Find the geometric locus $z in mathbb C$ so that $frac{z+2}{z(z+1)}in mathbb R$
$begingroup$
Find the geometric locus of the set of $z in mathbb C$ so that
$$frac{z+2}{z(z+1)}in mathbb R$$
Source: IME (Military Engineering Institute, Brazil, entrance exam, 1974)
My attempt: With the notation $z=a+bi$, the solution provided in a book is either the line $b=0$ or the circle $(a+2)^2+b^2=2$ but I could not find it, or not able to recognize this locus set from the algebraic development I did (or the solution or the statement has some mistake).
Hints and solutions are welcomed.
algebra-precalculus complex-numbers contest-math
asked Dec 16 '18 at 10:48
bluemasterbluemaster
1,663520
$endgroup$
add a comment |
$begingroup$
Find the geometric locus of the set of $z in mathbb C$ so that
$$frac{z+2}{z(z+1)}in mathbb R$$
Source: IME (Military Engineering Institute, Brazil, entrance exam, 1974)
My attempt: With the notation $z=a+bi$, the solution provided in a book is either the line $b=0$ or the circle $(a+2)^2+b^2=2$ but I could not find it, or not able to recognize this locus set from the algebraic development I did (or the solution or the statement has some mistake).
Hints and solutions are welcomed.
algebra-precalculus complex-numbers contest-math
asked Dec 16 '18 at 10:48
bluemasterbluemaster
1,663520
$endgroup$
add a comment |
$begingroup$
Find the geometric locus of the set of $z in mathbb C$ so that
$$frac{z+2}{z(z+1)}in mathbb R$$
Source: IME (Military Engineering Institute, Brazil, entrance exam, 1974)
My attempt: With the notation $z=a+bi$, the solution provided in a book is either the line $b=0$ or the circle $(a+2)^2+b^2=2$ but I could not find it, or not able to recognize this locus set from the algebraic development I did (or the solution or the statement has some mistake).
Hints and solutions are welcomed.
algebra-precalculus complex-numbers contest-math
asked Dec 16 '18 at 10:48
bluemasterbluemaster
1,663520
$endgroup$
Find the geometric locus of the set of $z in mathbb C$ so that
$$frac{z+2}{z(z+1)}in mathbb R$$
Source: IME (Military Engineering Institute, Brazil, entrance exam, 1974)
My attempt: With the notation $z=a+bi$, the solution provided in a book is either the line $b=0$ or the circle $(a+2)^2+b^2=2$ but I could not find it, or not able to recognize this locus set from the algebraic development I did (or the solution or the statement has some mistake).
Hints and solutions are welcomed.
algebra-precalculus complex-numbers contest-math
algebra-precalculus complex-numbers contest-math
asked Dec 16 '18 at 10:48
bluemasterbluemaster
1,663520
asked Dec 16 '18 at 10:48
bluemasterbluemaster
1,663520
asked Dec 16 '18 at 10:48
bluemasterbluemaster
1,663520
asked Dec 16 '18 at 10:48
bluemasterbluemaster
1,663520
asked Dec 16 '18 at 10:48
bluemasterbluemaster
1,663520
1,663520
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
A complex number $w$ is real if and only if it equals its conjugate $bar{w}$. So you need to solve
$$
frac{z+2}{z(z+1)}=frac{bar{z}+2}{bar{z}(bar{z}+1)}
$$
Cross multiplication gives
$$
zbar{z}^2+zbar{z}+2bar{z}^2+2bar{z}=z^2bar{z}+zbar{z}+2z^2+2z
$$
and we can transfer everything to the right-hand side:
$$
zbar{z}(z-bar{z})+2(z^2-bar{z}^2)+2(z-bar{z})=0
$$
This has the entire real line ($z=bar{z}$) as a solution, points $z=0$ and $z=-1$ excluded, of course; removing this factor we find all other solutions:
$$
zbar{z}+2(z+bar{z})+2=0 tag{*}
$$
Now we can substitute $z=x+yi$:
$$
x^2+y^2+4x+2=0
$$
that is,
$$
(x+2)^2+y^2=2
$$
which is a circle.
You can avoid passing to real coordinates by noticing that the equation (*) can be rewritten as $zbar{z}+2z+2bar{z}+4=2$, so $(z+2)(bar{z}+2)=2$ and finally
$$
|z+2|^2=2
$$
which is clearly a circle with center at $2$ and radius $sqrt{2}$.
answered Dec 16 '18 at 10:59
egregegreg
184k1486205
$endgroup$
add a comment |
$begingroup$
HINT: Multiply the fraction by
$$
frac{bar z(bar z-1)}{bar z(bar z-1)};.
$$
In this way the denominator will be real and you should only look at the numerator, splitting it into its real and imaginary part.
answered Dec 16 '18 at 10:53
JoeJoe
7,21421229
$endgroup$
add a comment |
$begingroup$
A little Partial Fraction Decomposition before rationalization of the denominator will reduce calculation.
$$dfrac{z+2}{z(z+1)}=dfrac1{z+1}+dfrac{2(z+1-z)}{z(z+1)}=dfrac2z-dfrac1{z+1}$$
Now the imaginary part of $dfrac2z$ is $-dfrac{2y}{x^2+y^2}$
and that of $dfrac1{z+1}$ is $-dfrac y{(x+1)^2+y^2}$
Finally, $dfrac{z+2}{z(z+1)}$ will be real if $-dfrac{2y}{x^2+y^2}=-dfrac y{(x+1)^2+y^2}$
$iff0=y(2x^2+2y^2+4x+2-x^2-y^2)=y{(x+2)^2+y^2-2}$
answered Dec 16 '18 at 12:32
lab bhattacharjeelab bhattacharjee
226k15158275
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add a comment |
$begingroup$
Using
$$
z = sqrt{x^2+y^2}e^{arctanfrac yx}
$$
we have
$$
frac{z+2}{z(z+1)} = frac{rho_1 e^{iphi_1}}{rho_2 e^{iphi_2}rho_3 e^{iphi_3}}
$$
and we seek for
$$
arctanfrac{y}{x+2}-arctanfrac yx -arctanfrac{y}{x+1} = 0
$$
then
$$
tanleft(arctanfrac{y}{x+2} -arctanfrac{y}{x+1}right) = frac yx
$$
or
$$
y(y^2+x^2+4x+2) = 0
$$
NOTE
$$
tan(a-b) = frac{tan a-tan b}{1+tan atan b}
$$
so
$$
tanleft(arctanfrac{y}{x+2} -arctanfrac{y}{x+1}right) = frac{y}{y^2+x^2+3x+2}
$$
etc.
answered Dec 16 '18 at 14:02
CesareoCesareo
9,3063517
$endgroup$
$begingroup$
How did you get that last step?
$endgroup$
– I like Serena
Dec 16 '18 at 14:37
2
$begingroup$
@IlikeSerena Please. See attached note.
$endgroup$
– Cesareo
Dec 16 '18 at 16:02
$begingroup$
@bluemaster Thanks for the hint. I have a real problem with addition (and subtraction)
$endgroup$
– Cesareo
Dec 19 '18 at 11:57
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A complex number $w$ is real if and only if it equals its conjugate $bar{w}$. So you need to solve
$$
frac{z+2}{z(z+1)}=frac{bar{z}+2}{bar{z}(bar{z}+1)}
$$
Cross multiplication gives
$$
zbar{z}^2+zbar{z}+2bar{z}^2+2bar{z}=z^2bar{z}+zbar{z}+2z^2+2z
$$
and we can transfer everything to the right-hand side:
$$
zbar{z}(z-bar{z})+2(z^2-bar{z}^2)+2(z-bar{z})=0
$$
This has the entire real line ($z=bar{z}$) as a solution, points $z=0$ and $z=-1$ excluded, of course; removing this factor we find all other solutions:
$$
zbar{z}+2(z+bar{z})+2=0 tag{*}
$$
Now we can substitute $z=x+yi$:
$$
x^2+y^2+4x+2=0
$$
that is,
$$
(x+2)^2+y^2=2
$$
which is a circle.
You can avoid passing to real coordinates by noticing that the equation (*) can be rewritten as $zbar{z}+2z+2bar{z}+4=2$, so $(z+2)(bar{z}+2)=2$ and finally
$$
|z+2|^2=2
$$
which is clearly a circle with center at $2$ and radius $sqrt{2}$.
answered Dec 16 '18 at 10:59
egregegreg
184k1486205
$endgroup$
add a comment |
$begingroup$
A complex number $w$ is real if and only if it equals its conjugate $bar{w}$. So you need to solve
$$
frac{z+2}{z(z+1)}=frac{bar{z}+2}{bar{z}(bar{z}+1)}
$$
Cross multiplication gives
$$
zbar{z}^2+zbar{z}+2bar{z}^2+2bar{z}=z^2bar{z}+zbar{z}+2z^2+2z
$$
and we can transfer everything to the right-hand side:
$$
zbar{z}(z-bar{z})+2(z^2-bar{z}^2)+2(z-bar{z})=0
$$
This has the entire real line ($z=bar{z}$) as a solution, points $z=0$ and $z=-1$ excluded, of course; removing this factor we find all other solutions:
$$
zbar{z}+2(z+bar{z})+2=0 tag{*}
$$
Now we can substitute $z=x+yi$:
$$
x^2+y^2+4x+2=0
$$
that is,
$$
(x+2)^2+y^2=2
$$
which is a circle.
You can avoid passing to real coordinates by noticing that the equation (*) can be rewritten as $zbar{z}+2z+2bar{z}+4=2$, so $(z+2)(bar{z}+2)=2$ and finally
$$
|z+2|^2=2
$$
which is clearly a circle with center at $2$ and radius $sqrt{2}$.
answered Dec 16 '18 at 10:59
egregegreg
184k1486205
$endgroup$
add a comment |
$begingroup$
A complex number $w$ is real if and only if it equals its conjugate $bar{w}$. So you need to solve
$$
frac{z+2}{z(z+1)}=frac{bar{z}+2}{bar{z}(bar{z}+1)}
$$
Cross multiplication gives
$$
zbar{z}^2+zbar{z}+2bar{z}^2+2bar{z}=z^2bar{z}+zbar{z}+2z^2+2z
$$
and we can transfer everything to the right-hand side:
$$
zbar{z}(z-bar{z})+2(z^2-bar{z}^2)+2(z-bar{z})=0
$$
This has the entire real line ($z=bar{z}$) as a solution, points $z=0$ and $z=-1$ excluded, of course; removing this factor we find all other solutions:
$$
zbar{z}+2(z+bar{z})+2=0 tag{*}
$$
Now we can substitute $z=x+yi$:
$$
x^2+y^2+4x+2=0
$$
that is,
$$
(x+2)^2+y^2=2
$$
which is a circle.
You can avoid passing to real coordinates by noticing that the equation (*) can be rewritten as $zbar{z}+2z+2bar{z}+4=2$, so $(z+2)(bar{z}+2)=2$ and finally
$$
|z+2|^2=2
$$
which is clearly a circle with center at $2$ and radius $sqrt{2}$.
answered Dec 16 '18 at 10:59
egregegreg
184k1486205
$endgroup$
A complex number $w$ is real if and only if it equals its conjugate $bar{w}$. So you need to solve
$$
frac{z+2}{z(z+1)}=frac{bar{z}+2}{bar{z}(bar{z}+1)}
$$
Cross multiplication gives
$$
zbar{z}^2+zbar{z}+2bar{z}^2+2bar{z}=z^2bar{z}+zbar{z}+2z^2+2z
$$
and we can transfer everything to the right-hand side:
$$
zbar{z}(z-bar{z})+2(z^2-bar{z}^2)+2(z-bar{z})=0
$$
This has the entire real line ($z=bar{z}$) as a solution, points $z=0$ and $z=-1$ excluded, of course; removing this factor we find all other solutions:
$$
zbar{z}+2(z+bar{z})+2=0 tag{*}
$$
Now we can substitute $z=x+yi$:
$$
x^2+y^2+4x+2=0
$$
that is,
$$
(x+2)^2+y^2=2
$$
which is a circle.
You can avoid passing to real coordinates by noticing that the equation (*) can be rewritten as $zbar{z}+2z+2bar{z}+4=2$, so $(z+2)(bar{z}+2)=2$ and finally
$$
|z+2|^2=2
$$
which is clearly a circle with center at $2$ and radius $sqrt{2}$.
answered Dec 16 '18 at 10:59
egregegreg
184k1486205
edited Dec 16 '18 at 12:17
answered Dec 16 '18 at 10:59
egregegreg
184k1486205
answered Dec 16 '18 at 10:59
egregegreg
184k1486205
answered Dec 16 '18 at 10:59
egregegreg
184k1486205
184k1486205
add a comment |
add a comment |
$begingroup$
HINT: Multiply the fraction by
$$
frac{bar z(bar z-1)}{bar z(bar z-1)};.
$$
In this way the denominator will be real and you should only look at the numerator, splitting it into its real and imaginary part.
answered Dec 16 '18 at 10:53
JoeJoe
7,21421229
$endgroup$
add a comment |
$begingroup$
HINT: Multiply the fraction by
$$
frac{bar z(bar z-1)}{bar z(bar z-1)};.
$$
In this way the denominator will be real and you should only look at the numerator, splitting it into its real and imaginary part.
answered Dec 16 '18 at 10:53
JoeJoe
7,21421229
$endgroup$
add a comment |
$begingroup$
HINT: Multiply the fraction by
$$
frac{bar z(bar z-1)}{bar z(bar z-1)};.
$$
In this way the denominator will be real and you should only look at the numerator, splitting it into its real and imaginary part.
answered Dec 16 '18 at 10:53
JoeJoe
7,21421229
$endgroup$
HINT: Multiply the fraction by
$$
frac{bar z(bar z-1)}{bar z(bar z-1)};.
$$
In this way the denominator will be real and you should only look at the numerator, splitting it into its real and imaginary part.
answered Dec 16 '18 at 10:53
JoeJoe
7,21421229
answered Dec 16 '18 at 10:53
JoeJoe
7,21421229
answered Dec 16 '18 at 10:53
JoeJoe
7,21421229
answered Dec 16 '18 at 10:53
JoeJoe
7,21421229
7,21421229
add a comment |
add a comment |
$begingroup$
A little Partial Fraction Decomposition before rationalization of the denominator will reduce calculation.
$$dfrac{z+2}{z(z+1)}=dfrac1{z+1}+dfrac{2(z+1-z)}{z(z+1)}=dfrac2z-dfrac1{z+1}$$
Now the imaginary part of $dfrac2z$ is $-dfrac{2y}{x^2+y^2}$
and that of $dfrac1{z+1}$ is $-dfrac y{(x+1)^2+y^2}$
Finally, $dfrac{z+2}{z(z+1)}$ will be real if $-dfrac{2y}{x^2+y^2}=-dfrac y{(x+1)^2+y^2}$
$iff0=y(2x^2+2y^2+4x+2-x^2-y^2)=y{(x+2)^2+y^2-2}$
answered Dec 16 '18 at 12:32
lab bhattacharjeelab bhattacharjee
226k15158275
$endgroup$
add a comment |
$begingroup$
A little Partial Fraction Decomposition before rationalization of the denominator will reduce calculation.
$$dfrac{z+2}{z(z+1)}=dfrac1{z+1}+dfrac{2(z+1-z)}{z(z+1)}=dfrac2z-dfrac1{z+1}$$
Now the imaginary part of $dfrac2z$ is $-dfrac{2y}{x^2+y^2}$
and that of $dfrac1{z+1}$ is $-dfrac y{(x+1)^2+y^2}$
Finally, $dfrac{z+2}{z(z+1)}$ will be real if $-dfrac{2y}{x^2+y^2}=-dfrac y{(x+1)^2+y^2}$
$iff0=y(2x^2+2y^2+4x+2-x^2-y^2)=y{(x+2)^2+y^2-2}$
answered Dec 16 '18 at 12:32
lab bhattacharjeelab bhattacharjee
226k15158275
$endgroup$
add a comment |
$begingroup$
A little Partial Fraction Decomposition before rationalization of the denominator will reduce calculation.
$$dfrac{z+2}{z(z+1)}=dfrac1{z+1}+dfrac{2(z+1-z)}{z(z+1)}=dfrac2z-dfrac1{z+1}$$
Now the imaginary part of $dfrac2z$ is $-dfrac{2y}{x^2+y^2}$
and that of $dfrac1{z+1}$ is $-dfrac y{(x+1)^2+y^2}$
Finally, $dfrac{z+2}{z(z+1)}$ will be real if $-dfrac{2y}{x^2+y^2}=-dfrac y{(x+1)^2+y^2}$
$iff0=y(2x^2+2y^2+4x+2-x^2-y^2)=y{(x+2)^2+y^2-2}$
answered Dec 16 '18 at 12:32
lab bhattacharjeelab bhattacharjee
226k15158275
$endgroup$
A little Partial Fraction Decomposition before rationalization of the denominator will reduce calculation.
$$dfrac{z+2}{z(z+1)}=dfrac1{z+1}+dfrac{2(z+1-z)}{z(z+1)}=dfrac2z-dfrac1{z+1}$$
Now the imaginary part of $dfrac2z$ is $-dfrac{2y}{x^2+y^2}$
and that of $dfrac1{z+1}$ is $-dfrac y{(x+1)^2+y^2}$
Finally, $dfrac{z+2}{z(z+1)}$ will be real if $-dfrac{2y}{x^2+y^2}=-dfrac y{(x+1)^2+y^2}$
$iff0=y(2x^2+2y^2+4x+2-x^2-y^2)=y{(x+2)^2+y^2-2}$
answered Dec 16 '18 at 12:32
lab bhattacharjeelab bhattacharjee
226k15158275
answered Dec 16 '18 at 12:32
lab bhattacharjeelab bhattacharjee
226k15158275
answered Dec 16 '18 at 12:32
lab bhattacharjeelab bhattacharjee
226k15158275
answered Dec 16 '18 at 12:32
lab bhattacharjeelab bhattacharjee
226k15158275
226k15158275
add a comment |
add a comment |
$begingroup$
Using
$$
z = sqrt{x^2+y^2}e^{arctanfrac yx}
$$
we have
$$
frac{z+2}{z(z+1)} = frac{rho_1 e^{iphi_1}}{rho_2 e^{iphi_2}rho_3 e^{iphi_3}}
$$
and we seek for
$$
arctanfrac{y}{x+2}-arctanfrac yx -arctanfrac{y}{x+1} = 0
$$
then
$$
tanleft(arctanfrac{y}{x+2} -arctanfrac{y}{x+1}right) = frac yx
$$
or
$$
y(y^2+x^2+4x+2) = 0
$$
NOTE
$$
tan(a-b) = frac{tan a-tan b}{1+tan atan b}
$$
so
$$
tanleft(arctanfrac{y}{x+2} -arctanfrac{y}{x+1}right) = frac{y}{y^2+x^2+3x+2}
$$
etc.
answered Dec 16 '18 at 14:02
CesareoCesareo
9,3063517
$endgroup$
$begingroup$
How did you get that last step?
$endgroup$
– I like Serena
Dec 16 '18 at 14:37
2
$begingroup$
@IlikeSerena Please. See attached note.
$endgroup$
– Cesareo
Dec 16 '18 at 16:02
$begingroup$
@bluemaster Thanks for the hint. I have a real problem with addition (and subtraction)
$endgroup$
– Cesareo
Dec 19 '18 at 11:57
add a comment |
$begingroup$
Using
$$
z = sqrt{x^2+y^2}e^{arctanfrac yx}
$$
we have
$$
frac{z+2}{z(z+1)} = frac{rho_1 e^{iphi_1}}{rho_2 e^{iphi_2}rho_3 e^{iphi_3}}
$$
and we seek for
$$
arctanfrac{y}{x+2}-arctanfrac yx -arctanfrac{y}{x+1} = 0
$$
then
$$
tanleft(arctanfrac{y}{x+2} -arctanfrac{y}{x+1}right) = frac yx
$$
or
$$
y(y^2+x^2+4x+2) = 0
$$
NOTE
$$
tan(a-b) = frac{tan a-tan b}{1+tan atan b}
$$
so
$$
tanleft(arctanfrac{y}{x+2} -arctanfrac{y}{x+1}right) = frac{y}{y^2+x^2+3x+2}
$$
etc.
answered Dec 16 '18 at 14:02
CesareoCesareo
9,3063517
$endgroup$
$begingroup$
How did you get that last step?
$endgroup$
– I like Serena
Dec 16 '18 at 14:37
2
$begingroup$
@IlikeSerena Please. See attached note.
$endgroup$
– Cesareo
Dec 16 '18 at 16:02
$begingroup$
@bluemaster Thanks for the hint. I have a real problem with addition (and subtraction)
$endgroup$
– Cesareo
Dec 19 '18 at 11:57
add a comment |
$begingroup$
Using
$$
z = sqrt{x^2+y^2}e^{arctanfrac yx}
$$
we have
$$
frac{z+2}{z(z+1)} = frac{rho_1 e^{iphi_1}}{rho_2 e^{iphi_2}rho_3 e^{iphi_3}}
$$
and we seek for
$$
arctanfrac{y}{x+2}-arctanfrac yx -arctanfrac{y}{x+1} = 0
$$
then
$$
tanleft(arctanfrac{y}{x+2} -arctanfrac{y}{x+1}right) = frac yx
$$
or
$$
y(y^2+x^2+4x+2) = 0
$$
NOTE
$$
tan(a-b) = frac{tan a-tan b}{1+tan atan b}
$$
so
$$
tanleft(arctanfrac{y}{x+2} -arctanfrac{y}{x+1}right) = frac{y}{y^2+x^2+3x+2}
$$
etc.
answered Dec 16 '18 at 14:02
CesareoCesareo
9,3063517
$endgroup$
Using
$$
z = sqrt{x^2+y^2}e^{arctanfrac yx}
$$
we have
$$
frac{z+2}{z(z+1)} = frac{rho_1 e^{iphi_1}}{rho_2 e^{iphi_2}rho_3 e^{iphi_3}}
$$
and we seek for
$$
arctanfrac{y}{x+2}-arctanfrac yx -arctanfrac{y}{x+1} = 0
$$
then
$$
tanleft(arctanfrac{y}{x+2} -arctanfrac{y}{x+1}right) = frac yx
$$
or
$$
y(y^2+x^2+4x+2) = 0
$$
NOTE
$$
tan(a-b) = frac{tan a-tan b}{1+tan atan b}
$$
so
$$
tanleft(arctanfrac{y}{x+2} -arctanfrac{y}{x+1}right) = frac{y}{y^2+x^2+3x+2}
$$
etc.
answered Dec 16 '18 at 14:02
CesareoCesareo
9,3063517
edited Dec 19 '18 at 11:53
answered Dec 16 '18 at 14:02
CesareoCesareo
9,3063517
answered Dec 16 '18 at 14:02
CesareoCesareo
9,3063517
answered Dec 16 '18 at 14:02
CesareoCesareo
9,3063517
9,3063517
$begingroup$
How did you get that last step?
$endgroup$
– I like Serena
Dec 16 '18 at 14:37
2
$begingroup$
@IlikeSerena Please. See attached note.
$endgroup$
– Cesareo
Dec 16 '18 at 16:02
$begingroup$
@bluemaster Thanks for the hint. I have a real problem with addition (and subtraction)
$endgroup$
– Cesareo
Dec 19 '18 at 11:57
add a comment |
$begingroup$
How did you get that last step?
$endgroup$
– I like Serena
Dec 16 '18 at 14:37
2
$begingroup$
@IlikeSerena Please. See attached note.
$endgroup$
– Cesareo
Dec 16 '18 at 16:02
$begingroup$
@bluemaster Thanks for the hint. I have a real problem with addition (and subtraction)
$endgroup$
– Cesareo
Dec 19 '18 at 11:57
$begingroup$
How did you get that last step?
$endgroup$
– I like Serena
Dec 16 '18 at 14:37
$begingroup$
How did you get that last step?
$endgroup$
– I like Serena
Dec 16 '18 at 14:37
2
2
$begingroup$
@IlikeSerena Please. See attached note.
$endgroup$
– Cesareo
Dec 16 '18 at 16:02
$begingroup$
@IlikeSerena Please. See attached note.
$endgroup$
– Cesareo
Dec 16 '18 at 16:02
$begingroup$
@bluemaster Thanks for the hint. I have a real problem with addition (and subtraction)
$endgroup$
– Cesareo
Dec 19 '18 at 11:57
$begingroup$
@bluemaster Thanks for the hint. I have a real problem with addition (and subtraction)
$endgroup$
– Cesareo
Dec 19 '18 at 11:57
add a comment |
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