Finding the image of this line under 1/z












3












$begingroup$


Let $T(z)=1/z$. Find the image of $y=2x+1$ under $T$.



I assume x and y are the real and imaginary part of z.



Basically what I need was let $w=1/(x+(2x+1)i)$ then multiplied by the complex conjugate we have that the real imaginary part of x.



We have



$$w=frac{x}{x^2+(2x+1)^2}+frac{(2x+1)}{x^2+(2x+1)^2}i$$



Then should we equate coefficient with the equation of a circle? I dont know how to proceed.



From university, I know that the image must be a circle or a line since the transformation is a Mobius transformation. How is an A level student meant to get this?



i just looked at another question. Do i need to find two points such that this line is the locus for those two points? Then i sub that in?






mobius-transformation complex-transformation







share|cite|improve this question











$endgroup$












  • $begingroup$
    Did I edit your post correctly?
    $endgroup$
    – Arbuja
    Apr 29 '17 at 20:38
















3












$begingroup$


Let $T(z)=1/z$. Find the image of $y=2x+1$ under $T$.



I assume x and y are the real and imaginary part of z.



Basically what I need was let $w=1/(x+(2x+1)i)$ then multiplied by the complex conjugate we have that the real imaginary part of x.



We have



$$w=frac{x}{x^2+(2x+1)^2}+frac{(2x+1)}{x^2+(2x+1)^2}i$$



Then should we equate coefficient with the equation of a circle? I dont know how to proceed.



From university, I know that the image must be a circle or a line since the transformation is a Mobius transformation. How is an A level student meant to get this?



i just looked at another question. Do i need to find two points such that this line is the locus for those two points? Then i sub that in?






mobius-transformation complex-transformation







share|cite|improve this question











$endgroup$












  • $begingroup$
    Did I edit your post correctly?
    $endgroup$
    – Arbuja
    Apr 29 '17 at 20:38














3












3








3


1



$begingroup$


Let $T(z)=1/z$. Find the image of $y=2x+1$ under $T$.



I assume x and y are the real and imaginary part of z.



Basically what I need was let $w=1/(x+(2x+1)i)$ then multiplied by the complex conjugate we have that the real imaginary part of x.



We have



$$w=frac{x}{x^2+(2x+1)^2}+frac{(2x+1)}{x^2+(2x+1)^2}i$$



Then should we equate coefficient with the equation of a circle? I dont know how to proceed.



From university, I know that the image must be a circle or a line since the transformation is a Mobius transformation. How is an A level student meant to get this?



i just looked at another question. Do i need to find two points such that this line is the locus for those two points? Then i sub that in?






mobius-transformation complex-transformation







share|cite|improve this question











$endgroup$




Let $T(z)=1/z$. Find the image of $y=2x+1$ under $T$.



I assume x and y are the real and imaginary part of z.



Basically what I need was let $w=1/(x+(2x+1)i)$ then multiplied by the complex conjugate we have that the real imaginary part of x.



We have



$$w=frac{x}{x^2+(2x+1)^2}+frac{(2x+1)}{x^2+(2x+1)^2}i$$



Then should we equate coefficient with the equation of a circle? I dont know how to proceed.



From university, I know that the image must be a circle or a line since the transformation is a Mobius transformation. How is an A level student meant to get this?



i just looked at another question. Do i need to find two points such that this line is the locus for those two points? Then i sub that in?






mobius-transformation complex-transformation




mobius-transformation complex-transformation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 5 at 0:04







Lost1

















asked Apr 29 '17 at 20:25









Lost1Lost1

5,59933371




5,59933371












  • $begingroup$
    Did I edit your post correctly?
    $endgroup$
    – Arbuja
    Apr 29 '17 at 20:38


















  • $begingroup$
    Did I edit your post correctly?
    $endgroup$
    – Arbuja
    Apr 29 '17 at 20:38
















$begingroup$
Did I edit your post correctly?
$endgroup$
– Arbuja
Apr 29 '17 at 20:38




$begingroup$
Did I edit your post correctly?
$endgroup$
– Arbuja
Apr 29 '17 at 20:38










1 Answer
1






active

oldest

votes


















3












$begingroup$

Let $w=frac 1zimplies z=frac 1w$



We write $w=u+iv$, so that $$z=x+iy=frac{1}{u+iv}=frac{u-iv}{u^2+v^2}$$



Then $$y=2x+1implies -frac{v}{u^2+v^2}=frac{2u}{u^2+v^2}+1$$



This is a circle $$u^2+v^2+2u+v=0$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Nicely done :).
    $endgroup$
    – Faraad Armwood
    Apr 29 '17 at 20:54










  • $begingroup$
    [+1] You can proceed by equivalence, i.e., replace the $implies$ arrow by an $iff$...
    $endgroup$
    – Jean Marie
    Apr 29 '17 at 21:06










  • $begingroup$
    Urgh i c. Express in term u v before substituting...
    $endgroup$
    – Lost1
    Apr 29 '17 at 21:20












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1 Answer
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1 Answer
1






active

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active

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3












$begingroup$

Let $w=frac 1zimplies z=frac 1w$



We write $w=u+iv$, so that $$z=x+iy=frac{1}{u+iv}=frac{u-iv}{u^2+v^2}$$



Then $$y=2x+1implies -frac{v}{u^2+v^2}=frac{2u}{u^2+v^2}+1$$



This is a circle $$u^2+v^2+2u+v=0$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Nicely done :).
    $endgroup$
    – Faraad Armwood
    Apr 29 '17 at 20:54










  • $begingroup$
    [+1] You can proceed by equivalence, i.e., replace the $implies$ arrow by an $iff$...
    $endgroup$
    – Jean Marie
    Apr 29 '17 at 21:06










  • $begingroup$
    Urgh i c. Express in term u v before substituting...
    $endgroup$
    – Lost1
    Apr 29 '17 at 21:20
















3












$begingroup$

Let $w=frac 1zimplies z=frac 1w$



We write $w=u+iv$, so that $$z=x+iy=frac{1}{u+iv}=frac{u-iv}{u^2+v^2}$$



Then $$y=2x+1implies -frac{v}{u^2+v^2}=frac{2u}{u^2+v^2}+1$$



This is a circle $$u^2+v^2+2u+v=0$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Nicely done :).
    $endgroup$
    – Faraad Armwood
    Apr 29 '17 at 20:54










  • $begingroup$
    [+1] You can proceed by equivalence, i.e., replace the $implies$ arrow by an $iff$...
    $endgroup$
    – Jean Marie
    Apr 29 '17 at 21:06










  • $begingroup$
    Urgh i c. Express in term u v before substituting...
    $endgroup$
    – Lost1
    Apr 29 '17 at 21:20














3












3








3





$begingroup$

Let $w=frac 1zimplies z=frac 1w$



We write $w=u+iv$, so that $$z=x+iy=frac{1}{u+iv}=frac{u-iv}{u^2+v^2}$$



Then $$y=2x+1implies -frac{v}{u^2+v^2}=frac{2u}{u^2+v^2}+1$$



This is a circle $$u^2+v^2+2u+v=0$$






share|cite|improve this answer









$endgroup$



Let $w=frac 1zimplies z=frac 1w$



We write $w=u+iv$, so that $$z=x+iy=frac{1}{u+iv}=frac{u-iv}{u^2+v^2}$$



Then $$y=2x+1implies -frac{v}{u^2+v^2}=frac{2u}{u^2+v^2}+1$$



This is a circle $$u^2+v^2+2u+v=0$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Apr 29 '17 at 20:51









David QuinnDavid Quinn

24.1k21141




24.1k21141












  • $begingroup$
    Nicely done :).
    $endgroup$
    – Faraad Armwood
    Apr 29 '17 at 20:54










  • $begingroup$
    [+1] You can proceed by equivalence, i.e., replace the $implies$ arrow by an $iff$...
    $endgroup$
    – Jean Marie
    Apr 29 '17 at 21:06










  • $begingroup$
    Urgh i c. Express in term u v before substituting...
    $endgroup$
    – Lost1
    Apr 29 '17 at 21:20


















  • $begingroup$
    Nicely done :).
    $endgroup$
    – Faraad Armwood
    Apr 29 '17 at 20:54










  • $begingroup$
    [+1] You can proceed by equivalence, i.e., replace the $implies$ arrow by an $iff$...
    $endgroup$
    – Jean Marie
    Apr 29 '17 at 21:06










  • $begingroup$
    Urgh i c. Express in term u v before substituting...
    $endgroup$
    – Lost1
    Apr 29 '17 at 21:20
















$begingroup$
Nicely done :).
$endgroup$
– Faraad Armwood
Apr 29 '17 at 20:54




$begingroup$
Nicely done :).
$endgroup$
– Faraad Armwood
Apr 29 '17 at 20:54












$begingroup$
[+1] You can proceed by equivalence, i.e., replace the $implies$ arrow by an $iff$...
$endgroup$
– Jean Marie
Apr 29 '17 at 21:06




$begingroup$
[+1] You can proceed by equivalence, i.e., replace the $implies$ arrow by an $iff$...
$endgroup$
– Jean Marie
Apr 29 '17 at 21:06












$begingroup$
Urgh i c. Express in term u v before substituting...
$endgroup$
– Lost1
Apr 29 '17 at 21:20




$begingroup$
Urgh i c. Express in term u v before substituting...
$endgroup$
– Lost1
Apr 29 '17 at 21:20


















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