Mixing
Greatest value of $|z|$ such that $|z-2i|le2$ and $ 0le arg(z+2)le 45^circ$
$begingroup$
I've to sketch the complex number $z$ such that it satisfy both the inequality
$|(z-2i)|le2$ and $ 0le arg(z+2)le 45^circ$
I was able to sketch and shade the region that satisfies both inequality, Please see the below link.
Sketch on Argand diagram
I've a problem in getting the greatest value of $|z|$ (maximum length of $z$ from $(0,0),$ it should come around 3.70)
complex-numbers
edited Sep 18 '18 at 13:14
Michael Hardy
1
asked Oct 11 '14 at 3:22
ArifArif
1261515
$endgroup$
add a comment |
$begingroup$
I've to sketch the complex number $z$ such that it satisfy both the inequality
$|(z-2i)|le2$ and $ 0le arg(z+2)le 45^circ$
I was able to sketch and shade the region that satisfies both inequality, Please see the below link.
Sketch on Argand diagram
I've a problem in getting the greatest value of $|z|$ (maximum length of $z$ from $(0,0),$ it should come around 3.70)
complex-numbers
edited Sep 18 '18 at 13:14
Michael Hardy
1
asked Oct 11 '14 at 3:22
ArifArif
1261515
$endgroup$
$begingroup$
looks ok to me. what is your problem with $|z|$?
$endgroup$
– David Holden
Oct 11 '14 at 3:28
$begingroup$
The place you've marked an x on seems good, call it $z$. As you indicated in the diagram, $z$ is the 45 degree point on the radius 2 circle, which should be $(sqrt{2},sqrt{2})$, but we've shifted it up by $2$, so $z = sqrt{2} + (2+sqrt{2})i$. Alternatively, writing $z = x+iy$, we have $y = x + 2$ as well as $y = sqrt{4-x^2}+2$.
$endgroup$
– Platehead
Oct 11 '14 at 3:37
$begingroup$
I want to find greatest length(its distance from (0,0) ) of $|z|$ it should be around 3.70 units, unable to understand how to work it out.
$endgroup$
– Arif
Oct 11 '14 at 4:26
add a comment |
$begingroup$
I've to sketch the complex number $z$ such that it satisfy both the inequality
$|(z-2i)|le2$ and $ 0le arg(z+2)le 45^circ$
I was able to sketch and shade the region that satisfies both inequality, Please see the below link.
Sketch on Argand diagram
I've a problem in getting the greatest value of $|z|$ (maximum length of $z$ from $(0,0),$ it should come around 3.70)
complex-numbers
edited Sep 18 '18 at 13:14
Michael Hardy
1
asked Oct 11 '14 at 3:22
ArifArif
1261515
$endgroup$
I've to sketch the complex number $z$ such that it satisfy both the inequality
$|(z-2i)|le2$ and $ 0le arg(z+2)le 45^circ$
I was able to sketch and shade the region that satisfies both inequality, Please see the below link.
Sketch on Argand diagram
I've a problem in getting the greatest value of $|z|$ (maximum length of $z$ from $(0,0),$ it should come around 3.70)
complex-numbers
complex-numbers
edited Sep 18 '18 at 13:14
Michael Hardy
1
asked Oct 11 '14 at 3:22
ArifArif
1261515
edited Sep 18 '18 at 13:14
Michael Hardy
1
asked Oct 11 '14 at 3:22
ArifArif
1261515
edited Sep 18 '18 at 13:14
Michael Hardy
1
edited Sep 18 '18 at 13:14
Michael Hardy
1
edited Sep 18 '18 at 13:14
Michael Hardy
1
1
asked Oct 11 '14 at 3:22
ArifArif
1261515
asked Oct 11 '14 at 3:22
ArifArif
1261515
asked Oct 11 '14 at 3:22
ArifArif
1261515
1261515
$begingroup$
looks ok to me. what is your problem with $|z|$?
$endgroup$
– David Holden
Oct 11 '14 at 3:28
$begingroup$
The place you've marked an x on seems good, call it $z$. As you indicated in the diagram, $z$ is the 45 degree point on the radius 2 circle, which should be $(sqrt{2},sqrt{2})$, but we've shifted it up by $2$, so $z = sqrt{2} + (2+sqrt{2})i$. Alternatively, writing $z = x+iy$, we have $y = x + 2$ as well as $y = sqrt{4-x^2}+2$.
$endgroup$
– Platehead
Oct 11 '14 at 3:37
$begingroup$
I want to find greatest length(its distance from (0,0) ) of $|z|$ it should be around 3.70 units, unable to understand how to work it out.
$endgroup$
– Arif
Oct 11 '14 at 4:26
add a comment |
$begingroup$
looks ok to me. what is your problem with $|z|$?
$endgroup$
– David Holden
Oct 11 '14 at 3:28
$begingroup$
The place you've marked an x on seems good, call it $z$. As you indicated in the diagram, $z$ is the 45 degree point on the radius 2 circle, which should be $(sqrt{2},sqrt{2})$, but we've shifted it up by $2$, so $z = sqrt{2} + (2+sqrt{2})i$. Alternatively, writing $z = x+iy$, we have $y = x + 2$ as well as $y = sqrt{4-x^2}+2$.
$endgroup$
– Platehead
Oct 11 '14 at 3:37
$begingroup$
I want to find greatest length(its distance from (0,0) ) of $|z|$ it should be around 3.70 units, unable to understand how to work it out.
$endgroup$
– Arif
Oct 11 '14 at 4:26
$begingroup$
looks ok to me. what is your problem with $|z|$?
$endgroup$
– David Holden
Oct 11 '14 at 3:28
$begingroup$
looks ok to me. what is your problem with $|z|$?
$endgroup$
– David Holden
Oct 11 '14 at 3:28
$begingroup$
The place you've marked an x on seems good, call it $z$. As you indicated in the diagram, $z$ is the 45 degree point on the radius 2 circle, which should be $(sqrt{2},sqrt{2})$, but we've shifted it up by $2$, so $z = sqrt{2} + (2+sqrt{2})i$. Alternatively, writing $z = x+iy$, we have $y = x + 2$ as well as $y = sqrt{4-x^2}+2$.
$endgroup$
– Platehead
Oct 11 '14 at 3:37
$begingroup$
The place you've marked an x on seems good, call it $z$. As you indicated in the diagram, $z$ is the 45 degree point on the radius 2 circle, which should be $(sqrt{2},sqrt{2})$, but we've shifted it up by $2$, so $z = sqrt{2} + (2+sqrt{2})i$. Alternatively, writing $z = x+iy$, we have $y = x + 2$ as well as $y = sqrt{4-x^2}+2$.
$endgroup$
– Platehead
Oct 11 '14 at 3:37
$begingroup$
I want to find greatest length(its distance from (0,0) ) of $|z|$ it should be around 3.70 units, unable to understand how to work it out.
$endgroup$
– Arif
Oct 11 '14 at 4:26
$begingroup$
I want to find greatest length(its distance from (0,0) ) of $|z|$ it should be around 3.70 units, unable to understand how to work it out.
$endgroup$
– Arif
Oct 11 '14 at 4:26
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
if you parametrize the circle of constraint it is $(2cos t,2sin t +2)$ with $t=0$ corresponding to the point $(2,2)$. so:
$$
frac18 |z|^2 = frac12 (cos^2 t +sin^2 t + 2 sin t +1 )= 1+ sin t
$$
answered Oct 11 '14 at 3:41
David HoldenDavid Holden
14.7k21224
$endgroup$
add a comment |
$begingroup$
Inspection of the figure indicates that the point $z_0:=sqrt{2}+i(2+sqrt{2})$ is the feasible point with largest absolute value: The closed disk $D_0$ with center $0$ and radius $|z_0|$ obviously contains the complete feasible region $F$, and $z_0in F$. It follows that
$$max_{zin F}|z|=|z_0|=sqrt{8+4sqrt{2}}doteq3.6955 .$$
answered Nov 14 '16 at 11:09
Christian BlatterChristian Blatter
172k7113326
$endgroup$
add a comment |
$begingroup$
Let $0+2i$ be C (the centre of the circle), let the point of furthest intersection (greatest $|z|$) of the 2 loci be A (You have indicated it on your diagram with a cross), construct the point B such that $triangle CAB$ has a right angle at B.
From here we can note that $angle CAB=frac{pi}{4}$(given - construction of CB is such that it is parallel to the x-axis, use corresponding angles), $AC=2$ (radii). Using trigonometry we can find that $BC=2cos(frac{pi}{4})=sqrt{2}$, similarly $AB=2sin(frac{pi}{4})=sqrt{2}$.
Applying our knowledge to the diagram (the Argand diagram), we see that the real part of the desired complex number is given by $BC$, $Re(z)=sqrt{2}$. The imaginary part is the sum of $AB$ and the height (distance) from the number $sqrt{2}$ to $sqrt{2}+2i$ (distance is $2$), thus $Im(z)=BC+2=2+sqrt{2}$.
$therefore$ $z=sqrt{2}+i(2+sqrt{2})$ and $|z|=2sqrt{2+sqrt{2}}$
answered Apr 3 '17 at 19:59
frog1944frog1944
1,28711322
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
if you parametrize the circle of constraint it is $(2cos t,2sin t +2)$ with $t=0$ corresponding to the point $(2,2)$. so:
$$
frac18 |z|^2 = frac12 (cos^2 t +sin^2 t + 2 sin t +1 )= 1+ sin t
$$
answered Oct 11 '14 at 3:41
David HoldenDavid Holden
14.7k21224
$endgroup$
add a comment |
$begingroup$
if you parametrize the circle of constraint it is $(2cos t,2sin t +2)$ with $t=0$ corresponding to the point $(2,2)$. so:
$$
frac18 |z|^2 = frac12 (cos^2 t +sin^2 t + 2 sin t +1 )= 1+ sin t
$$
answered Oct 11 '14 at 3:41
David HoldenDavid Holden
14.7k21224
$endgroup$
add a comment |
$begingroup$
if you parametrize the circle of constraint it is $(2cos t,2sin t +2)$ with $t=0$ corresponding to the point $(2,2)$. so:
$$
frac18 |z|^2 = frac12 (cos^2 t +sin^2 t + 2 sin t +1 )= 1+ sin t
$$
answered Oct 11 '14 at 3:41
David HoldenDavid Holden
14.7k21224
$endgroup$
if you parametrize the circle of constraint it is $(2cos t,2sin t +2)$ with $t=0$ corresponding to the point $(2,2)$. so:
$$
frac18 |z|^2 = frac12 (cos^2 t +sin^2 t + 2 sin t +1 )= 1+ sin t
$$
answered Oct 11 '14 at 3:41
David HoldenDavid Holden
14.7k21224
answered Oct 11 '14 at 3:41
David HoldenDavid Holden
14.7k21224
answered Oct 11 '14 at 3:41
David HoldenDavid Holden
14.7k21224
answered Oct 11 '14 at 3:41
David HoldenDavid Holden
14.7k21224
14.7k21224
add a comment |
add a comment |
$begingroup$
Inspection of the figure indicates that the point $z_0:=sqrt{2}+i(2+sqrt{2})$ is the feasible point with largest absolute value: The closed disk $D_0$ with center $0$ and radius $|z_0|$ obviously contains the complete feasible region $F$, and $z_0in F$. It follows that
$$max_{zin F}|z|=|z_0|=sqrt{8+4sqrt{2}}doteq3.6955 .$$
answered Nov 14 '16 at 11:09
Christian BlatterChristian Blatter
172k7113326
$endgroup$
add a comment |
$begingroup$
Inspection of the figure indicates that the point $z_0:=sqrt{2}+i(2+sqrt{2})$ is the feasible point with largest absolute value: The closed disk $D_0$ with center $0$ and radius $|z_0|$ obviously contains the complete feasible region $F$, and $z_0in F$. It follows that
$$max_{zin F}|z|=|z_0|=sqrt{8+4sqrt{2}}doteq3.6955 .$$
answered Nov 14 '16 at 11:09
Christian BlatterChristian Blatter
172k7113326
$endgroup$
add a comment |
$begingroup$
Inspection of the figure indicates that the point $z_0:=sqrt{2}+i(2+sqrt{2})$ is the feasible point with largest absolute value: The closed disk $D_0$ with center $0$ and radius $|z_0|$ obviously contains the complete feasible region $F$, and $z_0in F$. It follows that
$$max_{zin F}|z|=|z_0|=sqrt{8+4sqrt{2}}doteq3.6955 .$$
answered Nov 14 '16 at 11:09
Christian BlatterChristian Blatter
172k7113326
$endgroup$
Inspection of the figure indicates that the point $z_0:=sqrt{2}+i(2+sqrt{2})$ is the feasible point with largest absolute value: The closed disk $D_0$ with center $0$ and radius $|z_0|$ obviously contains the complete feasible region $F$, and $z_0in F$. It follows that
$$max_{zin F}|z|=|z_0|=sqrt{8+4sqrt{2}}doteq3.6955 .$$
answered Nov 14 '16 at 11:09
Christian BlatterChristian Blatter
172k7113326
answered Nov 14 '16 at 11:09
Christian BlatterChristian Blatter
172k7113326
answered Nov 14 '16 at 11:09
Christian BlatterChristian Blatter
172k7113326
answered Nov 14 '16 at 11:09
Christian BlatterChristian Blatter
172k7113326
172k7113326
add a comment |
add a comment |
$begingroup$
Let $0+2i$ be C (the centre of the circle), let the point of furthest intersection (greatest $|z|$) of the 2 loci be A (You have indicated it on your diagram with a cross), construct the point B such that $triangle CAB$ has a right angle at B.
From here we can note that $angle CAB=frac{pi}{4}$(given - construction of CB is such that it is parallel to the x-axis, use corresponding angles), $AC=2$ (radii). Using trigonometry we can find that $BC=2cos(frac{pi}{4})=sqrt{2}$, similarly $AB=2sin(frac{pi}{4})=sqrt{2}$.
Applying our knowledge to the diagram (the Argand diagram), we see that the real part of the desired complex number is given by $BC$, $Re(z)=sqrt{2}$. The imaginary part is the sum of $AB$ and the height (distance) from the number $sqrt{2}$ to $sqrt{2}+2i$ (distance is $2$), thus $Im(z)=BC+2=2+sqrt{2}$.
$therefore$ $z=sqrt{2}+i(2+sqrt{2})$ and $|z|=2sqrt{2+sqrt{2}}$
answered Apr 3 '17 at 19:59
frog1944frog1944
1,28711322
$endgroup$
add a comment |
$begingroup$
Let $0+2i$ be C (the centre of the circle), let the point of furthest intersection (greatest $|z|$) of the 2 loci be A (You have indicated it on your diagram with a cross), construct the point B such that $triangle CAB$ has a right angle at B.
From here we can note that $angle CAB=frac{pi}{4}$(given - construction of CB is such that it is parallel to the x-axis, use corresponding angles), $AC=2$ (radii). Using trigonometry we can find that $BC=2cos(frac{pi}{4})=sqrt{2}$, similarly $AB=2sin(frac{pi}{4})=sqrt{2}$.
Applying our knowledge to the diagram (the Argand diagram), we see that the real part of the desired complex number is given by $BC$, $Re(z)=sqrt{2}$. The imaginary part is the sum of $AB$ and the height (distance) from the number $sqrt{2}$ to $sqrt{2}+2i$ (distance is $2$), thus $Im(z)=BC+2=2+sqrt{2}$.
$therefore$ $z=sqrt{2}+i(2+sqrt{2})$ and $|z|=2sqrt{2+sqrt{2}}$
answered Apr 3 '17 at 19:59
frog1944frog1944
1,28711322
$endgroup$
add a comment |
$begingroup$
Let $0+2i$ be C (the centre of the circle), let the point of furthest intersection (greatest $|z|$) of the 2 loci be A (You have indicated it on your diagram with a cross), construct the point B such that $triangle CAB$ has a right angle at B.
From here we can note that $angle CAB=frac{pi}{4}$(given - construction of CB is such that it is parallel to the x-axis, use corresponding angles), $AC=2$ (radii). Using trigonometry we can find that $BC=2cos(frac{pi}{4})=sqrt{2}$, similarly $AB=2sin(frac{pi}{4})=sqrt{2}$.
Applying our knowledge to the diagram (the Argand diagram), we see that the real part of the desired complex number is given by $BC$, $Re(z)=sqrt{2}$. The imaginary part is the sum of $AB$ and the height (distance) from the number $sqrt{2}$ to $sqrt{2}+2i$ (distance is $2$), thus $Im(z)=BC+2=2+sqrt{2}$.
$therefore$ $z=sqrt{2}+i(2+sqrt{2})$ and $|z|=2sqrt{2+sqrt{2}}$
answered Apr 3 '17 at 19:59
frog1944frog1944
1,28711322
$endgroup$
Let $0+2i$ be C (the centre of the circle), let the point of furthest intersection (greatest $|z|$) of the 2 loci be A (You have indicated it on your diagram with a cross), construct the point B such that $triangle CAB$ has a right angle at B.
From here we can note that $angle CAB=frac{pi}{4}$(given - construction of CB is such that it is parallel to the x-axis, use corresponding angles), $AC=2$ (radii). Using trigonometry we can find that $BC=2cos(frac{pi}{4})=sqrt{2}$, similarly $AB=2sin(frac{pi}{4})=sqrt{2}$.
Applying our knowledge to the diagram (the Argand diagram), we see that the real part of the desired complex number is given by $BC$, $Re(z)=sqrt{2}$. The imaginary part is the sum of $AB$ and the height (distance) from the number $sqrt{2}$ to $sqrt{2}+2i$ (distance is $2$), thus $Im(z)=BC+2=2+sqrt{2}$.
$therefore$ $z=sqrt{2}+i(2+sqrt{2})$ and $|z|=2sqrt{2+sqrt{2}}$
answered Apr 3 '17 at 19:59
frog1944frog1944
1,28711322
answered Apr 3 '17 at 19:59
frog1944frog1944
1,28711322
answered Apr 3 '17 at 19:59
frog1944frog1944
1,28711322
answered Apr 3 '17 at 19:59
frog1944frog1944
1,28711322
1,28711322
add a comment |
add a comment |
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$begingroup$
looks ok to me. what is your problem with $|z|$?
$endgroup$
– David Holden
Oct 11 '14 at 3:28
$begingroup$
The place you've marked an x on seems good, call it $z$. As you indicated in the diagram, $z$ is the 45 degree point on the radius 2 circle, which should be $(sqrt{2},sqrt{2})$, but we've shifted it up by $2$, so $z = sqrt{2} + (2+sqrt{2})i$. Alternatively, writing $z = x+iy$, we have $y = x + 2$ as well as $y = sqrt{4-x^2}+2$.
$endgroup$
– Platehead
Oct 11 '14 at 3:37
$begingroup$
I want to find greatest length(its distance from (0,0) ) of $|z|$ it should be around 3.70 units, unable to understand how to work it out.
$endgroup$
– Arif
Oct 11 '14 at 4:26