Define f: C arrow R by f(z)=|z| z C. Then which of the following is false?

Correct Answer is : f(z12+z22)=f(z12)+f(z22)f(z₁²+z₂²)=f(z₁²)+f(z₂²)f(z12+z22)=f(z12)+f(z22) ∀z1,z2∈C z₁, z₂ C∀z1,z2∈C

f(z₁²+z₂²)=f(z₁²)+f(z₂²) z₁, z₂ C

Correct Answer is : f(z12+z22)=f(z12)+f(z22)f(z₁²+z₂²)=f(z₁²)+f(z₂²)f(z12+z22)=f(z12)+f(z22) ∀z1,z2∈C z₁, z₂ C∀z1,z2∈C

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AP EAPCET 25-Aug-2021 Shift 1 Paper

Section: Mathematics

Question : 10 of 160

Marks: +1, -0

Define

f: C \rightarrow R

f(z)=|z| \forall z \in C

. Then which of the following is false?

Solution:

Given,

f(z)=|z|

(a)

f(-z)=f(z)

is true

|z|=|-z|

(b)

f(\overline{z})=f(z)

is true

|\overline{z}|=|z|

(c)

f(z^{2})=[f(z)]^{2}

is true

|z^{n}|=|z|^{n}

∴

f(z^{2})=|z^{2}|=|z|^{2}=[f(z)]^{2}

(d)

f(z_{1}^{2}+z_{2}^{2})=f(z_{1}^{2})+f(z_{2}^{2})

\Rightarrow f(z_{1}^{2}+z_{2}^{2})=|z_{1}^{2}+z_{2}^{2}|

\Rightarrow f(z_{1}^{2})+f(z_{2}^{2})=|z_{1}^{2}|+|z_{2}|^{2}

\Rightarrow |z_{1}^{2}+z_{2}^{2}| \neq |z_{1}|^{2}+|z_{2}|^{2}

∴ It is false.

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