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How to prove $f(z)=z+frac{1}{z}$ maps circles with $rne 1$ onto ellipses?

1

$begingroup$

How to prove $f(z)=z+frac{1}{z}$ maps circles with $r> 1$ onto ellipses of the form $$frac{x^2}{(r+frac{1}{r})^2}+frac{y^2}{(r-frac{1}{r})^2}=1?$$ It is simple to understand the mapping to part. You just replace $z$ with $re^{Itheta}$ which simplifies to $$ k(theta)=f(re^{itheta})=re^{itheta}+frac{1}{re^{itheta}}=(r+frac{1}{r})cos(theta)+i(r-frac{1}{r})sin(theta)$$ which is the parameterization of the ellipse given above. However, I do not understand how it maps onto. I tried to produce a right inverse and got $h(y)=-iln(frac{y+sqrt{y^2-4}}{2r})$ and $g(y)=-iln(frac{y-sqrt{y^2-4}}{2r})$ which works but is not a real number. I believe this is a problem. How would I produce a correct right inverse or prove it is onto?

complex-analysis conic-sections

share|cite|improve this question

edited Mar 26 at 13:43

J. M. is a poor mathematician

61.3k5152291

asked Mar 26 at 11:05

Johnathon TaylorJohnathon Taylor

113

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Have you seen the parametrization $x=a cosphi, y=bsinphi,phiin[0,2pi),$ of the ellipse $$frac{x^2}{a^2}+frac{y^2}{b^2}=1?$$ It probably/possibly made an appearance in calculus when discussing parametrized curves?
  $endgroup$
  – Jyrki Lahtonen
  Mar 26 at 11:10
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your initial parameterization of the circle uses the full interval $theta in [0,2pi)$ - the same interval for the parameterization of a full ellipse.
  $endgroup$
  – Umberto P.
  Mar 26 at 11:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have seen this. I, also see that as we traverse the circle, we will traverse the ellipse in essentially the same way; however, that just doesn't seem like it follows to a formal argument.
  $endgroup$
  – Johnathon Taylor
  Mar 26 at 11:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If you know that all the points of that ellipse are of the form $x=acostheta, y=bsintheta$, then surely that is a formal argument for the mapping being onto!
  $endgroup$
  – Jyrki Lahtonen
  Mar 26 at 11:19
  
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Are the circles centered at the origin? If not, then the claim fails for any circle that touches the origin (which is mapped to infinity).
  $endgroup$
  – Oscar Lanzi
  Mar 26 at 11:26
  

  
  
  
  

 | 
show 2 more comments

1

$begingroup$

How to prove $f(z)=z+frac{1}{z}$ maps circles with $r> 1$ onto ellipses of the form $$frac{x^2}{(r+frac{1}{r})^2}+frac{y^2}{(r-frac{1}{r})^2}=1?$$ It is simple to understand the mapping to part. You just replace $z$ with $re^{Itheta}$ which simplifies to $$ k(theta)=f(re^{itheta})=re^{itheta}+frac{1}{re^{itheta}}=(r+frac{1}{r})cos(theta)+i(r-frac{1}{r})sin(theta)$$ which is the parameterization of the ellipse given above. However, I do not understand how it maps onto. I tried to produce a right inverse and got $h(y)=-iln(frac{y+sqrt{y^2-4}}{2r})$ and $g(y)=-iln(frac{y-sqrt{y^2-4}}{2r})$ which works but is not a real number. I believe this is a problem. How would I produce a correct right inverse or prove it is onto?

complex-analysis conic-sections

share|cite|improve this question

edited Mar 26 at 13:43

J. M. is a poor mathematician

61.3k5152291

asked Mar 26 at 11:05

Johnathon TaylorJohnathon Taylor

113

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Have you seen the parametrization $x=a cosphi, y=bsinphi,phiin[0,2pi),$ of the ellipse $$frac{x^2}{a^2}+frac{y^2}{b^2}=1?$$ It probably/possibly made an appearance in calculus when discussing parametrized curves?
  $endgroup$
  – Jyrki Lahtonen
  Mar 26 at 11:10
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your initial parameterization of the circle uses the full interval $theta in [0,2pi)$ - the same interval for the parameterization of a full ellipse.
  $endgroup$
  – Umberto P.
  Mar 26 at 11:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have seen this. I, also see that as we traverse the circle, we will traverse the ellipse in essentially the same way; however, that just doesn't seem like it follows to a formal argument.
  $endgroup$
  – Johnathon Taylor
  Mar 26 at 11:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If you know that all the points of that ellipse are of the form $x=acostheta, y=bsintheta$, then surely that is a formal argument for the mapping being onto!
  $endgroup$
  – Jyrki Lahtonen
  Mar 26 at 11:19
  
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Are the circles centered at the origin? If not, then the claim fails for any circle that touches the origin (which is mapped to infinity).
  $endgroup$
  – Oscar Lanzi
  Mar 26 at 11:26
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

How to prove $f(z)=z+frac{1}{z}$ maps circles with $r> 1$ onto ellipses of the form $$frac{x^2}{(r+frac{1}{r})^2}+frac{y^2}{(r-frac{1}{r})^2}=1?$$ It is simple to understand the mapping to part. You just replace $z$ with $re^{Itheta}$ which simplifies to $$ k(theta)=f(re^{itheta})=re^{itheta}+frac{1}{re^{itheta}}=(r+frac{1}{r})cos(theta)+i(r-frac{1}{r})sin(theta)$$ which is the parameterization of the ellipse given above. However, I do not understand how it maps onto. I tried to produce a right inverse and got $h(y)=-iln(frac{y+sqrt{y^2-4}}{2r})$ and $g(y)=-iln(frac{y-sqrt{y^2-4}}{2r})$ which works but is not a real number. I believe this is a problem. How would I produce a correct right inverse or prove it is onto?

complex-analysis conic-sections

share|cite|improve this question

edited Mar 26 at 13:43

J. M. is a poor mathematician

61.3k5152291

asked Mar 26 at 11:05

Johnathon TaylorJohnathon Taylor

113

$endgroup$

share|cite|improve this question

edited Mar 26 at 13:43

J. M. is a poor mathematician

61.3k5152291

asked Mar 26 at 11:05

Johnathon TaylorJohnathon Taylor

113

share|cite|improve this question

edited Mar 26 at 13:43

J. M. is a poor mathematician

61.3k5152291

asked Mar 26 at 11:05

Johnathon TaylorJohnathon Taylor

113

edited Mar 26 at 13:43

J. M. is a poor mathematician

61.3k5152291

edited Mar 26 at 13:43

J. M. is a poor mathematician

61.3k5152291

asked Mar 26 at 11:05

Johnathon TaylorJohnathon Taylor

113

asked Mar 26 at 11:05

Johnathon TaylorJohnathon Taylor

113

2 Answers
2

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oldest

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0

$begingroup$

Let $S(r)$ denote the circle with center $0$ and radius $r ne 1$ and $E(r)$ the ellipse defined in your question. You have shown that $I(r) = f(S(r)) subset E(r)$.

For $M subset mathbb{C}$ define $-M = { -z mid z in M}$. We have $S(r) = -S(r)$ and $E(r) = -E(r)$.

We have $f(-z) = -f(z)$ which implies $f(-M) = -f(M)$ for $M subset mathbb{C} setminus { 0 }$. Hence $I(r) = f(S(r)) = f(-S(r)) = -f(S(r)) = -I(r)$.

Now assume that $I(r) ne E(r)$. Choose $w in E(r) setminus I(r)$. We have $w notin I(r)$, hence also $-w notin -I(r) = I(r)$. Therefore $I(r) subset E'(r) = E(r) setminus { w,-w }$. The set $E'(r)$ has two connected components $E_i(r)$ which satisfy $E_1(r) = -E_2(r)$.

$I(r)$ is the continuous image of the connected set $S(r)$, hence it is connected and we conclude that $I(r) subset E_i(r)$ either for $i=1$ or $i=2$, wl.o.g. for $i=1$. But then also $I(r) = -I(r) subset - E_1(r) = E_2(r)$ which is impossible.

Therefore we must have $I(r) = E(r)$.

share|cite|improve this answer

answered Mar 26 at 13:38

Paul FrostPaul Frost

12.9k31035

$endgroup$

add a comment | 

0

$begingroup$

Let $z=x+iy$ with $x^2+y^2=r^2$.

We have

$$z+frac1z=x+iy+frac{x-iy}{x^2+y^2}=xleft(1+frac1{r^2}right)+iyleft(1-frac1{r^2}right)=u+iv.$$

From this,

$$left(frac{u}{1+dfrac1{r^2}}right)^2+left(frac{v}{1-dfrac1{r^2}}right)^2=r^2.$$

Note that the transformation is a mere scaling.

share|cite|improve this answer

answered Mar 26 at 13:49

Yves DaoustYves Daoust

133k676232

$endgroup$

add a comment | 

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0

$begingroup$

Let $S(r)$ denote the circle with center $0$ and radius $r ne 1$ and $E(r)$ the ellipse defined in your question. You have shown that $I(r) = f(S(r)) subset E(r)$.

For $M subset mathbb{C}$ define $-M = { -z mid z in M}$. We have $S(r) = -S(r)$ and $E(r) = -E(r)$.

We have $f(-z) = -f(z)$ which implies $f(-M) = -f(M)$ for $M subset mathbb{C} setminus { 0 }$. Hence $I(r) = f(S(r)) = f(-S(r)) = -f(S(r)) = -I(r)$.

Now assume that $I(r) ne E(r)$. Choose $w in E(r) setminus I(r)$. We have $w notin I(r)$, hence also $-w notin -I(r) = I(r)$. Therefore $I(r) subset E'(r) = E(r) setminus { w,-w }$. The set $E'(r)$ has two connected components $E_i(r)$ which satisfy $E_1(r) = -E_2(r)$.

$I(r)$ is the continuous image of the connected set $S(r)$, hence it is connected and we conclude that $I(r) subset E_i(r)$ either for $i=1$ or $i=2$, wl.o.g. for $i=1$. But then also $I(r) = -I(r) subset - E_1(r) = E_2(r)$ which is impossible.

Therefore we must have $I(r) = E(r)$.

share|cite|improve this answer

answered Mar 26 at 13:38

Paul FrostPaul Frost

12.9k31035

$endgroup$

add a comment | 

0

$begingroup$

Let $S(r)$ denote the circle with center $0$ and radius $r ne 1$ and $E(r)$ the ellipse defined in your question. You have shown that $I(r) = f(S(r)) subset E(r)$.

For $M subset mathbb{C}$ define $-M = { -z mid z in M}$. We have $S(r) = -S(r)$ and $E(r) = -E(r)$.

We have $f(-z) = -f(z)$ which implies $f(-M) = -f(M)$ for $M subset mathbb{C} setminus { 0 }$. Hence $I(r) = f(S(r)) = f(-S(r)) = -f(S(r)) = -I(r)$.

Now assume that $I(r) ne E(r)$. Choose $w in E(r) setminus I(r)$. We have $w notin I(r)$, hence also $-w notin -I(r) = I(r)$. Therefore $I(r) subset E'(r) = E(r) setminus { w,-w }$. The set $E'(r)$ has two connected components $E_i(r)$ which satisfy $E_1(r) = -E_2(r)$.

$I(r)$ is the continuous image of the connected set $S(r)$, hence it is connected and we conclude that $I(r) subset E_i(r)$ either for $i=1$ or $i=2$, wl.o.g. for $i=1$. But then also $I(r) = -I(r) subset - E_1(r) = E_2(r)$ which is impossible.

Therefore we must have $I(r) = E(r)$.

share|cite|improve this answer

answered Mar 26 at 13:38

Paul FrostPaul Frost

12.9k31035

$endgroup$

add a comment | 

$begingroup$

Let $S(r)$ denote the circle with center $0$ and radius $r ne 1$ and $E(r)$ the ellipse defined in your question. You have shown that $I(r) = f(S(r)) subset E(r)$.

For $M subset mathbb{C}$ define $-M = { -z mid z in M}$. We have $S(r) = -S(r)$ and $E(r) = -E(r)$.

We have $f(-z) = -f(z)$ which implies $f(-M) = -f(M)$ for $M subset mathbb{C} setminus { 0 }$. Hence $I(r) = f(S(r)) = f(-S(r)) = -f(S(r)) = -I(r)$.

Now assume that $I(r) ne E(r)$. Choose $w in E(r) setminus I(r)$. We have $w notin I(r)$, hence also $-w notin -I(r) = I(r)$. Therefore $I(r) subset E'(r) = E(r) setminus { w,-w }$. The set $E'(r)$ has two connected components $E_i(r)$ which satisfy $E_1(r) = -E_2(r)$.

$I(r)$ is the continuous image of the connected set $S(r)$, hence it is connected and we conclude that $I(r) subset E_i(r)$ either for $i=1$ or $i=2$, wl.o.g. for $i=1$. But then also $I(r) = -I(r) subset - E_1(r) = E_2(r)$ which is impossible.

Therefore we must have $I(r) = E(r)$.

share|cite|improve this answer

answered Mar 26 at 13:38

Paul FrostPaul Frost

12.9k31035

$endgroup$

share|cite|improve this answer

answered Mar 26 at 13:38

Paul FrostPaul Frost

12.9k31035

answered Mar 26 at 13:38

Paul FrostPaul Frost

12.9k31035

answered Mar 26 at 13:38

Paul FrostPaul Frost

12.9k31035

0

$begingroup$

Let $z=x+iy$ with $x^2+y^2=r^2$.

We have

$$z+frac1z=x+iy+frac{x-iy}{x^2+y^2}=xleft(1+frac1{r^2}right)+iyleft(1-frac1{r^2}right)=u+iv.$$

From this,

$$left(frac{u}{1+dfrac1{r^2}}right)^2+left(frac{v}{1-dfrac1{r^2}}right)^2=r^2.$$

Note that the transformation is a mere scaling.

share|cite|improve this answer

answered Mar 26 at 13:49

Yves DaoustYves Daoust

133k676232

$endgroup$

add a comment | 

0

$begingroup$

Let $z=x+iy$ with $x^2+y^2=r^2$.

We have

$$z+frac1z=x+iy+frac{x-iy}{x^2+y^2}=xleft(1+frac1{r^2}right)+iyleft(1-frac1{r^2}right)=u+iv.$$

From this,

$$left(frac{u}{1+dfrac1{r^2}}right)^2+left(frac{v}{1-dfrac1{r^2}}right)^2=r^2.$$

Note that the transformation is a mere scaling.

share|cite|improve this answer

answered Mar 26 at 13:49

Yves DaoustYves Daoust

133k676232

$endgroup$

add a comment | 

$begingroup$

Let $z=x+iy$ with $x^2+y^2=r^2$.

We have

$$z+frac1z=x+iy+frac{x-iy}{x^2+y^2}=xleft(1+frac1{r^2}right)+iyleft(1-frac1{r^2}right)=u+iv.$$

From this,

$$left(frac{u}{1+dfrac1{r^2}}right)^2+left(frac{v}{1-dfrac1{r^2}}right)^2=r^2.$$

Note that the transformation is a mere scaling.

share|cite|improve this answer

answered Mar 26 at 13:49

Yves DaoustYves Daoust

133k676232

$endgroup$

share|cite|improve this answer

answered Mar 26 at 13:49

Yves DaoustYves Daoust

133k676232

answered Mar 26 at 13:49

Yves DaoustYves Daoust

133k676232

answered Mar 26 at 13:49

Yves DaoustYves Daoust

133k676232