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Q.

If $α + i β = \tan^{- 1} z, z = x + i y$ and α is constant then the locus of z is______

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a

$x^{2} + y^{2} + 2 x \cot 2 α = 1$

b

$\cot 2 α (x^{2} + y^{2}) = 1 + x$

c

$x^{2} + y^{2} + 2 x \tan 2 α = 1$

d

$x^{2} + y^{2} + 2 y \tan 2 α = 1$

answer is A.

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Detailed Solution

$\begin{array}{l} \tan (α + i β) = x + i y \\ \tan (α - i β) = x - i y \\ \tan 2 α = \tan (α + i β + α - i β) \\ = \frac{x + i y + x - i y}{1 - (x + i y) (x - i y)} = \frac{2 x}{1 - x^{2} - y^{2}} \\ 1 - x^{2} - y^{2} = 2 x \cot 2 α \\ x^{2} + y^{2} + 2 x \cot 2 α = 1 \end{array}$

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If α+iβ=tan−1z, z=x+iy and α is constant then the locus of z is______