Get Answers to all your Questions

header-bg qa

Let C be the set of complex numbers. Prove that the mapping f: C \rightarrow R is given by f (z) = \vert z \vert , \forall z \in C, is neither one-one nor onto.
 

Answers (1)

Here, f: C \rightarrow R is given by f (z) = \vert z \vert , \forall z \in C

If we assume z = 4 + 3i

Then,

f(4 + 3i) = \vert 4 + 3i \vert = \sqrt (4\textsuperscript{2} + 3\textsuperscript{2})= \sqrt 25 = 5\\

Similarly, for z = 4 - 3i

f(4 - 3i) = \vert 4 - 3i \vert = \sqrt (4\textsuperscript{2} + 3\textsuperscript{2}) = \sqrt 25 = 5\\

Therefore, it is clear that f(z) is many-one.

So,   \vert z \vert \geq 0, \forall z \in C,\\

However, in the question R is the co-domain given.

Hence, f(z) is not onto. So, f(z) is neither one-one nor onto.

Posted by

infoexpert22

View full answer