Let C be the set of complex numbers- Prove that the mapping f - C - R given by f z - z - - z - C- is neither one-one nor onto-

The mapping f:C → R is given byf(z)=|z|, ∀ z ∈ C Now, f(1)=|1|=1 f( 1)=| 1|=1 f(1)=f( 1) Clearly, f is not one one. Also, f(z) is not onto as there is no ...

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Standard XII

Mathematics

Definition of Function

Let C be the ...

Question

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

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Solution

The mapping $f : C \to R$ is given by $f (z) = | z |, \forall z \in C$

Now, $f (1) = | 1 | = 1 f (- 1) = | - 1 | = 1 f (1) = f (- 1)$
Clearly, $f$ is not one-one.

Also, $f (z)$ is not onto as there is no pre-image for any negative element of $R$ under the mapping $f (z)$ .

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Let C be the set of complex numbers. Prove that the mapping f:C→R given by f(z)=|z|, ∀z∈C, is neither one-one nor onto.

The mapping f:C→R is given byf(z)=|z|,∀z∈C Now, f(1)=|1|=1f(−1)=|−1|=1f(1)=f(−1) Clearly, f is not one-one. Also, f(z) is not onto as there is no pre-image for any negative element of R under the mapping f(z).

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

The mapping $f : C \to R$ is given by $f (z) = | z |, \forall z \in C$

Now, $f (1) = | 1 | = 1 f (- 1) = | - 1 | = 1 f (1) = f (- 1)$
Clearly, $f$ is not one-one.

Also, $f (z)$ is not onto as there is no pre-image for any negative element of $R$ under the mapping $f (z)$ .