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The point of intersection the curves arg (z-i+2)=\frac{\pi}{6}  & arg (z+4-3i)=-\frac{\pi}{4} is given by

Option: 1

(-2+i)
 


Option: 2

2-i

 


Option: 3

2+i


Option: 4

none of these


Answers (1)

best_answer

 

Definition of Argument/Amplitude of z in Complex Numbers -

\theta =tan^{-1}|\frac{y}{x}|, z\neq 0

\boldsymbol{\theta,\pi-\theta,-\pi+\theta,-\theta} are Principal Argument if z lies in first, second, third or fourth quadrant respectively.

- wherein

 

 

arg(z-i+2)=\frac{\pi}{6}\; \; \; \;\Rightarrow \tan \frac{\pi}{6}=\frac{y-1}{x+2}

\Rightarrow x-\sqrt{3y}=-(\sqrt{3}+2),x>-2,y>1\; \; \; \; \; \; \; \; ..........(i)

arg(z+4-3i)=-\frac{\pi}{4}\Rightarrow \tan \left ( -\frac{\pi}{4} \right )=\frac{y-3}{x+4}

\Rightarrow y+x=-1,x>-4,y<3\; \; \; \; \; \; \; \; \; \; .......(ii)

So, there is no point of intersection.

 

Posted by

Rishabh

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