Difference between $\mathbb{R}$-linear and $\mathbb{C}$-linear maps Suppose we have a $f\colon \mathbb{C} \to \mathbb{C} $. I am confused when people say $f$ is $\mathbb{R}$-linear map or $\mathbb{C}$-linear map. What is the difference between these two ?
 A: *

*A $\mathbb{C}$-linear map $f : \mathbb{C} \to \mathbb{C}$ needs to fulfill $f(x+y) = f(x) + f(y)$ for all $x,y \in \mathbb{C}$ and $f(ax) = af(x)$ for all $a,x \in \mathbb{C}$.

*A $\mathbb{R}$-linear map $f : \mathbb{C} \to \mathbb{C}$ needs to fulfill $f(x+y) = f(x) + f(y)$ for all $x,y \in \mathbb{C}$ and $f(ax) = af(x)$ for all $a \in \mathbb{R}$ and $x \in \mathbb{C}$.
So every $\mathbb{C}$-linear map as above is $\mathbb{R}$-linear, but not conversely. Consider for instance the complex conjugation $f(a+ib) = a-ib$, which is $\mathbb{R}$-linear, but not $\mathbb{C}$-linear.
A: Suppose $V$ is a vector space over two fields $F$ and $K$. Then a map $f:V\to V$ is said to be  $F$-linear if $f$ is a linear transformation over the field $F$ when $V$ is regarded as a vector space over $F$. Also a map $f:V\to V$ is said to be  $K$-linear if $f$ is a linear transformation over the field $K$ and in this case $V$ is regarded as a vector space over $K$. 
Now clearly a $\mathbb C$-linear map $f:\mathbb C\to \mathbb C$ is $\mathbb R$-linear as $\mathbb R$ can be treated as a sub-field of $\mathbb C$. But the converse is not true i.e. a $\mathbb R$-linear map $f:\mathbb C \to\mathbb C$ need not be a $\mathbb C$-linear map (since $f(z)=\bar{z}$ is the required map which is $\mathbb R$-linear but not $\mathbb C$-linear). But you can prove the following: 


a $\mathbb R$-linear map $f:\mathbb C\to \mathbb C$ is $\mathbb C$-linear if $f(i)=if(1)$.


