How do I show that the Cauchy-Riemann Equations hold for any polynomial?

Question

I'm self-studying complex analysis and trying to show that the Cauchy-Riemann Equations hold for any complex polynomial $$f(z) = a_n z^n + \dots + a_1 z + a_0$$ but I'm unsure how to actually get a general expression for the real and imaginary parts of $f(z)$.

My attempt has been to suppose that $z = x + iy$, then $f(z)$ can be written as $$f(z) = a_n (x + iy)^n + \dots + a_1 (x + iy) + a_0 = \sum_{k = 0}^{n} a_k (x+iy)^k$$

Now, using the binomial we can write $$ (x+iy)^k = \sum_{j=0}^{k} \binom{k}{j}x^{k-j}(iy)^{j} $$

Thus, we can express $f(z)$ as a double sum: $$f(z) = \sum_{k = 0}^{n} a_k \left( \sum_{j=0}^{k} \binom{k}{j}x^{k-j}(iy)^{j} \right)$$

But I don't know how to proceed further, and separate the real and imaginary parts. Any hints or suggestions would be appreciated!

Answer 1

If we write $f(z)=u(z)+iv(z)$ we can calculate $f^{\prime}(z)$ depending on the way we approach the limit. Either along the real axis \begin{align*} f^{\prime}(z)&=\frac{\partial f}{\partial x}=\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}\tag{1}\\ \end{align*} or along the imaginary axis \begin{align*} f^{\prime}(z)&=-i\frac{\partial f}{\partial y}=-i\frac{\partial u}{\partial y}+\frac{\partial v}{\partial y}\tag{2}\\ \end{align*} We obtain be equating (1) and (2) the CR equation in the real form as \begin{align*} \frac{\partial u}{\partial x}=\frac{\partial v}{\partial y},\quad \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x} \end{align*} Here we use from (1) and (2) the complex form and get \begin{align*} \color{blue}{\frac{\partial f}{\partial x}=-i\frac{\partial f}{\partial y}}\tag{3} \end{align*}

We consider \begin{align*} f(z) =f(x+iy)= \sum_{k = 0}^{n} a_k (x+iy)^k\tag{4} \end{align*}

We apply the left-hand side of (3) to (4) and we obtain \begin{align*} \color{blue}{\frac{\partial f}{\partial x}} &=\frac{\partial }{\partial x}\sum_{k = 0}^{n} a_k (x+iy)^k\\ &=\sum_{k = 0}^{n}\frac{\partial }{\partial x}a_k (x+iy)^k\\ &\,\,\color{blue}{=\sum_{k = 1}^{n}ka_k (x+iy)^{k-1}}\\ \end{align*} Now we apply the right-hand side of (3) to (4) and we obtain \begin{align*} \color{blue}{-i\frac{\partial f}{\partial y}}&=-i\frac{\partial }{\partial y}\sum_{k = 0}^{n} a_k (x+iy)^k\\ &=-i\sum_{k = 0}^{n}\frac{\partial }{\partial y}a_k (x+iy)^k\\ &\,\,\color{blue}{=\sum_{k = 1}^{n}ka_k (x+iy)^{k-1}}\\ \end{align*}

Both calculations show the same result according to the claim.