If -i -tan -1 Z- Z-x-i y and - is constant- then the locus of Z is

Let z = x + iy where x,y ϵ R ∴α + i β = tan^ 1(x + iy) ⇒α i β = tan^ 1(x iy) ⇒ 2 α = tan^ 1 (x+iy) + tan^ 1(x iy) =tan^ 12x/1 x^2 y^2 ∴ x^2 + y^2 + 2x ...

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Byju's Answer

Standard XII

Mathematics

Geometrical Representation of a Complex Number

If α+i β=tan ...

Question

If $α + i β = t a n^{- 1} Z, Z = x + i y a n d α$ is constant, then the locus of Z is

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Solution

The correct option is A
$L e t z = x + i y w h e r e x, y ϵ R ∴ α + i β = t a n^{- 1} (x + i y) \Rightarrow α - i β = t a n^{- 1} (x - i y) \Rightarrow 2 α = t a n^{- 1} (x + i y) + t a n^{- 1} (x - i y) = t a n^{- 1} \frac{2 x}{1 - x^{2} - y^{2}} ∴ x^{2} + y^{2} + 2 x c o t 2 α = 1$

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Similar questions

Q. If

α + i β = t a n^{- 1} (z), z = x + i y

and

α

is constant then the locus of

z

is:

Q. If

α + i β = {tan}^{- 1} z

z = x + i y

and

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is constant then the locus of

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α + i β = {tan}^{- 1} z, α = \frac{π}{4}

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Q. If

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Q. If

e^{x + i y} = α + i β,

then

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∴ {log}_{e} (x + i y) = {log}_{e} (r e^{i θ})

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be the argument of

x + i y

i^{α + i β)} = α + i β,

then

α^{2} + β^{2}

equals

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

The correct option is A Let z=x+iy where x,y ϵ R∴α+iβ=tan−1(x+iy)⇒α−iβ=tan−1(x−iy)⇒2α=tan−1(x+iy)+tan−1(x−iy)=tan−12x1−x2−y2∴x2+y2+2x cot 2α=1

If $α + i β = t a n^{- 1} Z, Z = x + i y a n d α$ is constant, then the locus of Z is

The correct option is A
$L e t z = x + i y w h e r e x, y ϵ R ∴ α + i β = t a n^{- 1} (x + i y) \Rightarrow α - i β = t a n^{- 1} (x - i y) \Rightarrow 2 α = t a n^{- 1} (x + i y) + t a n^{- 1} (x - i y) = t a n^{- 1} \frac{2 x}{1 - x^{2} - y^{2}} ∴ x^{2} + y^{2} + 2 x c o t 2 α = 1$