Let f(z) be continuous on all of mathbb C and satisfy…

Let $f(z)$ be continuous on all of $\mathbb C$ and satisfy $\ f(z+z')=f(z)+f(z')\ $ and $\ f(zz')=f(z)f(z')\ $ for all $z,z'$. Prove that $f\equiv0$, or $f(z)=z$ for all $z$, or $f(z)=\bar z$ for all $z$.

Step 1 · $f(0)=0$ and $f(1)\in\{0,1\}$

Setting $z=z'=0$ in additivity: $f(0)=2f(0)\Rightarrow f(0)=0$. Setting $z=z'=1$ in multiplicativity: $f(1)=f(1)^2\Rightarrow f(1)\big(f(1)-1\big)=0$, so $f(1)=0$ or $f(1)=1$.

Step 2 · Trivial case $f(1)=0$

For any $z$: $f(z)=f(z\cdot1)=f(z)f(1)=0$. So $f\equiv0$ on $\mathbb C$.

Step 3 · Case $f(1)=1$: pin down $f$ on $\mathbb Q$

By repeated additivity, $f(n)=nf(1)=n$ for every positive integer $n$; and $0=f(0)=f(n)+f(-n)$ gives $f(-n)=-n$. For $p/q\in\mathbb Q$ ($q\neq0$): $$ p=f(p)=f\!\left(q\cdot\frac pq\right)=f(q)f\!\left(\frac pq\right)=q\,f\!\left(\frac pq\right)\ \Longrightarrow\ f\!\left(\frac pq\right)=\frac pq $$ So $f(r)=r$ for every rational $r$.

Step 4 · Continuity extends this to all reals

The identity map $x\mapsto x$ and $f$ are both continuous on $\mathbb R$ and agree on the dense subset $\mathbb Q$, so they agree everywhere: $$ f(x)=x\qquad\text{for all }x\in\mathbb R $$

Step 5 · Pin down $f(i)$

Since $i^2=-1$: $$ f(i)^2=f(i^2)=f(-1)=-1 $$ (using Step 4, as $-1$ is real), so $f(i)=i$ or $f(i)=-i$.

Step 6 · Determine $f$ on all of $\mathbb C$

For $z=x+iy$ with $x,y\in\mathbb R$: $$ f(z)=f(x)+f(iy)=f(x)+f(y)f(i)=x+y\,f(i) $$ using additivity, multiplicativity ($f(iy)=f(y\cdot i)=f(y)f(i)$), and Step 4 ($f(x)=x$, $f(y)=y$).

If $f(i)=i$: $\ f(z)=x+iy=z$. If $f(i)=-i$: $\ f(z)=x-iy=\bar z$.

Verification

Both $\operatorname{id}(z)=z$ and $\operatorname{conj}(z)=\bar z$ are continuous, and each trivially satisfies both functional equations ($\overline{z+z'}=\bar z+\bar z'$ and $\overline{zz'}=\bar z\,\bar z'$); together with the zero map, these are exactly the three candidates identified above.