Answer :

We have,

f : C R given by f (z) = |z|, z C

In order to prove that f is one-one, it is sufficient to prove that f(z1)=f(z2) z1=z2 z1, z2 C .

Let z1 = 2+3i and z2 = 2-3i are two distinct complex numbers.


f(z1) = |z1| = |2+3i | = = √ 13

f(z2) = |z2| = |2-3i | = = √ 13

here, we observe that f(z1) = f(z2) but z1 ≠ z2

This shows that different element of C may have the same value in R.

Thus, f(z) is not one-one.

f is onto if every element of R is the f-image of some element of C.

We have, f(z) = |z|, z C and |z| ≥ 0

We observe that negative real numbers in R do not have their pre-images in C.

Thus, f is not onto.

Hence, f(z) is neither one-one nor onto.

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