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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.
Dear StudentGiven, f: C → R such that f (z) = |z|,∀z∈C z = 6 + 8iThen,f(6+8i)=|6+8i|=√(6^2+8^2)=√100=10 for z = 6 - 8if(6-8i)=|6-8i|=√(6^2+8^2)=√100=10Hence, f (z) is many-one.Also,|z|≥ 0,∀z∈CBut the co-domain given is ‘R’Therefore, f(z) is not ontoThanks
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