Why does $f = u+iv$ holomorphic $\implies$ $-if = -iu + v$ holomorphic? If we multiply both sides of a holomorphic complex-valued function $f = u(x,y) + iv(x,y)$ by $-i$, why is it true that the resulting equation is also holomorphic?
 A: Because (complex) differentiation is a linear operation (over $\mathbf{C}$), so if $f$ is (complex) differentiable (i.e. holomorphic), then so is $\alpha f$ for all $\alpha\in\mathbf{C}$ (i.e. $\alpha f$ is also holomorphic). Your question consider $\alpha=-i$.
Edit
If we denote complex differentiation by $'$, then linearity of $'$ is stated by the properties
$$
(f+g)'=f'+g'\quad\text{and}\quad(\alpha f)'=\alpha f'
$$
where $f$ and $g$ are arbitrary differentiable functions and $\alpha\in\mathbf{C}$.
A: Let $g(z) = -iz$. $g$ is just multiplication by a constant and therefore holomorphic. 
$g(f(z))=-iu+v$ must be holomorphic because the composition of two holomorphic functions is holomorphic. 
A: Because the product of two holomorphic function is holomorphic.
A: Consider $f=u+iv$ if $f$ is holomorphic then $u_x=v_y$ and $u_y=-v_x$. Consider $g=-if=-v-iu$. Notive $(-v)_x = (-u)_y$ and $(-v)_y = -(-u)_x$ by application of the Cauchy-Riemann equations for $f$. Furthermore as $f$ is holomorphic we know $u,v$ and by extension $-u,-v$ are continuously differentiable hence $g$ is holomorphic on the same domain on which $f$ was holomorphic.
