Find the image of the region ${zinmathbb{C}: 0leq Re(z) leq 1/2,; -pi leq Im(z) leq pi}$ under the mapping...
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I have no clue if I am doing this correctly, just basing it off what my teacher did for a similar example.
I look at image of $y=a$
$u(x,y) = e^x cos{a}$
$v(x,y) = e^x sin{a}$
$$y=-pi , f(z) = u+iv = -e^x <0$$
$$y=pi , f(z) = u+iv = -e^x <0$$
So this is below the x-axis, I am confused on how to apply this to the Re(z) boundary if this method is even correct.
complex-analysis
edited Nov 16 at 12:58
amWhy
191k27223438
asked Nov 16 at 11:36
jaiidong
104
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up vote
0
down vote
favorite
I have no clue if I am doing this correctly, just basing it off what my teacher did for a similar example.
I look at image of $y=a$
$u(x,y) = e^x cos{a}$
$v(x,y) = e^x sin{a}$
$$y=-pi , f(z) = u+iv = -e^x <0$$
$$y=pi , f(z) = u+iv = -e^x <0$$
So this is below the x-axis, I am confused on how to apply this to the Re(z) boundary if this method is even correct.
complex-analysis
edited Nov 16 at 12:58
amWhy
191k27223438
asked Nov 16 at 11:36
jaiidong
104
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have no clue if I am doing this correctly, just basing it off what my teacher did for a similar example.
I look at image of $y=a$
$u(x,y) = e^x cos{a}$
$v(x,y) = e^x sin{a}$
$$y=-pi , f(z) = u+iv = -e^x <0$$
$$y=pi , f(z) = u+iv = -e^x <0$$
So this is below the x-axis, I am confused on how to apply this to the Re(z) boundary if this method is even correct.
complex-analysis
edited Nov 16 at 12:58
amWhy
191k27223438
asked Nov 16 at 11:36
jaiidong
104
I have no clue if I am doing this correctly, just basing it off what my teacher did for a similar example.
I look at image of $y=a$
$u(x,y) = e^x cos{a}$
$v(x,y) = e^x sin{a}$
$$y=-pi , f(z) = u+iv = -e^x <0$$
$$y=pi , f(z) = u+iv = -e^x <0$$
So this is below the x-axis, I am confused on how to apply this to the Re(z) boundary if this method is even correct.
complex-analysis
complex-analysis
edited Nov 16 at 12:58
amWhy
191k27223438
asked Nov 16 at 11:36
jaiidong
104
edited Nov 16 at 12:58
amWhy
191k27223438
asked Nov 16 at 11:36
jaiidong
104
edited Nov 16 at 12:58
amWhy
191k27223438
edited Nov 16 at 12:58
amWhy
191k27223438
edited Nov 16 at 12:58
amWhy
191k27223438
191k27223438
asked Nov 16 at 11:36
jaiidong
104
asked Nov 16 at 11:36
jaiidong
104
asked Nov 16 at 11:36
jaiidong
104
104
add a comment |
add a comment |
1 Answer
1
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1
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Let $z = a+ib in mathbb{C}$.
Then $Re(z) = a$, $Im(z) = b$ and $0 leq a leq 1/2$, $-pi leq b leq pi$.
We know that $e^z = e^{a+ib} = e^a(cos(b) + isin(b))$.
So, if we fix $a = 0$, for example, we have $e^a = 1$ and we are left with $cos(b) + isin(b)$, where $-pi leq b leq pi$.
So, if this is the range of $b$, what figure you obtain?
Now note that for every other $a$ you fix, you will obtain the same figure, but with a small difference.
What's this difference?
answered Nov 16 at 12:35
Bias of Priene
286112
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Let $z = a+ib in mathbb{C}$.
Then $Re(z) = a$, $Im(z) = b$ and $0 leq a leq 1/2$, $-pi leq b leq pi$.
We know that $e^z = e^{a+ib} = e^a(cos(b) + isin(b))$.
So, if we fix $a = 0$, for example, we have $e^a = 1$ and we are left with $cos(b) + isin(b)$, where $-pi leq b leq pi$.
So, if this is the range of $b$, what figure you obtain?
Now note that for every other $a$ you fix, you will obtain the same figure, but with a small difference.
What's this difference?
answered Nov 16 at 12:35
Bias of Priene
286112
add a comment |
up vote
1
down vote
Let $z = a+ib in mathbb{C}$.
Then $Re(z) = a$, $Im(z) = b$ and $0 leq a leq 1/2$, $-pi leq b leq pi$.
We know that $e^z = e^{a+ib} = e^a(cos(b) + isin(b))$.
So, if we fix $a = 0$, for example, we have $e^a = 1$ and we are left with $cos(b) + isin(b)$, where $-pi leq b leq pi$.
So, if this is the range of $b$, what figure you obtain?
Now note that for every other $a$ you fix, you will obtain the same figure, but with a small difference.
What's this difference?
answered Nov 16 at 12:35
Bias of Priene
286112
add a comment |
up vote
1
down vote
up vote
1
down vote
Let $z = a+ib in mathbb{C}$.
Then $Re(z) = a$, $Im(z) = b$ and $0 leq a leq 1/2$, $-pi leq b leq pi$.
We know that $e^z = e^{a+ib} = e^a(cos(b) + isin(b))$.
So, if we fix $a = 0$, for example, we have $e^a = 1$ and we are left with $cos(b) + isin(b)$, where $-pi leq b leq pi$.
So, if this is the range of $b$, what figure you obtain?
Now note that for every other $a$ you fix, you will obtain the same figure, but with a small difference.
What's this difference?
answered Nov 16 at 12:35
Bias of Priene
286112
Let $z = a+ib in mathbb{C}$.
Then $Re(z) = a$, $Im(z) = b$ and $0 leq a leq 1/2$, $-pi leq b leq pi$.
We know that $e^z = e^{a+ib} = e^a(cos(b) + isin(b))$.
So, if we fix $a = 0$, for example, we have $e^a = 1$ and we are left with $cos(b) + isin(b)$, where $-pi leq b leq pi$.
So, if this is the range of $b$, what figure you obtain?
Now note that for every other $a$ you fix, you will obtain the same figure, but with a small difference.
What's this difference?
answered Nov 16 at 12:35
Bias of Priene
286112
answered Nov 16 at 12:35
Bias of Priene
286112
answered Nov 16 at 12:35
Bias of Priene
286112
answered Nov 16 at 12:35
Bias of Priene
286112
286112
add a comment |
add a comment |
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