Let $f(z) = u(x,y) + iv(x,y)$ be an holomorphic function and $2u(x,y) + v(x,y) = 5$. Show that $f$ is constant

Question

I understand that it suffices to show that $f'(z)=0$

What I don't understand though, is how to take a partial derivative from $2u(x,y) + v(x,y) = 5$ ...

Another solution I thought of is,

If $2u(x,y) + v(x,y) = 5$, and 5 is a real number. Then 2 $y1$ and $y2$ is equal to $0$, and 2 $x1$ and $x2$ is equal to $5$.

meaning that if I take the derivative of 2 $x1$, its always going to be zero because 5 is a constant. $\frac{du}{dx} = 0$ and since it's analytic, by CRE it implies that $\frac{dv}{dy}$ is also zero.

Likewise if I take the derivative of $x2$, $\frac{dv}{dx}$ is also going to be $0$, which again by CRE implies that $-\frac{dv}{dx} = \frac{du}{dy}$.

So it's all $0$, meaning that $f$ is a constant.

Any hints? or suggestions in regards to this solution?

Answer 1

The image of such an analytic function would be a subset of a line in the plane. I don't know where you are in your study of complex variable, but such a function would not be an open mapping [if it is nonconstant], so it cannot be analytic.

Answer 2

Taking partial derivatives of $$2u+v=5$$ gives $$2u_x+v_x=0\\2u_y+v_y=0$$ Use the Cauchy-Riemann equations to express the partials with respect to $y$ in terms of the partials with respect to $x$, and conclude that all fo them are $0$.

Answer 3

"I understand that it suffices to show that $f′(z)=0$". This is true only for domains in the complex plane. For a counterexample, take $f \equiv 1$ in $B_1 (0)$ and $f \equiv 0$ in $B_1 (3)$. $f$ is only locally constant. Other than that, @saulspatz already gave the answer I was about to give you.