Mixing
point cloud in complex plane
$begingroup$
I want to draw the point cloud represented by the following term.
$$M_{4}=left{z in mathbb{C} : | z-1left|=frac{1}{2}right| z-j|right}$$
$j$ equals $i$, the imaginary square root of $-1$.
I have made several attempts to get a solution for the equation. This is the one that looks the most promising. I don't really have an approach on how to continue. I think I have to generate a $j$, but I haven't seen the right way to do so. Thank you in advance.
$ |x+j y-1|=frac{1}{2}|x+j y-j| $
$ |(x-1)+jy|=frac{1}{2}|x+(j y-j)| $
$ sqrt{(x-1)^2+jy^2}=frac{1}{4}sqrt{x^2+(j y-j)^2} $
$ (x-1)^2+jy^2=frac{1}{4}(x^2+(j y-j)^2) $
$ x^2 -2x +1 -y^2=frac{1}{4}(x^2+(jy^2 -2jy^2 +j^2) $
$ x^2 -2x +1 -y^2=frac{1}{4}(x^2-y^2 +2y^2 -1) $
$ frac{3} {4}x^2 - 2x + frac{5}{4}y^2 - frac{1}{2}y = 0$
complex-numbers
edited Jan 28 at 22:26
hardmath
29.2k953101
asked Jan 25 at 22:42
maxiangelomaxiangelo
12
$endgroup$
add a comment |
$begingroup$
I want to draw the point cloud represented by the following term.
$$M_{4}=left{z in mathbb{C} : | z-1left|=frac{1}{2}right| z-j|right}$$
$j$ equals $i$, the imaginary square root of $-1$.
I have made several attempts to get a solution for the equation. This is the one that looks the most promising. I don't really have an approach on how to continue. I think I have to generate a $j$, but I haven't seen the right way to do so. Thank you in advance.
$ |x+j y-1|=frac{1}{2}|x+j y-j| $
$ |(x-1)+jy|=frac{1}{2}|x+(j y-j)| $
$ sqrt{(x-1)^2+jy^2}=frac{1}{4}sqrt{x^2+(j y-j)^2} $
$ (x-1)^2+jy^2=frac{1}{4}(x^2+(j y-j)^2) $
$ x^2 -2x +1 -y^2=frac{1}{4}(x^2+(jy^2 -2jy^2 +j^2) $
$ x^2 -2x +1 -y^2=frac{1}{4}(x^2-y^2 +2y^2 -1) $
$ frac{3} {4}x^2 - 2x + frac{5}{4}y^2 - frac{1}{2}y = 0$
complex-numbers
edited Jan 28 at 22:26
hardmath
29.2k953101
asked Jan 25 at 22:42
maxiangelomaxiangelo
12
$endgroup$
$begingroup$
I have tried to edit your post. Please review to check if I unintentionally changed your meaning. A particular point to clarify is what you meant by "generating a $j$", which is a constant if you mean the square root of $-1$. Perhaps you meant "generating a $z$" instead?
$endgroup$
– hardmath
Jan 28 at 22:29
add a comment |
$begingroup$
I want to draw the point cloud represented by the following term.
$$M_{4}=left{z in mathbb{C} : | z-1left|=frac{1}{2}right| z-j|right}$$
$j$ equals $i$, the imaginary square root of $-1$.
I have made several attempts to get a solution for the equation. This is the one that looks the most promising. I don't really have an approach on how to continue. I think I have to generate a $j$, but I haven't seen the right way to do so. Thank you in advance.
$ |x+j y-1|=frac{1}{2}|x+j y-j| $
$ |(x-1)+jy|=frac{1}{2}|x+(j y-j)| $
$ sqrt{(x-1)^2+jy^2}=frac{1}{4}sqrt{x^2+(j y-j)^2} $
$ (x-1)^2+jy^2=frac{1}{4}(x^2+(j y-j)^2) $
$ x^2 -2x +1 -y^2=frac{1}{4}(x^2+(jy^2 -2jy^2 +j^2) $
$ x^2 -2x +1 -y^2=frac{1}{4}(x^2-y^2 +2y^2 -1) $
$ frac{3} {4}x^2 - 2x + frac{5}{4}y^2 - frac{1}{2}y = 0$
complex-numbers
edited Jan 28 at 22:26
hardmath
29.2k953101
asked Jan 25 at 22:42
maxiangelomaxiangelo
12
$endgroup$
I want to draw the point cloud represented by the following term.
$$M_{4}=left{z in mathbb{C} : | z-1left|=frac{1}{2}right| z-j|right}$$
$j$ equals $i$, the imaginary square root of $-1$.
I have made several attempts to get a solution for the equation. This is the one that looks the most promising. I don't really have an approach on how to continue. I think I have to generate a $j$, but I haven't seen the right way to do so. Thank you in advance.
$ |x+j y-1|=frac{1}{2}|x+j y-j| $
$ |(x-1)+jy|=frac{1}{2}|x+(j y-j)| $
$ sqrt{(x-1)^2+jy^2}=frac{1}{4}sqrt{x^2+(j y-j)^2} $
$ (x-1)^2+jy^2=frac{1}{4}(x^2+(j y-j)^2) $
$ x^2 -2x +1 -y^2=frac{1}{4}(x^2+(jy^2 -2jy^2 +j^2) $
$ x^2 -2x +1 -y^2=frac{1}{4}(x^2-y^2 +2y^2 -1) $
$ frac{3} {4}x^2 - 2x + frac{5}{4}y^2 - frac{1}{2}y = 0$
complex-numbers
complex-numbers
edited Jan 28 at 22:26
hardmath
29.2k953101
asked Jan 25 at 22:42
maxiangelomaxiangelo
12
edited Jan 28 at 22:26
hardmath
29.2k953101
asked Jan 25 at 22:42
maxiangelomaxiangelo
12
edited Jan 28 at 22:26
hardmath
29.2k953101
edited Jan 28 at 22:26
hardmath
29.2k953101
edited Jan 28 at 22:26
hardmath
29.2k953101
29.2k953101
asked Jan 25 at 22:42
maxiangelomaxiangelo
12
asked Jan 25 at 22:42
maxiangelomaxiangelo
12
asked Jan 25 at 22:42
maxiangelomaxiangelo
12
12
$begingroup$
I have tried to edit your post. Please review to check if I unintentionally changed your meaning. A particular point to clarify is what you meant by "generating a $j$", which is a constant if you mean the square root of $-1$. Perhaps you meant "generating a $z$" instead?
$endgroup$
– hardmath
Jan 28 at 22:29
add a comment |
$begingroup$
I have tried to edit your post. Please review to check if I unintentionally changed your meaning. A particular point to clarify is what you meant by "generating a $j$", which is a constant if you mean the square root of $-1$. Perhaps you meant "generating a $z$" instead?
$endgroup$
– hardmath
Jan 28 at 22:29
$begingroup$
I have tried to edit your post. Please review to check if I unintentionally changed your meaning. A particular point to clarify is what you meant by "generating a $j$", which is a constant if you mean the square root of $-1$. Perhaps you meant "generating a $z$" instead?
$endgroup$
– hardmath
Jan 28 at 22:29
$begingroup$
I have tried to edit your post. Please review to check if I unintentionally changed your meaning. A particular point to clarify is what you meant by "generating a $j$", which is a constant if you mean the square root of $-1$. Perhaps you meant "generating a $z$" instead?
$endgroup$
– hardmath
Jan 28 at 22:29
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I am sorry if I am giving you a bum steer. I believe by point cloud we are looking for the locus of points described by the above equation. I don't know TEX. Six steps to solution. First, multiply both sides by of (z+i). Simplify. Second, add (iz)^2 to both sides. Simplify. Third, add the imaginary unit to both sides and simplify. Fourth, bring all terms involving z over to the left hand side of the equation and factor out z. Fifth, divide both sides by the constant coefficient of z. Sixth, and lastly, rationalize the denominator on the right hand side by multiplying top and bottom by (1-i). The answer I get is z equals modulus of [(3+i)/4]. I am not certain this is the right answer. Sorry if I am wrong.
answered Jan 27 at 6:19
Richard Anthony BaumRichard Anthony Baum
413
$endgroup$
$begingroup$
@hardmath thanks for the clarification. I was off base, not knowing my material.
$endgroup$
– Richard Anthony Baum
Jan 30 at 6:22
$begingroup$
In the third line of your solution, the third term on the left hand side should read (iy)^2
$endgroup$
– Richard Anthony Baum
Feb 20 at 11:53
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I am sorry if I am giving you a bum steer. I believe by point cloud we are looking for the locus of points described by the above equation. I don't know TEX. Six steps to solution. First, multiply both sides by of (z+i). Simplify. Second, add (iz)^2 to both sides. Simplify. Third, add the imaginary unit to both sides and simplify. Fourth, bring all terms involving z over to the left hand side of the equation and factor out z. Fifth, divide both sides by the constant coefficient of z. Sixth, and lastly, rationalize the denominator on the right hand side by multiplying top and bottom by (1-i). The answer I get is z equals modulus of [(3+i)/4]. I am not certain this is the right answer. Sorry if I am wrong.
answered Jan 27 at 6:19
Richard Anthony BaumRichard Anthony Baum
413
$endgroup$
$begingroup$
@hardmath thanks for the clarification. I was off base, not knowing my material.
$endgroup$
– Richard Anthony Baum
Jan 30 at 6:22
$begingroup$
In the third line of your solution, the third term on the left hand side should read (iy)^2
$endgroup$
– Richard Anthony Baum
Feb 20 at 11:53
add a comment |
$begingroup$
I am sorry if I am giving you a bum steer. I believe by point cloud we are looking for the locus of points described by the above equation. I don't know TEX. Six steps to solution. First, multiply both sides by of (z+i). Simplify. Second, add (iz)^2 to both sides. Simplify. Third, add the imaginary unit to both sides and simplify. Fourth, bring all terms involving z over to the left hand side of the equation and factor out z. Fifth, divide both sides by the constant coefficient of z. Sixth, and lastly, rationalize the denominator on the right hand side by multiplying top and bottom by (1-i). The answer I get is z equals modulus of [(3+i)/4]. I am not certain this is the right answer. Sorry if I am wrong.
answered Jan 27 at 6:19
Richard Anthony BaumRichard Anthony Baum
413
$endgroup$
$begingroup$
@hardmath thanks for the clarification. I was off base, not knowing my material.
$endgroup$
– Richard Anthony Baum
Jan 30 at 6:22
$begingroup$
In the third line of your solution, the third term on the left hand side should read (iy)^2
$endgroup$
– Richard Anthony Baum
Feb 20 at 11:53
add a comment |
$begingroup$
I am sorry if I am giving you a bum steer. I believe by point cloud we are looking for the locus of points described by the above equation. I don't know TEX. Six steps to solution. First, multiply both sides by of (z+i). Simplify. Second, add (iz)^2 to both sides. Simplify. Third, add the imaginary unit to both sides and simplify. Fourth, bring all terms involving z over to the left hand side of the equation and factor out z. Fifth, divide both sides by the constant coefficient of z. Sixth, and lastly, rationalize the denominator on the right hand side by multiplying top and bottom by (1-i). The answer I get is z equals modulus of [(3+i)/4]. I am not certain this is the right answer. Sorry if I am wrong.
answered Jan 27 at 6:19
Richard Anthony BaumRichard Anthony Baum
413
$endgroup$
I am sorry if I am giving you a bum steer. I believe by point cloud we are looking for the locus of points described by the above equation. I don't know TEX. Six steps to solution. First, multiply both sides by of (z+i). Simplify. Second, add (iz)^2 to both sides. Simplify. Third, add the imaginary unit to both sides and simplify. Fourth, bring all terms involving z over to the left hand side of the equation and factor out z. Fifth, divide both sides by the constant coefficient of z. Sixth, and lastly, rationalize the denominator on the right hand side by multiplying top and bottom by (1-i). The answer I get is z equals modulus of [(3+i)/4]. I am not certain this is the right answer. Sorry if I am wrong.
answered Jan 27 at 6:19
Richard Anthony BaumRichard Anthony Baum
413
edited Feb 14 at 6:41
answered Jan 27 at 6:19
Richard Anthony BaumRichard Anthony Baum
413
answered Jan 27 at 6:19
Richard Anthony BaumRichard Anthony Baum
413
answered Jan 27 at 6:19
Richard Anthony BaumRichard Anthony Baum
413
413
$begingroup$
@hardmath thanks for the clarification. I was off base, not knowing my material.
$endgroup$
– Richard Anthony Baum
Jan 30 at 6:22
$begingroup$
In the third line of your solution, the third term on the left hand side should read (iy)^2
$endgroup$
– Richard Anthony Baum
Feb 20 at 11:53
add a comment |
$begingroup$
@hardmath thanks for the clarification. I was off base, not knowing my material.
$endgroup$
– Richard Anthony Baum
Jan 30 at 6:22
$begingroup$
In the third line of your solution, the third term on the left hand side should read (iy)^2
$endgroup$
– Richard Anthony Baum
Feb 20 at 11:53
$begingroup$
@hardmath thanks for the clarification. I was off base, not knowing my material.
$endgroup$
– Richard Anthony Baum
Jan 30 at 6:22
$begingroup$
@hardmath thanks for the clarification. I was off base, not knowing my material.
$endgroup$
– Richard Anthony Baum
Jan 30 at 6:22
$begingroup$
In the third line of your solution, the third term on the left hand side should read (iy)^2
$endgroup$
– Richard Anthony Baum
Feb 20 at 11:53
$begingroup$
In the third line of your solution, the third term on the left hand side should read (iy)^2
$endgroup$
– Richard Anthony Baum
Feb 20 at 11:53
add a comment |
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$begingroup$
I have tried to edit your post. Please review to check if I unintentionally changed your meaning. A particular point to clarify is what you meant by "generating a $j$", which is a constant if you mean the square root of $-1$. Perhaps you meant "generating a $z$" instead?
$endgroup$
– hardmath
Jan 28 at 22:29