Mixing
Find the complex numbers $z$ such that $w=frac{2z-1}{2+iz}$ is real
$begingroup$
Find the complex numbers $z$ such that $w=dfrac{2z-1}{2+iz}$ is real.
I have been trying to separate the imaginary part from the real one, so I can cancel the imaginary one.
Trouble is that, by manipulating $w$, it just seems to get much worse.
Any path to follow?
complex-numbers
edited Feb 3 at 17:27
Did
248k23227466
asked Jan 30 at 4:49
Sebas Martinez SantosSebas Martinez Santos
667
$endgroup$
add a comment |
$begingroup$
Find the complex numbers $z$ such that $w=dfrac{2z-1}{2+iz}$ is real.
I have been trying to separate the imaginary part from the real one, so I can cancel the imaginary one.
Trouble is that, by manipulating $w$, it just seems to get much worse.
Any path to follow?
complex-numbers
edited Feb 3 at 17:27
Did
248k23227466
asked Jan 30 at 4:49
Sebas Martinez SantosSebas Martinez Santos
667
$endgroup$
2
$begingroup$
Hint: let $z=x+iy$...
$endgroup$
– Sean Roberson
Jan 30 at 4:53
$begingroup$
Thanks. I already tried it but doesn't seem to get any simpler. I tried to multiply the $w$ by its conjugate
$endgroup$
– Sebas Martinez Santos
Jan 30 at 5:07
$begingroup$
$$iff w=bar w$$
$endgroup$
– lab bhattacharjee
Jan 30 at 5:18
add a comment |
$begingroup$
Find the complex numbers $z$ such that $w=dfrac{2z-1}{2+iz}$ is real.
I have been trying to separate the imaginary part from the real one, so I can cancel the imaginary one.
Trouble is that, by manipulating $w$, it just seems to get much worse.
Any path to follow?
complex-numbers
edited Feb 3 at 17:27
Did
248k23227466
asked Jan 30 at 4:49
Sebas Martinez SantosSebas Martinez Santos
667
$endgroup$
Find the complex numbers $z$ such that $w=dfrac{2z-1}{2+iz}$ is real.
I have been trying to separate the imaginary part from the real one, so I can cancel the imaginary one.
Trouble is that, by manipulating $w$, it just seems to get much worse.
Any path to follow?
complex-numbers
complex-numbers
edited Feb 3 at 17:27
Did
248k23227466
asked Jan 30 at 4:49
Sebas Martinez SantosSebas Martinez Santos
667
edited Feb 3 at 17:27
Did
248k23227466
asked Jan 30 at 4:49
Sebas Martinez SantosSebas Martinez Santos
667
edited Feb 3 at 17:27
Did
248k23227466
edited Feb 3 at 17:27
Did
248k23227466
edited Feb 3 at 17:27
Did
248k23227466
248k23227466
asked Jan 30 at 4:49
Sebas Martinez SantosSebas Martinez Santos
667
asked Jan 30 at 4:49
Sebas Martinez SantosSebas Martinez Santos
667
asked Jan 30 at 4:49
Sebas Martinez SantosSebas Martinez Santos
667
667
2
$begingroup$
Hint: let $z=x+iy$...
$endgroup$
– Sean Roberson
Jan 30 at 4:53
$begingroup$
Thanks. I already tried it but doesn't seem to get any simpler. I tried to multiply the $w$ by its conjugate
$endgroup$
– Sebas Martinez Santos
Jan 30 at 5:07
$begingroup$
$$iff w=bar w$$
$endgroup$
– lab bhattacharjee
Jan 30 at 5:18
add a comment |
2
$begingroup$
Hint: let $z=x+iy$...
$endgroup$
– Sean Roberson
Jan 30 at 4:53
$begingroup$
Thanks. I already tried it but doesn't seem to get any simpler. I tried to multiply the $w$ by its conjugate
$endgroup$
– Sebas Martinez Santos
Jan 30 at 5:07
$begingroup$
$$iff w=bar w$$
$endgroup$
– lab bhattacharjee
Jan 30 at 5:18
2
2
$begingroup$
Hint: let $z=x+iy$...
$endgroup$
– Sean Roberson
Jan 30 at 4:53
$begingroup$
Hint: let $z=x+iy$...
$endgroup$
– Sean Roberson
Jan 30 at 4:53
$begingroup$
Thanks. I already tried it but doesn't seem to get any simpler. I tried to multiply the $w$ by its conjugate
$endgroup$
– Sebas Martinez Santos
Jan 30 at 5:07
$begingroup$
Thanks. I already tried it but doesn't seem to get any simpler. I tried to multiply the $w$ by its conjugate
$endgroup$
– Sebas Martinez Santos
Jan 30 at 5:07
$begingroup$
$$iff w=bar w$$
$endgroup$
– lab bhattacharjee
Jan 30 at 5:18
$begingroup$
$$iff w=bar w$$
$endgroup$
– lab bhattacharjee
Jan 30 at 5:18
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$$w=frac{(2x-1)+iy}{(2-y)+ix}=frac{[(2x-1)+iy][(2-y)-ix]}{(2-y)^2+x^2}$$
Multiply by the numerator to get
$$(2x-1)(2-y)-i(2x^2-x)-i(2y-y^2)-xy$$
The imaginary part is $x-2x^2+y^2-2y$. Setting it equal to zero, you can conplete the squares on $x$ and $y$ to maybe get a familiar geometric shape....
$$-2left(x^2-frac{1}{2}x+frac{1}{16}right)+(y^2-2y+1)+frac{1}{8}-1=0$$
$$(y-1)^2-2left(x-frac{1}{4}right)^2=frac{7}{8}$$
answered Jan 30 at 5:16
Eleven-ElevenEleven-Eleven
5,78872759
$endgroup$
add a comment |
$begingroup$
HINT:
$$frac{2z-1}{2+iz}=frac{2z-1}{2+iz},frac{2-ibar z}{2- ibar z}=frac{4z-2-i2|z|^2-ibar z}{4+|z|^2+4text{Im}(z)}$$
Now, answer the question "what values of $z$ makes the numerator real?"
answered Jan 30 at 5:19
Mark ViolaMark Viola
134k1278176
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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votes
$begingroup$
$$w=frac{(2x-1)+iy}{(2-y)+ix}=frac{[(2x-1)+iy][(2-y)-ix]}{(2-y)^2+x^2}$$
Multiply by the numerator to get
$$(2x-1)(2-y)-i(2x^2-x)-i(2y-y^2)-xy$$
The imaginary part is $x-2x^2+y^2-2y$. Setting it equal to zero, you can conplete the squares on $x$ and $y$ to maybe get a familiar geometric shape....
$$-2left(x^2-frac{1}{2}x+frac{1}{16}right)+(y^2-2y+1)+frac{1}{8}-1=0$$
$$(y-1)^2-2left(x-frac{1}{4}right)^2=frac{7}{8}$$
answered Jan 30 at 5:16
Eleven-ElevenEleven-Eleven
5,78872759
$endgroup$
add a comment |
$begingroup$
$$w=frac{(2x-1)+iy}{(2-y)+ix}=frac{[(2x-1)+iy][(2-y)-ix]}{(2-y)^2+x^2}$$
Multiply by the numerator to get
$$(2x-1)(2-y)-i(2x^2-x)-i(2y-y^2)-xy$$
The imaginary part is $x-2x^2+y^2-2y$. Setting it equal to zero, you can conplete the squares on $x$ and $y$ to maybe get a familiar geometric shape....
$$-2left(x^2-frac{1}{2}x+frac{1}{16}right)+(y^2-2y+1)+frac{1}{8}-1=0$$
$$(y-1)^2-2left(x-frac{1}{4}right)^2=frac{7}{8}$$
answered Jan 30 at 5:16
Eleven-ElevenEleven-Eleven
5,78872759
$endgroup$
add a comment |
$begingroup$
$$w=frac{(2x-1)+iy}{(2-y)+ix}=frac{[(2x-1)+iy][(2-y)-ix]}{(2-y)^2+x^2}$$
Multiply by the numerator to get
$$(2x-1)(2-y)-i(2x^2-x)-i(2y-y^2)-xy$$
The imaginary part is $x-2x^2+y^2-2y$. Setting it equal to zero, you can conplete the squares on $x$ and $y$ to maybe get a familiar geometric shape....
$$-2left(x^2-frac{1}{2}x+frac{1}{16}right)+(y^2-2y+1)+frac{1}{8}-1=0$$
$$(y-1)^2-2left(x-frac{1}{4}right)^2=frac{7}{8}$$
answered Jan 30 at 5:16
Eleven-ElevenEleven-Eleven
5,78872759
$endgroup$
$$w=frac{(2x-1)+iy}{(2-y)+ix}=frac{[(2x-1)+iy][(2-y)-ix]}{(2-y)^2+x^2}$$
Multiply by the numerator to get
$$(2x-1)(2-y)-i(2x^2-x)-i(2y-y^2)-xy$$
The imaginary part is $x-2x^2+y^2-2y$. Setting it equal to zero, you can conplete the squares on $x$ and $y$ to maybe get a familiar geometric shape....
$$-2left(x^2-frac{1}{2}x+frac{1}{16}right)+(y^2-2y+1)+frac{1}{8}-1=0$$
$$(y-1)^2-2left(x-frac{1}{4}right)^2=frac{7}{8}$$
answered Jan 30 at 5:16
Eleven-ElevenEleven-Eleven
5,78872759
edited Jan 30 at 18:00
answered Jan 30 at 5:16
Eleven-ElevenEleven-Eleven
5,78872759
answered Jan 30 at 5:16
Eleven-ElevenEleven-Eleven
5,78872759
answered Jan 30 at 5:16
Eleven-ElevenEleven-Eleven
5,78872759
5,78872759
add a comment |
add a comment |
$begingroup$
HINT:
$$frac{2z-1}{2+iz}=frac{2z-1}{2+iz},frac{2-ibar z}{2- ibar z}=frac{4z-2-i2|z|^2-ibar z}{4+|z|^2+4text{Im}(z)}$$
Now, answer the question "what values of $z$ makes the numerator real?"
answered Jan 30 at 5:19
Mark ViolaMark Viola
134k1278176
$endgroup$
add a comment |
$begingroup$
HINT:
$$frac{2z-1}{2+iz}=frac{2z-1}{2+iz},frac{2-ibar z}{2- ibar z}=frac{4z-2-i2|z|^2-ibar z}{4+|z|^2+4text{Im}(z)}$$
Now, answer the question "what values of $z$ makes the numerator real?"
answered Jan 30 at 5:19
Mark ViolaMark Viola
134k1278176
$endgroup$
add a comment |
$begingroup$
HINT:
$$frac{2z-1}{2+iz}=frac{2z-1}{2+iz},frac{2-ibar z}{2- ibar z}=frac{4z-2-i2|z|^2-ibar z}{4+|z|^2+4text{Im}(z)}$$
Now, answer the question "what values of $z$ makes the numerator real?"
answered Jan 30 at 5:19
Mark ViolaMark Viola
134k1278176
$endgroup$
HINT:
$$frac{2z-1}{2+iz}=frac{2z-1}{2+iz},frac{2-ibar z}{2- ibar z}=frac{4z-2-i2|z|^2-ibar z}{4+|z|^2+4text{Im}(z)}$$
Now, answer the question "what values of $z$ makes the numerator real?"
answered Jan 30 at 5:19
Mark ViolaMark Viola
134k1278176
answered Jan 30 at 5:19
Mark ViolaMark Viola
134k1278176
answered Jan 30 at 5:19
Mark ViolaMark Viola
134k1278176
answered Jan 30 at 5:19
Mark ViolaMark Viola
134k1278176
134k1278176
add a comment |
add a comment |
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2
$begingroup$
Hint: let $z=x+iy$...
$endgroup$
– Sean Roberson
Jan 30 at 4:53
$begingroup$
Thanks. I already tried it but doesn't seem to get any simpler. I tried to multiply the $w$ by its conjugate
$endgroup$
– Sebas Martinez Santos
Jan 30 at 5:07
$begingroup$
$$iff w=bar w$$
$endgroup$
– lab bhattacharjee
Jan 30 at 5:18