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The locus of z such that { { z }^{ 2 } }{ z-1 } is always r

Question

The locus of $z$ such that $\frac{z ^{2}}{z - 1}$ is always real is

A

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

B

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

C

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

D

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

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Liam Discussed

The locus of

z

such that

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

is always real is

15 mins ago

Discuss this question LIVE

15 mins ago

Text solutionVerified

let

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

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

\frac{x ^{3} - x ^{2} - x ^{2} y i + y ^{2} x - y ^{2} - y ^{3} i - 2 x ^{2} y i + 2 x y i - 2 x y ^{2}}{( x - 1 ) ^{2} + y ^{2}}

\frac{x ^{3} - x ^{2} + y ^{2} x - y ^{2} - 2 x y ^{2} + ( 2 x y - y ^{3} - 2 x ^{2} y + y ^{2} ) i}{( x - 1 ) ^{2} + y ^{2}}

∵

the above term is purely real,

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

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

∴ y = 0

&

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

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Question Text	The locus of $z$ such that $\frac{z ^{2}}{z - 1}$ is always real is
Answer Type	Text solution:1
Upvotes	150