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Question: The locus of the point z satisfying arg \(\left( \frac{z - 1}{z + 1} \right) = k.\) (where k is non-...

The locus of the point z satisfying arg (z−1z+1)=k.\left( \frac{z - 1}{z + 1} \right) = k. (where k is non-zero) is

A

Circle with centre on y–axis

B

Circle with centre on x–axis

C

A straight line parallel to x–axis

D

A straight line making an angle 60° with x–axis

Answer

Circle with centre on y–axis

Explanation

Solution

Sol. a⥂rg(z−1z+1)=ka ⥂ rg\left( \frac{z - 1}{z + 1} \right) = k ⇒ a⥂rg[(x−1)+iy(x+1)+iy]=ka ⥂ rg\left\lbrack \frac{(x - 1) + iy}{(x + 1) + iy} \right\rbrack = k

⇒a⥂rg[(x−1)+iy]−a⥂rg[(x+1)+iy]=ka ⥂ rg\lbrack(x - 1) + iy\rbrack - a ⥂ rg\lbrack(x + 1) + iy\rbrack = k

⇒ tan⁡−1(yx−1)−tan⁡−1(yx+1)=k\mathbf{\tan}^{\mathbf{- 1}}\left( \frac{\mathbf{y}}{\mathbf{x - 1}} \right)\mathbf{-}\mathbf{\tan}^{\mathbf{- 1}}\left( \frac{\mathbf{y}}{\mathbf{x + 1}} \right)\mathbf{= k} ⇒tan⁡−1[yx−1−yx+11+y2x2−1]=k\mathbf{\tan}^{\mathbf{-}\mathbf{1}}\left\lbrack \frac{\frac{\mathbf{y}}{\mathbf{x}\mathbf{-}\mathbf{1}}\mathbf{-}\frac{\mathbf{y}}{\mathbf{x + 1}}}{\mathbf{1 +}\frac{\mathbf{y}^{\mathbf{2}}}{\mathbf{x}^{\mathbf{2}}\mathbf{-}\mathbf{1}}} \right\rbrack\mathbf{= k}

⇒ tan⁡k=y(x+1)−y(x−1)x2+y2−1=2yx2+y2−1\mathbf{\tan}\mathbf{k}\mathbf{=}\frac{\mathbf{y(x + 1) - y(x - 1)}}{\mathbf{x}^{\mathbf{2}}\mathbf{+}\mathbf{y}^{\mathbf{2}}\mathbf{- 1}}\mathbf{=}\frac{\mathbf{2y}}{\mathbf{x}^{\mathbf{2}}\mathbf{+}\mathbf{y}^{\mathbf{2}}\mathbf{-}\mathbf{1}}

⇒ 2ytan⁡k=x2+y2−1\frac{\mathbf{2y}}{\mathbf{\tan}\mathbf{k}}\mathbf{=}\mathbf{x}^{\mathbf{2}}\mathbf{+}\mathbf{y}^{\mathbf{2}}\mathbf{- 1} ⇒ x2+y2−2ytan⁡k−1=0x^{2} + y^{2} - \frac{2y}{\tan k} - 1 = 0

It is an equation of circle whose centre is (−g,−f)=(0,cot k)( - g, - f) = (\text{0,cot }k) on y–axis.