If z is a complex number, then $\left.\overline{\left(z^{-1}\right.}\right)(z)$ is equal to
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If $z$ is a complex number, then $\left.\overline{\left(z^{-1}\right.}\right)(z)$ is equal to

(A) 1

(B) -1

(C) 0

(D) None of these

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Best answer

SOLUTION —

Let $z=x+i y \Rightarrow \bar{z}=x-i y$ and

$\Rightarrow \overline{\left(z^{-1}\right)}=\frac{1}{x-t y}=\frac{x+i y}{x^2+y^2}$

$\therefore \quad \overline{\left(z^{-1}\right)} \bar{z}=\frac{x+i y}{x^2+y^2} \times(x-i y)=1$

So, The correct option will be (A).

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