. 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.

Dear Student

Given, f: C → R such that f (z) = |z|,∀z∈C
z = 6 + 8i
Then,
f(6+8i)=|6+8i|=√(6^2+8^2)=√100=10
for z = 6 - 8i

f(6-8i)=|6-8i|=√(6^2+8^2)=√100=10
Hence, f (z) is many-one.
Also,|z|≥ 0,∀z∈C
But the co-domain given is ‘R’

Therefore, f(z) is not onto

Thanks