Solve for $z$, which satisfy $arg(z-3-2i) = frac{pi}{6}$ and $arg(z-3-4i) = frac{2pi}{3}$.
solve for $z$, which satisfy $displaystyle arg(z-3-2i) = frac{pi
}{6}$ and $displaystylearg(z-3-4i) = frac{2pi}{3}$.
So I'm first assuming $z=x+iy$, then putting it in the place of $z$ and putting real parts together and imaginary parts together.
Then I'm using $tantheta = frac{text{imaginary part}}{text{real part}}$
$z-3-4i = (x-3)+(y-2)i$
Then,
$tan30° = frac{y-2}{x-3}$
$frac{1}{sqrt3} = frac{y-2}{x-3}$
I'm processing like this and my answer comes as $y=5/2$
And $x= 3+frac{5sqrt3}{2}-2sqrt3$
But the answer key says there's no such $z$ which satisfies this equation.
Is this the wrong way to solve this question, is my answer wrong or the answer key's?
complex-analysis complex-numbers
edited Nov 22 '18 at 4:37
Kemono Chen
2,621437
asked Nov 22 '18 at 3:53
Kaustuv Sawarn
465
|
show 1 more comment
solve for $z$, which satisfy $displaystyle arg(z-3-2i) = frac{pi
}{6}$ and $displaystylearg(z-3-4i) = frac{2pi}{3}$.
So I'm first assuming $z=x+iy$, then putting it in the place of $z$ and putting real parts together and imaginary parts together.
Then I'm using $tantheta = frac{text{imaginary part}}{text{real part}}$
$z-3-4i = (x-3)+(y-2)i$
Then,
$tan30° = frac{y-2}{x-3}$
$frac{1}{sqrt3} = frac{y-2}{x-3}$
I'm processing like this and my answer comes as $y=5/2$
And $x= 3+frac{5sqrt3}{2}-2sqrt3$
But the answer key says there's no such $z$ which satisfies this equation.
Is this the wrong way to solve this question, is my answer wrong or the answer key's?
complex-analysis complex-numbers
edited Nov 22 '18 at 4:37
Kemono Chen
2,621437
asked Nov 22 '18 at 3:53
Kaustuv Sawarn
465
Why are you equating the arguments of the $Arg$'s? Or should that $2$ be a $4$?
– Michael Burr
Nov 22 '18 at 4:00
1
It may be useful to draw a diagram.
– Kemono Chen
Nov 22 '18 at 4:04
@MichaelBurr i didn't get you. I'm not equating the arguments.
– Kaustuv Sawarn
Nov 22 '18 at 4:06
@KemonoChen but I get the answer this way, i want to know what am I doing wrong?
– Kaustuv Sawarn
Nov 22 '18 at 4:07
Your approach is wrong .
– Akash Roy
Nov 22 '18 at 4:07
|
show 1 more comment
solve for $z$, which satisfy $displaystyle arg(z-3-2i) = frac{pi
}{6}$ and $displaystylearg(z-3-4i) = frac{2pi}{3}$.
So I'm first assuming $z=x+iy$, then putting it in the place of $z$ and putting real parts together and imaginary parts together.
Then I'm using $tantheta = frac{text{imaginary part}}{text{real part}}$
$z-3-4i = (x-3)+(y-2)i$
Then,
$tan30° = frac{y-2}{x-3}$
$frac{1}{sqrt3} = frac{y-2}{x-3}$
I'm processing like this and my answer comes as $y=5/2$
And $x= 3+frac{5sqrt3}{2}-2sqrt3$
But the answer key says there's no such $z$ which satisfies this equation.
Is this the wrong way to solve this question, is my answer wrong or the answer key's?
complex-analysis complex-numbers
edited Nov 22 '18 at 4:37
Kemono Chen
2,621437
asked Nov 22 '18 at 3:53
Kaustuv Sawarn
465
solve for $z$, which satisfy $displaystyle arg(z-3-2i) = frac{pi
}{6}$ and $displaystylearg(z-3-4i) = frac{2pi}{3}$.
So I'm first assuming $z=x+iy$, then putting it in the place of $z$ and putting real parts together and imaginary parts together.
Then I'm using $tantheta = frac{text{imaginary part}}{text{real part}}$
$z-3-4i = (x-3)+(y-2)i$
Then,
$tan30° = frac{y-2}{x-3}$
$frac{1}{sqrt3} = frac{y-2}{x-3}$
I'm processing like this and my answer comes as $y=5/2$
And $x= 3+frac{5sqrt3}{2}-2sqrt3$
But the answer key says there's no such $z$ which satisfies this equation.
Is this the wrong way to solve this question, is my answer wrong or the answer key's?
complex-analysis complex-numbers
complex-analysis complex-numbers
edited Nov 22 '18 at 4:37
Kemono Chen
2,621437
asked Nov 22 '18 at 3:53
Kaustuv Sawarn
465
edited Nov 22 '18 at 4:37
Kemono Chen
2,621437
asked Nov 22 '18 at 3:53
Kaustuv Sawarn
465
edited Nov 22 '18 at 4:37
Kemono Chen
2,621437
edited Nov 22 '18 at 4:37
Kemono Chen
2,621437
edited Nov 22 '18 at 4:37
Kemono Chen
2,621437
2,621437
asked Nov 22 '18 at 3:53
Kaustuv Sawarn
465
asked Nov 22 '18 at 3:53
Kaustuv Sawarn
465
asked Nov 22 '18 at 3:53
Kaustuv Sawarn
465
465
Why are you equating the arguments of the $Arg$'s? Or should that $2$ be a $4$?
– Michael Burr
Nov 22 '18 at 4:00
1
It may be useful to draw a diagram.
– Kemono Chen
Nov 22 '18 at 4:04
@MichaelBurr i didn't get you. I'm not equating the arguments.
– Kaustuv Sawarn
Nov 22 '18 at 4:06
@KemonoChen but I get the answer this way, i want to know what am I doing wrong?
– Kaustuv Sawarn
Nov 22 '18 at 4:07
Your approach is wrong .
– Akash Roy
Nov 22 '18 at 4:07
|
show 1 more comment
Why are you equating the arguments of the $Arg$'s? Or should that $2$ be a $4$?
– Michael Burr
Nov 22 '18 at 4:00
1
It may be useful to draw a diagram.
– Kemono Chen
Nov 22 '18 at 4:04
@MichaelBurr i didn't get you. I'm not equating the arguments.
– Kaustuv Sawarn
Nov 22 '18 at 4:06
@KemonoChen but I get the answer this way, i want to know what am I doing wrong?
– Kaustuv Sawarn
Nov 22 '18 at 4:07
Your approach is wrong .
– Akash Roy
Nov 22 '18 at 4:07
Why are you equating the arguments of the $Arg$'s? Or should that $2$ be a $4$?
– Michael Burr
Nov 22 '18 at 4:00
Why are you equating the arguments of the $Arg$'s? Or should that $2$ be a $4$?
– Michael Burr
Nov 22 '18 at 4:00
1
1
It may be useful to draw a diagram.
– Kemono Chen
Nov 22 '18 at 4:04
It may be useful to draw a diagram.
– Kemono Chen
Nov 22 '18 at 4:04
@MichaelBurr i didn't get you. I'm not equating the arguments.
– Kaustuv Sawarn
Nov 22 '18 at 4:06
@MichaelBurr i didn't get you. I'm not equating the arguments.
– Kaustuv Sawarn
Nov 22 '18 at 4:06
@KemonoChen but I get the answer this way, i want to know what am I doing wrong?
– Kaustuv Sawarn
Nov 22 '18 at 4:07
@KemonoChen but I get the answer this way, i want to know what am I doing wrong?
– Kaustuv Sawarn
Nov 22 '18 at 4:07
Your approach is wrong .
– Akash Roy
Nov 22 '18 at 4:07
Your approach is wrong .
– Akash Roy
Nov 22 '18 at 4:07
|
show 1 more comment
1 Answer
1
active
oldest
votes
$$arg(z-3-2i)=frac{pi}{6}$$ is a line which originate from $(3,2)$ and making an angle of $30^circ$ with positive $x$ axis.
And $$arg(z-3-4i)=frac{2pi}{3}$$ is a line which originate from $(3,4)$ and making an angle of $120^circ$ with positive $x$ axis.
Now drawing These line in $x-y$ Coordinate axis.
You will get no point of Intersection.
So no $z$ which satisfy above these two equations.
answered Nov 22 '18 at 4:03
D Tiwari
5,4132630
Okay, but is my approach wrong? If it is, where is it wrong?
– Kaustuv Sawarn
Nov 22 '18 at 4:05
Actually you have take above line as infinite ray.
– D Tiwari
Nov 22 '18 at 4:06
add a comment |
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1 Answer
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$$arg(z-3-2i)=frac{pi}{6}$$ is a line which originate from $(3,2)$ and making an angle of $30^circ$ with positive $x$ axis.
And $$arg(z-3-4i)=frac{2pi}{3}$$ is a line which originate from $(3,4)$ and making an angle of $120^circ$ with positive $x$ axis.
Now drawing These line in $x-y$ Coordinate axis.
You will get no point of Intersection.
So no $z$ which satisfy above these two equations.
answered Nov 22 '18 at 4:03
D Tiwari
5,4132630
Okay, but is my approach wrong? If it is, where is it wrong?
– Kaustuv Sawarn
Nov 22 '18 at 4:05
Actually you have take above line as infinite ray.
– D Tiwari
Nov 22 '18 at 4:06
add a comment |
$$arg(z-3-2i)=frac{pi}{6}$$ is a line which originate from $(3,2)$ and making an angle of $30^circ$ with positive $x$ axis.
And $$arg(z-3-4i)=frac{2pi}{3}$$ is a line which originate from $(3,4)$ and making an angle of $120^circ$ with positive $x$ axis.
Now drawing These line in $x-y$ Coordinate axis.
You will get no point of Intersection.
So no $z$ which satisfy above these two equations.
answered Nov 22 '18 at 4:03
D Tiwari
5,4132630
Okay, but is my approach wrong? If it is, where is it wrong?
– Kaustuv Sawarn
Nov 22 '18 at 4:05
Actually you have take above line as infinite ray.
– D Tiwari
Nov 22 '18 at 4:06
add a comment |
$$arg(z-3-2i)=frac{pi}{6}$$ is a line which originate from $(3,2)$ and making an angle of $30^circ$ with positive $x$ axis.
And $$arg(z-3-4i)=frac{2pi}{3}$$ is a line which originate from $(3,4)$ and making an angle of $120^circ$ with positive $x$ axis.
Now drawing These line in $x-y$ Coordinate axis.
You will get no point of Intersection.
So no $z$ which satisfy above these two equations.
answered Nov 22 '18 at 4:03
D Tiwari
5,4132630
$$arg(z-3-2i)=frac{pi}{6}$$ is a line which originate from $(3,2)$ and making an angle of $30^circ$ with positive $x$ axis.
And $$arg(z-3-4i)=frac{2pi}{3}$$ is a line which originate from $(3,4)$ and making an angle of $120^circ$ with positive $x$ axis.
Now drawing These line in $x-y$ Coordinate axis.
You will get no point of Intersection.
So no $z$ which satisfy above these two equations.
answered Nov 22 '18 at 4:03
D Tiwari
5,4132630
answered Nov 22 '18 at 4:03
D Tiwari
5,4132630
answered Nov 22 '18 at 4:03
D Tiwari
5,4132630
answered Nov 22 '18 at 4:03
D Tiwari
5,4132630
5,4132630
Okay, but is my approach wrong? If it is, where is it wrong?
– Kaustuv Sawarn
Nov 22 '18 at 4:05
Actually you have take above line as infinite ray.
– D Tiwari
Nov 22 '18 at 4:06
add a comment |
Okay, but is my approach wrong? If it is, where is it wrong?
– Kaustuv Sawarn
Nov 22 '18 at 4:05
Actually you have take above line as infinite ray.
– D Tiwari
Nov 22 '18 at 4:06
Okay, but is my approach wrong? If it is, where is it wrong?
– Kaustuv Sawarn
Nov 22 '18 at 4:05
Okay, but is my approach wrong? If it is, where is it wrong?
– Kaustuv Sawarn
Nov 22 '18 at 4:05
Actually you have take above line as infinite ray.
– D Tiwari
Nov 22 '18 at 4:06
Actually you have take above line as infinite ray.
– D Tiwari
Nov 22 '18 at 4:06
add a comment |
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Why are you equating the arguments of the $Arg$'s? Or should that $2$ be a $4$?
– Michael Burr
Nov 22 '18 at 4:00
1
It may be useful to draw a diagram.
– Kemono Chen
Nov 22 '18 at 4:04
@MichaelBurr i didn't get you. I'm not equating the arguments.
– Kaustuv Sawarn
Nov 22 '18 at 4:06
@KemonoChen but I get the answer this way, i want to know what am I doing wrong?
– Kaustuv Sawarn
Nov 22 '18 at 4:07
Your approach is wrong .
– Akash Roy
Nov 22 '18 at 4:07