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Question: If \(\alpha + i\beta = \tan^{- 1}(z),z = x + iy\) and \(\alpha\) is constant, the locus of 'z' is...

If α+iβ=tan⁡−1(z),z=x+iy\alpha + i\beta = \tan^{- 1}(z),z = x + iy and α\alpha is constant, the locus of 'z' is

A

x2+y2+2xcot⁡2α=1x^{2} + y^{2} + 2x\cot 2\alpha = 1

B

cot⁡2α(x2+y2)=1+x\cot 2\alpha(x^{2} + y^{2}) = 1 + x

C

x2+y2+2ytan⁡2α=1x^{2} + y^{2} + 2y\tan 2\alpha = 1

D

x2+y2+2xsin⁡2α=1x^{2} + y^{2} + 2x\sin 2\alpha = 1

Answer

x2+y2+2xcot⁡2α=1x^{2} + y^{2} + 2x\cot 2\alpha = 1

Explanation

Solution

Sol. tan⁡(α+iβ)=x+iy\tan(\alpha + i\beta) = x + iy

∴ tan⁡(α−iβ)=x−iy\mathbf{\tan}\mathbf{(}\mathbf{\alpha}\mathbf{-}\mathbf{i\beta) = x}\mathbf{-}\mathbf{iy} (conjugate), α is a constant and β is known to be eliminated

tan⁡2α=tan⁡(α+iβ‾+α−iβ‾)\tan 2\alpha = \tan(\overline{\alpha + i\beta} + \overline{\alpha - i\beta})

⇒ tan⁡2α=x+iy+x−iy1−(x2+y2)\tan 2\alpha = \frac{x + iy + x - iy}{1 - (x^{2} + y^{2})}

⇒ 1−(x2+y2)=2xcot⁡2α1 - (x^{2} + y^{2}) = 2x\cot 2\alpha

∴ x2+y2+2xcot⁡2α=1.x^{2} + y^{2} + 2x\cot 2\alpha = 1.