Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Let C be the set of complex numbers. Prove that the mapping f: C → R is given by f (z) = |z|, ∀ z ∈ C, is neither one-one nor onto.

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

Answers (1)

Here, $f: C \rightarrow R_{\text {is given by }} f(z)=|z|, \forall z \in C$
If we assume $z=4+3 i$
Then, $f(4+3 i)=|4+3 i|=\sqrt{\left(4^2+3^2\right) }=\sqrt{2} 5=5$
Similarly, for $z=4-3 i$

$
f(4-3 i)=|4-3 i|=\sqrt{\left(4^2+3^2\right)}=\sqrt{2} 5=5
$

Therefore, it is clear that $\mathrm{f}(\mathrm{z})$ is many-one.
So, $|z| \geq 0, \forall z \in C$,

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

Similar Questions

Latest Question