The equation ≤ft(z-1/z+1)=π/4 represents a circle with [26 Aug 2021 Shift 1]

Correct Answer is : centre at (0,1)(0,1)(0,1) and radius 2{2}2

centre at (0,-1) and radius {2}

Correct Answer is : centre at (0,1)(0,1)(0,1) and radius 2{2}2

Test Index

Complex Numbers Part 1

Section: Mathematics

Question : 15 of 100

Marks: +1, -0

The equation

\arg\left(\frac{z-1}{z+1}\right)=\frac{\pi}{4}

Solution:

We have,

\arg\left(\frac{z-1}{z+1}\right)=\frac{\pi}{4}

\Rightarrow \arg(z-1)-\arg(z+1)=\frac{\pi}{4}

Let

z=x+iy

\arg[(x-1)+iy]-\arg[(x+1)+iy]=\frac{\pi}{4}

\Rightarrow \tan^{-1}\left(\frac{y}{x-1}\right)-\tan^{-1}\left(\frac{y}{x+1}\right)=\frac{\pi}{4}

\Rightarrow \left( \frac{ \frac{y}{x-1} - \frac{y}{x+1} }{ 1 + \frac{y}{x-1} \cdot \frac{y}{x+1} } \right) = \tan \frac{\pi}{4}

\Rightarrow \frac{ y(x+1)-y(x-1) }{ (x^{2}-1)+y^{2} } = 1

\Rightarrow 2y = x^{2}+y^{2}-1

\Rightarrow x^{2}+y^{2}-2y-1=0

\Rightarrow x^{2}+(y-1)^{2}=2

\Rightarrow x^{2}+(y-1)^{2}=(\sqrt{2})^{2}

Which is a circle with Centre

(0,1)

and Radius

=\sqrt{2}

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