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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