Example of a function $\varphi: \mathrm{C} \to \mathrm{C}$ such that $\varphi(z + w) = \varphi(z) + \varphi(w)$ but $\varphi$ is not homogeneous

Question

Context: This is the exercise 9 of chapter 3.A of Linear Algebra Done Right by Axler. We want to find such a complex function $\varphi$, that $\varphi$ is additive but non-homogeneous, i.e. $\forall z, w \in \mathrm{C}: \varphi(z + w) = \varphi(z) + \varphi(w)$, but $\exists c, z \in \mathrm{C}: \varphi(cz) \neq c\varphi(z)$. The function $\varphi$ being additive means that for any $z \in \mathrm{C}$ we know that for any point $z$ and change $z'$, the image of a point and its change in coordinates $z + z'$ is equal to the sum of the image of the point $z$ and the image of the said change $z'$. But $\varphi$ being non-homogeneous means that the scaling of some points does not match the scaled version of the image of the said points.

Question: Any hint is welcome, as right now I have no clue on how to start to construct the function $\varphi$.

Possibly helpful: Overview of basic facts about Cauchy functional equation. — Martin R, Oct 20 '21 at 16:17
Complex conjugation is not $\Bbb C$-homogeneous. See this https://math.stackexchange.com/a/1836869/591889 — Sumanta, Oct 20 '21 at 16:18
$f(z+w) = f(z) + f(w)$ is NOT linearity. It is called additivity. Linearity is indeed additivity + homogenity. — amsmath, Oct 20 '21 at 17:13

score 2 · Accepted Answer · answered Oct 20 '21 at 18:41

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The function $f : \mathbb C \rightarrow \mathbb C, f(z) = z^*$ is additive, but not homogeneous, because $f(iz) = i^*z^* = -iz^* \neq iz^*$. This function is known as complex conjugation.

answered Oct 20 '21 at 18:41

Angel

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Example of a function $\varphi: \mathrm{C} \to \mathrm{C}$ such that $\varphi(z + w) = \varphi(z) + \varphi(w)$ but $\varphi$ is not homogeneous

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