[proofplan] We identify multiplication by $i$ on $\mathbb{C}$ with a real linear map $J:\mathbb{R}^2\to\mathbb{R}^2$. A real linear map is complex-linear exactly when it commutes with this map $J$. Computing the two compositions $A\circ J$ and $J\circ A$ gives the equations $c=-b$ and $d=a$, and these same equations identify $A$ with multiplication by $a+ib$. [/proofplan] [step:Translate complex-linearity into commutation with multiplication by $i$] Let \begin{align*} J: \mathbb{R}^2 &\to \mathbb{R}^2 \\ (x,y) &\mapsto (-y,x) \end{align*} denote the real linear map corresponding, under the identification $\mathbb{C}\cong\mathbb{R}^2$, to multiplication by $i$. Because $A$ is already assumed to be real-linear, $A$ is complex-linear if and only if it respects multiplication by $i$, meaning \begin{align*} A(J(x,y))=J(A(x,y)) \end{align*} for every $(x,y)\in\mathbb{R}^2$. [guided] Under the identification $\mathbb{C}\cong\mathbb{R}^2$, the complex number $x+iy$ is represented by the vector $(x,y)$. Multiplication by $i$ sends \begin{align*} i(x+iy)=-y+ix, \end{align*} which corresponds to the vector $(-y,x)$. Therefore we define \begin{align*} J: \mathbb{R}^2 &\to \mathbb{R}^2 \\ (x,y) &\mapsto (-y,x). \end{align*} A complex-linear map is a map that is linear over $\mathbb{C}$, not merely over $\mathbb{R}$. Since $A$ is already real-linear by hypothesis, the only additional scalar multiplication that must be checked is multiplication by $i$: every complex scalar $\alpha+ i\beta$ acts as $\alpha$ times the identity plus $\beta$ times multiplication by $i$. Thus $A$ is complex-linear exactly when \begin{align*} A(J(x,y))=J(A(x,y)) \end{align*} for every $(x,y)\in\mathbb{R}^2$. [/guided] [/step] [step:Compute the commutation equations for the matrix of $A$] For every $(x,y)\in\mathbb{R}^2$, the given matrix representation gives \begin{align*} A(x,y)=(ax+cy,bx+dy). \end{align*} Therefore \begin{align*} A(J(x,y)) &=A(-y,x) \\ &=(c x-a y,d x-b y), \end{align*} while \begin{align*} J(A(x,y)) &=J(ax+cy,bx+dy) \\ &=(-b x-d y,a x+c y). \end{align*} Hence $A(J(x,y))=J(A(x,y))$ for every $(x,y)\in\mathbb{R}^2$ if and only if \begin{align*} c x-a y&=-b x-d y,\\ d x-b y&=a x+c y \end{align*} for every $(x,y)\in\mathbb{R}^2$. Comparing coefficients of $x$ and $y$ in these two polynomial identities over $\mathbb{R}$ gives \begin{align*} c=-b,\qquad -a=-d,\qquad d=a,\qquad -b=c. \end{align*} Equivalently, \begin{align*} a=d,\qquad c=-b. \end{align*} [/step] [step:Identify the resulting map as multiplication by $a+ib$] Assume first that $A$ is complex-linear. By the preceding step, its matrix coefficients satisfy $a=d$ and $c=-b$. Conversely, assume that $a=d$ and $c=-b$. Then for every $(x,y)\in\mathbb{R}^2$, \begin{align*} A(x,y)=(ax-by,bx+ay). \end{align*} Under the identification $(x,y)\leftrightarrow x+iy$, multiplication by $a+ib$ gives \begin{align*} (a+ib)(x+iy) &=(ax-by)+i(bx+ay), \end{align*} which corresponds exactly to $(ax-by,bx+ay)$. Therefore $A$ is multiplication by $a+ib$, and in particular $A$ is complex-linear. This proves both directions and the final identification. [/step]