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प्रश्न
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.
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उत्तर
Given, f: C → R such that f(z) = |z|, ∀ z ∈ C
Now, let take z = 6 + 8i
Then,
f(6 + 8i) = |6 + 8i|
= `sqrt(6^2 + 8^2)`
= `sqrt(100)`
= 10
And, for z = 6 – 8i
f(6 – 8i) = |6 – 8i|
= `sqrt(6^2 - 8^2)`
= `sqrt(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.
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