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Let C be the set of complex numbers.Prove that the mapping $f\; :\;C \rightarrow R\;given\;by\;f(z)=|\;z\;|,\forall\; z \in C,$is neither one-one nor onto.

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  • A function $f: A \rightarrow B$ where for every $x1, x2 \in X, f(x1) = f(x2) \Rightarrow x1 = x2$ is called a one-one or injective function
  • A function$ f : X \rightarrow Y$ is said to be onto or surjective, if every element of Y is the image of some element of X under f, i.e., for every $y \in Y$, there exists an element x in X such that $f(x) = y$.
Given $f(z)=|z|\;for\; all\; Z \in c$
Let $z_1=x+iy\;and\;z_2=x-iy \qquad x,y \in R$
Step1: Injective or One-One function:
$|z_1|=\sqrt {x^2+y^2}\qquad;\qquad|z_2|=\sqrt {x^2+y^2}$
But $z_1 \neq z_1$
Hence $f(z)$ is not a one-one function
Step 2: Surjective or On-to function:
Let $y \in R$
Let $y=- \sqrt 2 $
There does not exists any no. z in c such that $|z|=-\sqrt 2$
Therefore f is not onto
Solution: Therefore f is neither one-one nor onto
Hence proved


answered Mar 2, 2013 by meena.p
edited Mar 26, 2013 by meena.p

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