CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

Assertion :The locus of the point $$z$$ satisfying the condition $$\displaystyle arg\frac { z-1 }{ z+1 } =\frac { \pi  }{ 3 } $$ is a parabola  Reason: $$\displaystyle arg\frac { { z }_{ 1 } }{ { z }_{ 2 } } =arg{ z }_{ 1 }-arg{ z }_{ 2 }$$


A
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
loader
B
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
loader
C
Assertion is correct but Reason is incorrect
loader
D
Both Assertion and Reason are incorrect
loader

Solution

The correct option is D Both Assertion and Reason are incorrect
We have, $$\displaystyle arg\frac { z-1 }{ z+1 } =\frac { \pi  }{ 3 } \Rightarrow arg\frac { x+iy-1 }{ x+iy+1 } =\frac { \pi  }{ 3 } $$   $$[$$ putting $$z=x+iy]$$
$$\displaystyle \Rightarrow \tan ^{ -1 }{ \frac { y }{ x-1 }  } -\tan ^{ -1 }{ \frac { y }{ x+1 }  } =\frac { \pi  }{ 3 } \quad \quad \quad \left( \because arg\frac { { z }_{ 1 } }{ { z }_{ 2 } } =arg{ z }_{ 1 }-arg{ z }_{ 2 } \right) $$
$$\displaystyle \Rightarrow \tan ^{ -1 }{ \frac { \frac { y }{ x-1 } -\frac { y }{ x+1 }  }{ 1+\frac { { y }^{ 2 } }{ { x }^{ 2 }-1 }  }  } =\frac { \pi  }{ 3 } \Rightarrow \frac { 2y }{ { x }^{ 2 }+{ y }^{ 2 }-1 } =\tan { \frac { \pi  }{ 3 }  } =\sqrt { 3 } $$
$$\displaystyle \Rightarrow { x }^{ 2 }+{ y }^{ 2 }-\frac { 2 }{ \sqrt { 3 }  } y-1=0,$$ which is a circle.
$$arg\dfrac{z_1}{z_2}=arg z_1-arg z_2+2\pi n$$ .... n=0 if $$ -\pi/2 <argz_1-argz_2<\pi/2$$

Mathematics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image