Determine the complex number Z that satisfy $|z-3-3i|=1$ and that has maximum absolute value
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I'm having a hard time solving following question:
Determine the complex number Z that satisfy $|z-3-3i|=1$ and that has maximum absolute value.
Z should be written on the form z=x+iy.
I have determined with the help of the triangle inequality that $|z|=1+sqrt{18}$.
This is the point where i run into problem. I don't know how to determine z on the form $z=x+iy$, using the information above.
If someone could give me a hint i would be very thankful.
complex-numbers
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up vote
-1
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I'm having a hard time solving following question:
Determine the complex number Z that satisfy $|z-3-3i|=1$ and that has maximum absolute value.
Z should be written on the form z=x+iy.
I have determined with the help of the triangle inequality that $|z|=1+sqrt{18}$.
This is the point where i run into problem. I don't know how to determine z on the form $z=x+iy$, using the information above.
If someone could give me a hint i would be very thankful.
complex-numbers
2
This is the point on the circle $|z-(3+3i)|=1$ that lies furthest from the origin.
– saulspatz
Nov 15 at 16:58
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
I'm having a hard time solving following question:
Determine the complex number Z that satisfy $|z-3-3i|=1$ and that has maximum absolute value.
Z should be written on the form z=x+iy.
I have determined with the help of the triangle inequality that $|z|=1+sqrt{18}$.
This is the point where i run into problem. I don't know how to determine z on the form $z=x+iy$, using the information above.
If someone could give me a hint i would be very thankful.
complex-numbers
I'm having a hard time solving following question:
Determine the complex number Z that satisfy $|z-3-3i|=1$ and that has maximum absolute value.
Z should be written on the form z=x+iy.
I have determined with the help of the triangle inequality that $|z|=1+sqrt{18}$.
This is the point where i run into problem. I don't know how to determine z on the form $z=x+iy$, using the information above.
If someone could give me a hint i would be very thankful.
complex-numbers
complex-numbers
edited Nov 15 at 17:22
greedoid
34.6k114489
34.6k114489
asked Nov 15 at 16:55
Fosorf
13
13
2
This is the point on the circle $|z-(3+3i)|=1$ that lies furthest from the origin.
– saulspatz
Nov 15 at 16:58
add a comment |
2
This is the point on the circle $|z-(3+3i)|=1$ that lies furthest from the origin.
– saulspatz
Nov 15 at 16:58
2
2
This is the point on the circle $|z-(3+3i)|=1$ that lies furthest from the origin.
– saulspatz
Nov 15 at 16:58
This is the point on the circle $|z-(3+3i)|=1$ that lies furthest from the origin.
– saulspatz
Nov 15 at 16:58
add a comment |
2 Answers
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Think of the complex numbers as a real coordinate plane. The equation $|z-3-3i|=1$ is basically a circle of radius $1$ with the center at $(3, 3)$. What point on the circumference is farthest away from the origin?
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Any number with magnitude $1$ can be written as $e^{it}$ for real $t$, so your $z$ must be of the form $z= z_0+e^{it}$ where $z_0=3+3i$.
Note that we can rewrite $z_0$ as $3sqrt{2}e^{ipi/4}$. Let $mathbf v_0$ denote the vector from $0$ to $z_0$, and $mathbf v_1$ the vector from $0$ to $e^{it}$. You are looking to maximize $|mathbf v_0+mathbf v_1|$.
The vector $mathbf v_0$ always points in the direction of the angle $pi/4$. As we let $t$ vary, $mathbf v_1$ swings around, pointing in the direction of the angle $t$. The sum of $mathbf v_0$ and $mathbf v_1$ has the greatest magnitude (namely $|mathbf v_0| + |mathbf v_1|$) when the two vectors point in the same direction. That is, when $t=pi/4$.
So $|z|$ is maximized by taking $z=3+3i+e^{ipi/4}=(3+frac12sqrt{2})+(3+frac12sqrt{2})i=frac{6+sqrt{2}}{2} + frac{6+sqrt{2}}{2}i$.
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Think of the complex numbers as a real coordinate plane. The equation $|z-3-3i|=1$ is basically a circle of radius $1$ with the center at $(3, 3)$. What point on the circumference is farthest away from the origin?
add a comment |
up vote
0
down vote
Think of the complex numbers as a real coordinate plane. The equation $|z-3-3i|=1$ is basically a circle of radius $1$ with the center at $(3, 3)$. What point on the circumference is farthest away from the origin?
add a comment |
up vote
0
down vote
up vote
0
down vote
Think of the complex numbers as a real coordinate plane. The equation $|z-3-3i|=1$ is basically a circle of radius $1$ with the center at $(3, 3)$. What point on the circumference is farthest away from the origin?
Think of the complex numbers as a real coordinate plane. The equation $|z-3-3i|=1$ is basically a circle of radius $1$ with the center at $(3, 3)$. What point on the circumference is farthest away from the origin?
answered Nov 15 at 17:01
gt6989b
32k22351
32k22351
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Any number with magnitude $1$ can be written as $e^{it}$ for real $t$, so your $z$ must be of the form $z= z_0+e^{it}$ where $z_0=3+3i$.
Note that we can rewrite $z_0$ as $3sqrt{2}e^{ipi/4}$. Let $mathbf v_0$ denote the vector from $0$ to $z_0$, and $mathbf v_1$ the vector from $0$ to $e^{it}$. You are looking to maximize $|mathbf v_0+mathbf v_1|$.
The vector $mathbf v_0$ always points in the direction of the angle $pi/4$. As we let $t$ vary, $mathbf v_1$ swings around, pointing in the direction of the angle $t$. The sum of $mathbf v_0$ and $mathbf v_1$ has the greatest magnitude (namely $|mathbf v_0| + |mathbf v_1|$) when the two vectors point in the same direction. That is, when $t=pi/4$.
So $|z|$ is maximized by taking $z=3+3i+e^{ipi/4}=(3+frac12sqrt{2})+(3+frac12sqrt{2})i=frac{6+sqrt{2}}{2} + frac{6+sqrt{2}}{2}i$.
add a comment |
up vote
0
down vote
Any number with magnitude $1$ can be written as $e^{it}$ for real $t$, so your $z$ must be of the form $z= z_0+e^{it}$ where $z_0=3+3i$.
Note that we can rewrite $z_0$ as $3sqrt{2}e^{ipi/4}$. Let $mathbf v_0$ denote the vector from $0$ to $z_0$, and $mathbf v_1$ the vector from $0$ to $e^{it}$. You are looking to maximize $|mathbf v_0+mathbf v_1|$.
The vector $mathbf v_0$ always points in the direction of the angle $pi/4$. As we let $t$ vary, $mathbf v_1$ swings around, pointing in the direction of the angle $t$. The sum of $mathbf v_0$ and $mathbf v_1$ has the greatest magnitude (namely $|mathbf v_0| + |mathbf v_1|$) when the two vectors point in the same direction. That is, when $t=pi/4$.
So $|z|$ is maximized by taking $z=3+3i+e^{ipi/4}=(3+frac12sqrt{2})+(3+frac12sqrt{2})i=frac{6+sqrt{2}}{2} + frac{6+sqrt{2}}{2}i$.
add a comment |
up vote
0
down vote
up vote
0
down vote
Any number with magnitude $1$ can be written as $e^{it}$ for real $t$, so your $z$ must be of the form $z= z_0+e^{it}$ where $z_0=3+3i$.
Note that we can rewrite $z_0$ as $3sqrt{2}e^{ipi/4}$. Let $mathbf v_0$ denote the vector from $0$ to $z_0$, and $mathbf v_1$ the vector from $0$ to $e^{it}$. You are looking to maximize $|mathbf v_0+mathbf v_1|$.
The vector $mathbf v_0$ always points in the direction of the angle $pi/4$. As we let $t$ vary, $mathbf v_1$ swings around, pointing in the direction of the angle $t$. The sum of $mathbf v_0$ and $mathbf v_1$ has the greatest magnitude (namely $|mathbf v_0| + |mathbf v_1|$) when the two vectors point in the same direction. That is, when $t=pi/4$.
So $|z|$ is maximized by taking $z=3+3i+e^{ipi/4}=(3+frac12sqrt{2})+(3+frac12sqrt{2})i=frac{6+sqrt{2}}{2} + frac{6+sqrt{2}}{2}i$.
Any number with magnitude $1$ can be written as $e^{it}$ for real $t$, so your $z$ must be of the form $z= z_0+e^{it}$ where $z_0=3+3i$.
Note that we can rewrite $z_0$ as $3sqrt{2}e^{ipi/4}$. Let $mathbf v_0$ denote the vector from $0$ to $z_0$, and $mathbf v_1$ the vector from $0$ to $e^{it}$. You are looking to maximize $|mathbf v_0+mathbf v_1|$.
The vector $mathbf v_0$ always points in the direction of the angle $pi/4$. As we let $t$ vary, $mathbf v_1$ swings around, pointing in the direction of the angle $t$. The sum of $mathbf v_0$ and $mathbf v_1$ has the greatest magnitude (namely $|mathbf v_0| + |mathbf v_1|$) when the two vectors point in the same direction. That is, when $t=pi/4$.
So $|z|$ is maximized by taking $z=3+3i+e^{ipi/4}=(3+frac12sqrt{2})+(3+frac12sqrt{2})i=frac{6+sqrt{2}}{2} + frac{6+sqrt{2}}{2}i$.
answered Nov 15 at 17:35
MPW
29.5k11856
29.5k11856
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This is the point on the circle $|z-(3+3i)|=1$ that lies furthest from the origin.
– saulspatz
Nov 15 at 16:58