0

Guest

Courses New

. 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.

. 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. 

Grade:12

1 Answers

Harshit Singh
askIITians Faculty 5964 Points
one year ago
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

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

Other Related Questions on Analytical Geometry

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Register ICAT

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More