Finding the image of this line under 1/z
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
asked Apr 29 '17 at 20:25
Lost1
5,54433369
add a comment |
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
asked Apr 29 '17 at 20:25
Lost1
5,54433369
Did I edit your post correctly?
– Arbuja
Apr 29 '17 at 20:38
add a comment |
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
asked Apr 29 '17 at 20:25
Lost1
5,54433369
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
asked Apr 29 '17 at 20:25
Lost1
5,54433369
asked Apr 29 '17 at 20:25
Lost1
5,54433369
edited yesterday
asked Apr 29 '17 at 20:25
Lost1
5,54433369
asked Apr 29 '17 at 20:25
Lost1
5,54433369
asked Apr 29 '17 at 20:25
Lost1
5,54433369
5,54433369
Did I edit your post correctly?
– Arbuja
Apr 29 '17 at 20:38
add a comment |
Did I edit your post correctly?
– Arbuja
Apr 29 '17 at 20:38
Did I edit your post correctly?
– Arbuja
Apr 29 '17 at 20:38
Did I edit your post correctly?
– Arbuja
Apr 29 '17 at 20:38
add a comment |
1 Answer
1
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oldest
votes
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 $$u^2+v^2+2u+v=0$$
answered Apr 29 '17 at 20:51
David Quinn
23.9k21140
Nicely done :).
– Faraad Armwood
Apr 29 '17 at 20:54
[+1] You can proceed by equivalence, i.e., replace the $implies$ arrow by an $iff$...
– Jean Marie
Apr 29 '17 at 21:06
Urgh i c. Express in term u v before substituting...
– Lost1
Apr 29 '17 at 21:20
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 $$u^2+v^2+2u+v=0$$
answered Apr 29 '17 at 20:51
David Quinn
23.9k21140
Nicely done :).
– Faraad Armwood
Apr 29 '17 at 20:54
[+1] You can proceed by equivalence, i.e., replace the $implies$ arrow by an $iff$...
– Jean Marie
Apr 29 '17 at 21:06
Urgh i c. Express in term u v before substituting...
– Lost1
Apr 29 '17 at 21:20
add a comment |
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 $$u^2+v^2+2u+v=0$$
answered Apr 29 '17 at 20:51
David Quinn
23.9k21140
Nicely done :).
– Faraad Armwood
Apr 29 '17 at 20:54
[+1] You can proceed by equivalence, i.e., replace the $implies$ arrow by an $iff$...
– Jean Marie
Apr 29 '17 at 21:06
Urgh i c. Express in term u v before substituting...
– Lost1
Apr 29 '17 at 21:20
add a comment |
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 $$u^2+v^2+2u+v=0$$
answered Apr 29 '17 at 20:51
David Quinn
23.9k21140
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 $$u^2+v^2+2u+v=0$$
answered Apr 29 '17 at 20:51
David Quinn
23.9k21140
answered Apr 29 '17 at 20:51
David Quinn
23.9k21140
answered Apr 29 '17 at 20:51
David Quinn
23.9k21140
answered Apr 29 '17 at 20:51
David Quinn
23.9k21140
23.9k21140
Nicely done :).
– Faraad Armwood
Apr 29 '17 at 20:54
[+1] You can proceed by equivalence, i.e., replace the $implies$ arrow by an $iff$...
– Jean Marie
Apr 29 '17 at 21:06
Urgh i c. Express in term u v before substituting...
– Lost1
Apr 29 '17 at 21:20
add a comment |
Nicely done :).
– Faraad Armwood
Apr 29 '17 at 20:54
[+1] You can proceed by equivalence, i.e., replace the $implies$ arrow by an $iff$...
– Jean Marie
Apr 29 '17 at 21:06
Urgh i c. Express in term u v before substituting...
– Lost1
Apr 29 '17 at 21:20
Nicely done :).
– Faraad Armwood
Apr 29 '17 at 20:54
Nicely done :).
– Faraad Armwood
Apr 29 '17 at 20:54
[+1] You can proceed by equivalence, i.e., replace the $implies$ arrow by an $iff$...
– Jean Marie
Apr 29 '17 at 21:06
[+1] You can proceed by equivalence, i.e., replace the $implies$ arrow by an $iff$...
– Jean Marie
Apr 29 '17 at 21:06
Urgh i c. Express in term u v before substituting...
– Lost1
Apr 29 '17 at 21:20
Urgh i c. Express in term u v before substituting...
– Lost1
Apr 29 '17 at 21:20
add a comment |
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Did I edit your post correctly?
– Arbuja
Apr 29 '17 at 20:38