How to show that $f_1(z) = -f_2(z)$, where $f_1$ and $f_2$ are the square root functions with different...
$begingroup$
Let $f_1(z)=sqrt{r}e^{itheta/2}, (r>0, 0<theta<pi)$ and $f_2(z)=sqrt{r}e^{itheta/2}, (r>0, pi<theta<5pi/2)$. How to show $f_1(z)=-f_2(z)$ in the first quadrant?
If we set $z_0=re^{itheta}, (r>0, 0<theta<pi)$, then we can plug it into $f_1(z)$, but what about $f_2(z)$? The angle of $z_0$ is not in $(pi, 5pi/2)$. Is it correct to add $pi$ to the argument of $z_0$, then plug it into $f_2$?
complex-numbers
asked Nov 30 '18 at 17:37
user398843user398843
629215
$endgroup$
add a comment |
$begingroup$
Let $f_1(z)=sqrt{r}e^{itheta/2}, (r>0, 0<theta<pi)$ and $f_2(z)=sqrt{r}e^{itheta/2}, (r>0, pi<theta<5pi/2)$. How to show $f_1(z)=-f_2(z)$ in the first quadrant?
If we set $z_0=re^{itheta}, (r>0, 0<theta<pi)$, then we can plug it into $f_1(z)$, but what about $f_2(z)$? The angle of $z_0$ is not in $(pi, 5pi/2)$. Is it correct to add $pi$ to the argument of $z_0$, then plug it into $f_2$?
complex-numbers
asked Nov 30 '18 at 17:37
user398843user398843
629215
$endgroup$
$begingroup$
This is part of an exercise on the book. The exercise guides us to see that if $f_2$ is an analytic continuation of $f_1$, and $f_3$ is an analytic continuation of $f_2$, then it is not necessarily true that $f_1=f_3$ in the intersection of the domains of $f_1$ and $f_3$. But it seems to me that the intersection of the domains of $f_1$ and $f_3$ is empty.
$endgroup$
– user398843
Nov 30 '18 at 18:12
add a comment |
$begingroup$
Let $f_1(z)=sqrt{r}e^{itheta/2}, (r>0, 0<theta<pi)$ and $f_2(z)=sqrt{r}e^{itheta/2}, (r>0, pi<theta<5pi/2)$. How to show $f_1(z)=-f_2(z)$ in the first quadrant?
If we set $z_0=re^{itheta}, (r>0, 0<theta<pi)$, then we can plug it into $f_1(z)$, but what about $f_2(z)$? The angle of $z_0$ is not in $(pi, 5pi/2)$. Is it correct to add $pi$ to the argument of $z_0$, then plug it into $f_2$?
complex-numbers
asked Nov 30 '18 at 17:37
user398843user398843
629215
$endgroup$
Let $f_1(z)=sqrt{r}e^{itheta/2}, (r>0, 0<theta<pi)$ and $f_2(z)=sqrt{r}e^{itheta/2}, (r>0, pi<theta<5pi/2)$. How to show $f_1(z)=-f_2(z)$ in the first quadrant?
If we set $z_0=re^{itheta}, (r>0, 0<theta<pi)$, then we can plug it into $f_1(z)$, but what about $f_2(z)$? The angle of $z_0$ is not in $(pi, 5pi/2)$. Is it correct to add $pi$ to the argument of $z_0$, then plug it into $f_2$?
complex-numbers
complex-numbers
asked Nov 30 '18 at 17:37
user398843user398843
629215
asked Nov 30 '18 at 17:37
user398843user398843
629215
edited Nov 30 '18 at 18:19
user398843
asked Nov 30 '18 at 17:37
user398843user398843
629215
asked Nov 30 '18 at 17:37
user398843user398843
629215
asked Nov 30 '18 at 17:37
user398843user398843
629215
629215
$begingroup$
This is part of an exercise on the book. The exercise guides us to see that if $f_2$ is an analytic continuation of $f_1$, and $f_3$ is an analytic continuation of $f_2$, then it is not necessarily true that $f_1=f_3$ in the intersection of the domains of $f_1$ and $f_3$. But it seems to me that the intersection of the domains of $f_1$ and $f_3$ is empty.
$endgroup$
– user398843
Nov 30 '18 at 18:12
add a comment |
$begingroup$
This is part of an exercise on the book. The exercise guides us to see that if $f_2$ is an analytic continuation of $f_1$, and $f_3$ is an analytic continuation of $f_2$, then it is not necessarily true that $f_1=f_3$ in the intersection of the domains of $f_1$ and $f_3$. But it seems to me that the intersection of the domains of $f_1$ and $f_3$ is empty.
$endgroup$
– user398843
Nov 30 '18 at 18:12
$begingroup$
This is part of an exercise on the book. The exercise guides us to see that if $f_2$ is an analytic continuation of $f_1$, and $f_3$ is an analytic continuation of $f_2$, then it is not necessarily true that $f_1=f_3$ in the intersection of the domains of $f_1$ and $f_3$. But it seems to me that the intersection of the domains of $f_1$ and $f_3$ is empty.
$endgroup$
– user398843
Nov 30 '18 at 18:12
$begingroup$
This is part of an exercise on the book. The exercise guides us to see that if $f_2$ is an analytic continuation of $f_1$, and $f_3$ is an analytic continuation of $f_2$, then it is not necessarily true that $f_1=f_3$ in the intersection of the domains of $f_1$ and $f_3$. But it seems to me that the intersection of the domains of $f_1$ and $f_3$ is empty.
$endgroup$
– user398843
Nov 30 '18 at 18:12
add a comment |
2 Answers
2
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To get $f_2(x)$ let $z_0=re^{i(theta +2pi) }$. Then $f_2(z_0)=sqrt{r}e^{i(theta/2 +pi) }=-sqrt{r}e^{itheta/2}=-f_1(z_0)$. You can do this because $e^{2npi i}=1$
for all $n$, while $e^{pi i}=-1$.
answered Nov 30 '18 at 18:19
herb steinbergherb steinberg
2,5032310
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add a comment |
$begingroup$
Let $z=x+iy (x,yin mathbb R)$. Then $z=x+iy=re^{iTheta+2npi}$ ($pi<Theta le pi$, $nin mathbb Z$). Hence, $z=re^{iTheta}$ is only one of an infinite number of possiblities for the exponential form of $x+iy$.
answered Nov 30 '18 at 18:37
user398843user398843
629215
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
To get $f_2(x)$ let $z_0=re^{i(theta +2pi) }$. Then $f_2(z_0)=sqrt{r}e^{i(theta/2 +pi) }=-sqrt{r}e^{itheta/2}=-f_1(z_0)$. You can do this because $e^{2npi i}=1$
for all $n$, while $e^{pi i}=-1$.
answered Nov 30 '18 at 18:19
herb steinbergherb steinberg
2,5032310
$endgroup$
add a comment |
$begingroup$
To get $f_2(x)$ let $z_0=re^{i(theta +2pi) }$. Then $f_2(z_0)=sqrt{r}e^{i(theta/2 +pi) }=-sqrt{r}e^{itheta/2}=-f_1(z_0)$. You can do this because $e^{2npi i}=1$
for all $n$, while $e^{pi i}=-1$.
answered Nov 30 '18 at 18:19
herb steinbergherb steinberg
2,5032310
$endgroup$
add a comment |
$begingroup$
To get $f_2(x)$ let $z_0=re^{i(theta +2pi) }$. Then $f_2(z_0)=sqrt{r}e^{i(theta/2 +pi) }=-sqrt{r}e^{itheta/2}=-f_1(z_0)$. You can do this because $e^{2npi i}=1$
for all $n$, while $e^{pi i}=-1$.
answered Nov 30 '18 at 18:19
herb steinbergherb steinberg
2,5032310
$endgroup$
To get $f_2(x)$ let $z_0=re^{i(theta +2pi) }$. Then $f_2(z_0)=sqrt{r}e^{i(theta/2 +pi) }=-sqrt{r}e^{itheta/2}=-f_1(z_0)$. You can do this because $e^{2npi i}=1$
for all $n$, while $e^{pi i}=-1$.
answered Nov 30 '18 at 18:19
herb steinbergherb steinberg
2,5032310
edited Nov 30 '18 at 18:38
answered Nov 30 '18 at 18:19
herb steinbergherb steinberg
2,5032310
answered Nov 30 '18 at 18:19
herb steinbergherb steinberg
2,5032310
answered Nov 30 '18 at 18:19
herb steinbergherb steinberg
2,5032310
2,5032310
add a comment |
add a comment |
$begingroup$
Let $z=x+iy (x,yin mathbb R)$. Then $z=x+iy=re^{iTheta+2npi}$ ($pi<Theta le pi$, $nin mathbb Z$). Hence, $z=re^{iTheta}$ is only one of an infinite number of possiblities for the exponential form of $x+iy$.
answered Nov 30 '18 at 18:37
user398843user398843
629215
$endgroup$
add a comment |
$begingroup$
Let $z=x+iy (x,yin mathbb R)$. Then $z=x+iy=re^{iTheta+2npi}$ ($pi<Theta le pi$, $nin mathbb Z$). Hence, $z=re^{iTheta}$ is only one of an infinite number of possiblities for the exponential form of $x+iy$.
answered Nov 30 '18 at 18:37
user398843user398843
629215
$endgroup$
add a comment |
$begingroup$
Let $z=x+iy (x,yin mathbb R)$. Then $z=x+iy=re^{iTheta+2npi}$ ($pi<Theta le pi$, $nin mathbb Z$). Hence, $z=re^{iTheta}$ is only one of an infinite number of possiblities for the exponential form of $x+iy$.
answered Nov 30 '18 at 18:37
user398843user398843
629215
$endgroup$
Let $z=x+iy (x,yin mathbb R)$. Then $z=x+iy=re^{iTheta+2npi}$ ($pi<Theta le pi$, $nin mathbb Z$). Hence, $z=re^{iTheta}$ is only one of an infinite number of possiblities for the exponential form of $x+iy$.
answered Nov 30 '18 at 18:37
user398843user398843
629215
answered Nov 30 '18 at 18:37
user398843user398843
629215
answered Nov 30 '18 at 18:37
user398843user398843
629215
answered Nov 30 '18 at 18:37
user398843user398843
629215
629215
add a comment |
add a comment |
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$begingroup$
This is part of an exercise on the book. The exercise guides us to see that if $f_2$ is an analytic continuation of $f_1$, and $f_3$ is an analytic continuation of $f_2$, then it is not necessarily true that $f_1=f_3$ in the intersection of the domains of $f_1$ and $f_3$. But it seems to me that the intersection of the domains of $f_1$ and $f_3$ is empty.
$endgroup$
– user398843
Nov 30 '18 at 18:12