Harmonic functions proof explanation

Question

I need help understanding a step in a proof.

Theorem: The imaginary and real part of a regular function on $G$ are harmonic functions on $G$ Proof: Let $f(z) = u + iv$ for $u = u(x,y)$ and $v=v(x,y)$ be regular functions on G. Then we have: $$f'(z) = \frac{\delta u}{\delta x} + i \frac{\delta v}{\delta y} = \frac{\delta v}{\delta y} \color{red}{ - i \frac{\delta u}{\delta y}}$$ $$f''(z) = \frac{\delta^2 u}{ \delta x^2} + i \frac{\delta^2 v}{\delta y^2} = -\frac{\delta^2 u}{\delta y^2} \color{red}{- i \frac{\delta^2v}{\delta y^2}}...$$ I do not understand where the red terms come from.

Answer 1

I think there is a mistake in the first equality of your formula. $$f'(z)=\lim_{t \to 0} \frac{f(z+t)-f(z)}{t}$$

Chosing $t \in \mathbb R$ gives $$f'(z)=\lim_{t \to 0} \frac{u(x+t,y)+iv(x+t,y)}{t}=\frac{\delta u}{\delta x} + i \frac{\delta v}{{ \delta \color{red} x}}$$

Now, C-R gives the first equality.

Now derivating one more time, you should get $$f''(z) = \frac{\delta^2 u}{ \delta x^2} + i \frac{\delta^2 v}{\delta \color{red} x^2}$$

Answer 2

Those are called the C-R equations.A complex fuction is diffrentiable only if it satisfies u_x=v_y and u_y=-v_x.You can see the proofs here. http://mathworld.wolfram.com/Cauchy-RiemannEquations.html