# to show $f$ is a polynomial in $z$

given that $f$ is an analytic function with real part $u$ is a polynomial in the variable of $x,y$, $z=x+iy$, we need to show $f$ is a polynomial in $z$, I am kind of puzzled to see the problem, first I would like to say:any hint please

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If $f(x,y) = u(x,y)+iv(x,y)$ is analytic in some region of the complex plane, then $u_x = v_y$ and $u_y = -v_x$ -- the Cauchy-Riemann equations.

If $u$ is a function of $x$ and $y$, then $u_x$ and $u_y$ are both non-zero.

If $u(x,y)$ is a polynomial in $x$ and $y$, then we write

$$u(x,y) = a_0 + a_1 x^1 + a_2 x^2 + \cdots + a_k y^1 + a_{k+1}y^2 + \cdots + a_n x^1y^1 + a_{n+1} x^1y^2 + a_{n+2}x^2 y^1 + \cdots$$

Differentiate this to obtain $u_x$ and $u_y$. From there, you can obtain $v_x$ and $v_y$.

Integrate either $v_x$ with respect to $x$, or $v_y$ with respect to $y$, and show that the resulting $v(x,y)$ is a polynomial.

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When ${\rm Re}(f)=u$ is a polynomial in $x$ and $y$ then so is ${\rm Re}(f')=u_x$. It follows that there is an $n\geq0$ with ${\rm Re}(f^{(n)})\equiv0$, or $f^{(n)}(z)\equiv0$. Therefore $f$ has to be a polynomial in $z$.

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Let $f=u+vi$. By Cauchy-Riemann equation $$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$$ Calculate $\frac{\partial u}{\partial x}$ and find antiderivative of $\frac{\partial u}{\partial x}$ by $y$.

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