Define f: mathbb{C} rightarrow mathbb{R} by f | vedclass

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(D) Given $f(z) = |z|$.$(a)$ $f(-z) = |-z| = |z| = f(z)$,which is true.$(b)$ $f(\bar{z}) = |\bar{z}| = |z| = f(z)$,which is true.$(c)$ $f(z^2) = |z^2|

Vedclass · Accepted Answer

$f(z_1^2 + z_2^2) = f(z_1^2) + f(z_2^2), \forall z_1, z_2 \in \mathbb{C}$

Vedclass · Answer

(D) Given $f(z) = |z|$.$(a)$ $f(-z) = |-z| = |z| = f(z)$,which is true.$(b)$ $f(\bar{z}) = |\bar{z}| = |z| = f(z)$,which is true.$(c)$ $f(z^2) = |z^2| = |z|^2 = (f(z))^2$,which is true.$(d)$ $f(z_1^2 + z_2^2) = |z_1^2 + z_2^2|$ and $f(z_1^2) + f(z_2^2) = |z_1^2| + |z_2^2| = |z_1|^2 + |z_2|^2$.By the triangle inequality,$|z_1^2 + z_2^2| \leq |z_1^2| + |z_2^2| = |z_1|^2 + |z_2|^2$. Equality does not hold in general.Therefore,$f(z_1^2 + z_2^2) = f(z_1^2) + f(z_2^2)$ is false.

Define $f: \mathbb{C} \rightarrow \mathbb{R}$ by $f(z) = |z|, \forall z \in \mathbb{C}$. Then,which of the following is false?

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