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Let $\mathrm{C}$ 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 oneone nor onto.
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The mapping $f: C \rightarrow R$
Given, $f(z)=|z|, \forall \mathrm{z} \in \mathrm{C}$
$$
\begin{aligned}
&\Rightarrow \mathrm{f}(1)=|1|=1 \\
&\Rightarrow \mathrm{f}(-1)=|-1|=1 \\
&\Rightarrow \mathrm{f}(1)=\mathrm{f}(-1)
\end{aligned}
$$
But $1 \neq-1$
So, $\mathrm{f}(\mathrm{z})$ is not one-one. Also, $\mathrm{f}(\mathrm{z})$ is not onto as there is no pre-image for any negative element of $\mathrm{R}$ under the mapping $f(z)$
Given, $f(z)=|z|, \forall \mathrm{z} \in \mathrm{C}$
$$
\begin{aligned}
&\Rightarrow \mathrm{f}(1)=|1|=1 \\
&\Rightarrow \mathrm{f}(-1)=|-1|=1 \\
&\Rightarrow \mathrm{f}(1)=\mathrm{f}(-1)
\end{aligned}
$$
But $1 \neq-1$
So, $\mathrm{f}(\mathrm{z})$ is not one-one. Also, $\mathrm{f}(\mathrm{z})$ is not onto as there is no pre-image for any negative element of $\mathrm{R}$ under the mapping $f(z)$
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