Does there exist an analytic function whose real part is given by $u(x, y) = x^2 + y^2$, where $z = x + iy$?

My attempt: I think yes because it will satisfy the Cauchy-Riemann equation; we know that every function that satisfies the Cauchy Riemann equation is analytic.

Is this true?

  • 1
    $\begingroup$ You need two functions to test the CR equations. What is your candidate for the imaginary part? $\endgroup$
    – lulu
    Jan 8, 2020 at 20:38
  • $\begingroup$ may be $0$@lulu $\endgroup$
    – jasmine
    Jan 8, 2020 at 20:39
  • 2
    $\begingroup$ There can't be an analytic function which only takes real values (other than constants). $\endgroup$
    – lulu
    Jan 8, 2020 at 20:41

2 Answers 2


There is not an analytic function with such property because $u$ is not harmonic: $u_{xx}=2$ and $u_{yy}=2$ so $u_{xx}+u_{yy}=4\neq 0$.

There is a result that the real and the imaginary parts of an analytic function must be harmonic (by the way, this follows from Cauchy-Riemann equations).


Any such analytic function $f=u+iv$ must satisfy

$$2x=\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$$


$$2y=\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$$

The first equation gives $v=2xy+a(y)$ but the second equation gives $v=-2xy+b(x)$. You can then see that there is no such $v$.


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