Jul 25 at 19:13
Marcos Paulo
447
add a comment |Â
up vote
2
down vote
favorite
My question is the following:
Prove that the image of the disk $D=z$ under the function $f(z)=cos z$ is contained in the disk $D'=w in mathbbC : $. (Hint: $e^leq|e^z|leq e^$, for all $z in mathbbC$)
By the definition, we have
$$
|w|=|cos z|=frac12left|e^zi+e^-ziright|leqfrac12(|e^zi|+|e^-zi|).
$$
Now, using the hint,
beginequation
|w|leqfrac12(e^+e^).
endequation
And then, as $zin D$, we have $|zi|=|z||i|=|z|leq1$. Therefore
$$
|w|leqfrac12(e^+e^)leqfrac12(e+e^-1)=frace^2+12e.
$$
But I am not sure that I may use the hint to conclude that $|e^-zi|leq e^$. Can anyone have another solution to give me or even how to prove this inequality?
complex-analysis
asked Jul 25 at 19:13
Marcos Paulo
447
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
My question is the following:
Prove that the image of the disk $D=z$ under the function $f(z)=cos z$ is contained in the disk $D'=w in mathbbC : $. (Hint: $e^leq|e^z|leq e^$, for all $z in mathbbC$)
By the definition, we have
$$
|w|=|cos z|=frac12left|e^zi+e^-ziright|leqfrac12(|e^zi|+|e^-zi|).
$$
Now, using the hint,
beginequation
|w|leqfrac12(e^+e^).
endequation
And then, as $zin D$, we have $|zi|=|z||i|=|z|leq1$. Therefore
$$
|w|leqfrac12(e^+e^)leqfrac12(e+e^-1)=frace^2+12e.
$$
But I am not sure that I may use the hint to conclude that $|e^-zi|leq e^$. Can anyone have another solution to give me or even how to prove this inequality?
complex-analysis
asked Jul 25 at 19:13
Marcos Paulo
447
My question is the following:
Prove that the image of the disk $D=z$ under the function $f(z)=cos z$ is contained in the disk $D'=w in mathbbC : $. (Hint: $e^leq|e^z|leq e^$, for all $z in mathbbC$)
By the definition, we have
$$
|w|=|cos z|=frac12left|e^zi+e^-ziright|leqfrac12(|e^zi|+|e^-zi|).
$$
Now, using the hint,
beginequation
|w|leqfrac12(e^+e^).
endequation
And then, as $zin D$, we have $|zi|=|z||i|=|z|leq1$. Therefore
$$
|w|leqfrac12(e^+e^)leqfrac12(e+e^-1)=frace^2+12e.
$$
But I am not sure that I may use the hint to conclude that $|e^-zi|leq e^$. Can anyone have another solution to give me or even how to prove this inequality?
complex-analysis
asked Jul 25 at 19:13
Marcos Paulo
447
asked Jul 25 at 19:13
Marcos Paulo
447
asked Jul 25 at 19:13
Marcos Paulo
447
asked Jul 25 at 19:13
Marcos Paulo
447
447
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
I'm not sure the hint is useful. Write $z=x+iy.$ Then
$$tag 1 |cos z| = |e^iz+ e^-iz|/2 le (|e^iz|+ |e^-iz|)/2 = (e^-y+e^y)/2.$$
Now $e^-y+e^y$ is even and increases on $[0,1].$ Since $|z|le 1 $ implies $yle 1,$ the right side of $(1)$ is bounded above by $(e^-1+e^1)/2 = (1+e^2)/(2e).$
answered Jul 25 at 20:52
zhw.
65.6k42870
I made an error in my first answer. It should be fine now.
– zhw.
Jul 25 at 21:37
Could you please give me more details about the final part of your argument? I can't understand why the right side of (1) is bounded $(e^-1+e^1)/2$. I understand that $yleq1$ implies $e^yleq e^1$, but how can this imply $e^-y leq e^-1$?
– Marcos Paulo
Jul 30 at 13:05
add a comment |Â
up vote
0
down vote
I was about to ask if you could have done something wrong when I saw you wrote $frac12(e^-y+e^y)$, but I remembered that $yleq1$ implies that $2yleq2$ and $e^2y leq e^2$, and therefore
$$|w|leq frac12(e^-y+e^y)=frace^-y2(1+e^2y)leqfrac1+e^22e$$
I got convinced
answered Jul 26 at 3:46
Marcos Paulo
447
How did you get the last inequality?
– zhw.
Jul 27 at 3:59
$e^-y=frac1e^y<frac1e$
– Marcos Paulo
Jul 27 at 14:58
What if $y=-1$ or $y=0?$
– zhw.
Jul 27 at 15:36
You're right :/
– Marcos Paulo
Jul 27 at 15:46
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
I'm not sure the hint is useful. Write $z=x+iy.$ Then
$$tag 1 |cos z| = |e^iz+ e^-iz|/2 le (|e^iz|+ |e^-iz|)/2 = (e^-y+e^y)/2.$$
Now $e^-y+e^y$ is even and increases on $[0,1].$ Since $|z|le 1 $ implies $yle 1,$ the right side of $(1)$ is bounded above by $(e^-1+e^1)/2 = (1+e^2)/(2e).$
answered Jul 25 at 20:52
zhw.
65.6k42870
I made an error in my first answer. It should be fine now.
– zhw.
Jul 25 at 21:37
Could you please give me more details about the final part of your argument? I can't understand why the right side of (1) is bounded $(e^-1+e^1)/2$. I understand that $yleq1$ implies $e^yleq e^1$, but how can this imply $e^-y leq e^-1$?
– Marcos Paulo
Jul 30 at 13:05
add a comment |Â
up vote
2
down vote
accepted
I'm not sure the hint is useful. Write $z=x+iy.$ Then
$$tag 1 |cos z| = |e^iz+ e^-iz|/2 le (|e^iz|+ |e^-iz|)/2 = (e^-y+e^y)/2.$$
Now $e^-y+e^y$ is even and increases on $[0,1].$ Since $|z|le 1 $ implies $yle 1,$ the right side of $(1)$ is bounded above by $(e^-1+e^1)/2 = (1+e^2)/(2e).$
answered Jul 25 at 20:52
zhw.
65.6k42870
I made an error in my first answer. It should be fine now.
– zhw.
Jul 25 at 21:37
Could you please give me more details about the final part of your argument? I can't understand why the right side of (1) is bounded $(e^-1+e^1)/2$. I understand that $yleq1$ implies $e^yleq e^1$, but how can this imply $e^-y leq e^-1$?
– Marcos Paulo
Jul 30 at 13:05
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
I'm not sure the hint is useful. Write $z=x+iy.$ Then
$$tag 1 |cos z| = |e^iz+ e^-iz|/2 le (|e^iz|+ |e^-iz|)/2 = (e^-y+e^y)/2.$$
Now $e^-y+e^y$ is even and increases on $[0,1].$ Since $|z|le 1 $ implies $yle 1,$ the right side of $(1)$ is bounded above by $(e^-1+e^1)/2 = (1+e^2)/(2e).$
answered Jul 25 at 20:52
zhw.
65.6k42870
I'm not sure the hint is useful. Write $z=x+iy.$ Then
$$tag 1 |cos z| = |e^iz+ e^-iz|/2 le (|e^iz|+ |e^-iz|)/2 = (e^-y+e^y)/2.$$
Now $e^-y+e^y$ is even and increases on $[0,1].$ Since $|z|le 1 $ implies $yle 1,$ the right side of $(1)$ is bounded above by $(e^-1+e^1)/2 = (1+e^2)/(2e).$
answered Jul 25 at 20:52
zhw.
65.6k42870
edited Jul 25 at 21:35
answered Jul 25 at 20:52
zhw.
65.6k42870
answered Jul 25 at 20:52
zhw.
65.6k42870
answered Jul 25 at 20:52
zhw.
65.6k42870
65.6k42870
I made an error in my first answer. It should be fine now.
– zhw.
Jul 25 at 21:37
Could you please give me more details about the final part of your argument? I can't understand why the right side of (1) is bounded $(e^-1+e^1)/2$. I understand that $yleq1$ implies $e^yleq e^1$, but how can this imply $e^-y leq e^-1$?
– Marcos Paulo
Jul 30 at 13:05
add a comment |Â
I made an error in my first answer. It should be fine now.
– zhw.
Jul 25 at 21:37
Could you please give me more details about the final part of your argument? I can't understand why the right side of (1) is bounded $(e^-1+e^1)/2$. I understand that $yleq1$ implies $e^yleq e^1$, but how can this imply $e^-y leq e^-1$?
– Marcos Paulo
Jul 30 at 13:05
I made an error in my first answer. It should be fine now.
– zhw.
Jul 25 at 21:37
I made an error in my first answer. It should be fine now.
– zhw.
Jul 25 at 21:37
Could you please give me more details about the final part of your argument? I can't understand why the right side of (1) is bounded $(e^-1+e^1)/2$. I understand that $yleq1$ implies $e^yleq e^1$, but how can this imply $e^-y leq e^-1$?
– Marcos Paulo
Jul 30 at 13:05
Could you please give me more details about the final part of your argument? I can't understand why the right side of (1) is bounded $(e^-1+e^1)/2$. I understand that $yleq1$ implies $e^yleq e^1$, but how can this imply $e^-y leq e^-1$?
– Marcos Paulo
Jul 30 at 13:05
add a comment |Â
up vote
0
down vote
I was about to ask if you could have done something wrong when I saw you wrote $frac12(e^-y+e^y)$, but I remembered that $yleq1$ implies that $2yleq2$ and $e^2y leq e^2$, and therefore
$$|w|leq frac12(e^-y+e^y)=frace^-y2(1+e^2y)leqfrac1+e^22e$$
I got convinced
answered Jul 26 at 3:46
Marcos Paulo
447
How did you get the last inequality?
– zhw.
Jul 27 at 3:59
$e^-y=frac1e^y<frac1e$
– Marcos Paulo
Jul 27 at 14:58
What if $y=-1$ or $y=0?$
– zhw.
Jul 27 at 15:36
You're right :/
– Marcos Paulo
Jul 27 at 15:46
add a comment |Â
up vote
0
down vote
I was about to ask if you could have done something wrong when I saw you wrote $frac12(e^-y+e^y)$, but I remembered that $yleq1$ implies that $2yleq2$ and $e^2y leq e^2$, and therefore
$$|w|leq frac12(e^-y+e^y)=frace^-y2(1+e^2y)leqfrac1+e^22e$$
I got convinced
answered Jul 26 at 3:46
Marcos Paulo
447
How did you get the last inequality?
– zhw.
Jul 27 at 3:59
$e^-y=frac1e^y<frac1e$
– Marcos Paulo
Jul 27 at 14:58
What if $y=-1$ or $y=0?$
– zhw.
Jul 27 at 15:36
You're right :/
– Marcos Paulo
Jul 27 at 15:46
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I was about to ask if you could have done something wrong when I saw you wrote $frac12(e^-y+e^y)$, but I remembered that $yleq1$ implies that $2yleq2$ and $e^2y leq e^2$, and therefore
$$|w|leq frac12(e^-y+e^y)=frace^-y2(1+e^2y)leqfrac1+e^22e$$
I got convinced
answered Jul 26 at 3:46
Marcos Paulo
447
I was about to ask if you could have done something wrong when I saw you wrote $frac12(e^-y+e^y)$, but I remembered that $yleq1$ implies that $2yleq2$ and $e^2y leq e^2$, and therefore
$$|w|leq frac12(e^-y+e^y)=frace^-y2(1+e^2y)leqfrac1+e^22e$$
I got convinced
answered Jul 26 at 3:46
Marcos Paulo
447
answered Jul 26 at 3:46
Marcos Paulo
447
answered Jul 26 at 3:46
Marcos Paulo
447
answered Jul 26 at 3:46
Marcos Paulo
447
447
How did you get the last inequality?
– zhw.
Jul 27 at 3:59
$e^-y=frac1e^y<frac1e$
– Marcos Paulo
Jul 27 at 14:58
What if $y=-1$ or $y=0?$
– zhw.
Jul 27 at 15:36
You're right :/
– Marcos Paulo
Jul 27 at 15:46
add a comment |Â
How did you get the last inequality?
– zhw.
Jul 27 at 3:59
$e^-y=frac1e^y<frac1e$
– Marcos Paulo
Jul 27 at 14:58
What if $y=-1$ or $y=0?$
– zhw.
Jul 27 at 15:36
You're right :/
– Marcos Paulo
Jul 27 at 15:46
How did you get the last inequality?
– zhw.
Jul 27 at 3:59
How did you get the last inequality?
– zhw.
Jul 27 at 3:59
$e^-y=frac1e^y<frac1e$
– Marcos Paulo
Jul 27 at 14:58
$e^-y=frac1e^y<frac1e$
– Marcos Paulo
Jul 27 at 14:58
What if $y=-1$ or $y=0?$
– zhw.
Jul 27 at 15:36
What if $y=-1$ or $y=0?$
– zhw.
Jul 27 at 15:36
You're right :/
– Marcos Paulo
Jul 27 at 15:46
You're right :/
– Marcos Paulo
Jul 27 at 15:46
add a comment |Â
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