Consider a function f:C↦R. As far as I know, for this function f(x,y)=u(x,y)+iv(x,y) to be analytic, it must satisfy the Cauchy-Riemann conditions, which state:∂u∂x=∂v∂y∂u∂y=−∂v∂xOf course, v(x,y)=0. This implies that u(x,y)=C where C is constant, and thus the only real function of a complex variable is f(z)=C.This cannot be right, surely?

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Consider a function f:C↦R. As far as I know, for this function f(x,y)=u(x,y)+iv(x,y)  to be analytic, it must satisfy the Cauchy-Riemann conditions, which state:∂u∂x=∂v∂y∂u∂y=−∂v∂xOf course, v(x,y)=0. This implies that u(x,y)=C where C is constant, and thus the only real function of a complex variable is f(z)=C.This cannot be right, surely?

Marquis Ibarra · Accepted Answer

Your proof is right! Here is a different proof: to say a function is holomorphic at a point z0 is to say that the limitlimh→0f(z0+h)−f(z0)hexists where h is approaching 0 in the complex plane. If f takes only real values, the quotient will be purely imaginary if h approaches 0 along the imaginary axis and purely real if h approaches 0 along the real axis, so for the limit to exist at z0 the derivative must be zero and since z0 was arbitrary, f is constant.What&#039;s going on here is that the condition that the limit has to exist when h can be any complex number going to 0 is much stronger than the analogous condition in real variables.

Consider a function \(\displaystyle{f}:{\mathbb{{{C}}}}\mapsto{\mathbb{{{R}}}}\). As far as

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