Solve for $z$, which satisfy $arg(z-3-2i) = frac{pi}{6}$ and $arg(z-3-4i) = frac{2pi}{3}$.












2














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







share|cite|improve this question
























  • 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
















2














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







share|cite|improve this question
























  • 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














2












2








2


0





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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 22 '18 at 4:37









Kemono Chen

2,621437




2,621437










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


















  • 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










1 Answer
1






active

oldest

votes


















2














$$arg(z-3-2i)=frac{pi}{6}$$ 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.






share|cite|improve this answer





















  • 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











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









2














$$arg(z-3-2i)=frac{pi}{6}$$ 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.






share|cite|improve this answer





















  • 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
















2














$$arg(z-3-2i)=frac{pi}{6}$$ 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.






share|cite|improve this answer





















  • 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














2












2








2






$$arg(z-3-2i)=frac{pi}{6}$$ 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.






share|cite|improve this answer












$$arg(z-3-2i)=frac{pi}{6}$$ 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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • 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


















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