# Showing $f=u+iv$ satisfies $f'(x)=u_x(x,0)-iu_y(x,0)$ for real $x$.

Suppose $u,v$ are harmonic and satisfy the Cauchy-Riemann equations in $\mathbb{R^2}$. Show that $f=u+iv$ satisfies $f'(x)=u_x(x,0)-iu_y(x,0)$ for real $x$.

I don't really understand what the question is asking. I can see how $f=u+iv$ can be written as $f'(x)=u_x(x,0)-iu_y(x,0)$. Since $u,v$ are harmonic, am I suppose to use the fact $$u_{xx}+u_{yy}=0?$$A hint would be very helpful.

$$f(x,y)=u(x,y)+iv(x,y)$$ satisfies the Cauchy-Riemann equations if $$u_x=v_y, u_y=-v_x.$$

Thus

$$f_x(x,y)=u_x(x,y)+iv_x(x,y)=u_x(x,y)-iu_y(x,y).$$

• do you mean $u_x(x,y)-iu_y(x,y)?$
– user557493
Commented Aug 15, 2018 at 1:12
• @Bell Yes. Thank you for noticing the typo.
– mfl
Commented Aug 15, 2018 at 1:13
• Then the "harmonic" is redundant……
– xbh
Commented Aug 15, 2018 at 1:14
• All good. Does this conclude the solution?
– user557493
Commented Aug 15, 2018 at 1:15
• In my opinion the solution is complete. @xbh "Harmonic" is a consequence of Cauchy-Riemann equations. If $f=u+iv$ and $u,v$ satisfy Cauchy-Riemann equations then $f$ is holomorphic and $u,v$ are harmonic.
– mfl
Commented Aug 15, 2018 at 1:18