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If z 1 and {z^{2}}{z-1} is real, then the point represented

Question

If $z \neq = 1$ and $\frac{z ^{2}}{z - 1}$ is real, then the point represented by the complex number $z$ lies:

either on the real axis or on a circle passing through the origin.

on a circle with centre at the origin.

either on the real axis or on a circle not passing through the origin.

on the imaginary axis

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Daniel Discussed

z \neq = 1

and

\frac{z ^{2}}{z - 1}

is real, then the point represented by the complex number

z

lies:

15 mins ago

Discuss this question LIVE

15 mins ago

Text solutionVerified

z \neq = 1

and

\frac{z ^{2}}{z - 1}

is real so imaginary Part is 0

Lets say

z = x + i y

So expression

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

Rationalisation

\to \frac{x ^{2} - y ^{2} + 2 i x y}{( x - 1 ) + i y} \times \frac{( x - 1 ) - i y}{( x - 1 ) - i y}

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

(Imaginary Part

= 0

)

\Rightarrow 2 x y (x - 1) - x^{2} y + y^{3} = 0

\Rightarrow 2 x^{2} y - 2 x y - x^{2} y + y^{3} = 0

\Rightarrow x^{2} y + y^{3} - 2 x y = 0

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

\Rightarrow y = 0

x^{2} + y^{2} - 2 x = 0

\Rightarrow Real axis Line : y = 0 or Circle : x^{2} + y^{2} - 2 x = 0 \Rightarrow (x - 1)^{2} + (y - 0)^{2} = 1^{2}

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Question Text	If $z \neq = 1$ and $\frac{z ^{2}}{z - 1}$ is real, then the point represented by the complex number $z$ lies:
Answer Type	Text solution:1
Upvotes	150