I am trying to find out the absolute value of a Maclaurin power series of the below type: $f(x) = a_0 + a_1 x+a_2 x^2+ \dots + a_n x^n$, where $x$ is a complex number. I am interested to know the vakue of $|f(x)| = |a_0 + a_1 x+a_2 x^2+ \dots + a_n x^n|$.

I have tried looking around and found something relevant here. But, I am not sure if this relates to my problem and also what is the name of the equation $|\lambda(z)| = 1 + Re(a_n z^n) + O(z^n)$ and meaning of the term $O(z^n)$. Please let me know if there is any guidance in this regard. Thanks!


1 Answer 1



The Maclaurin series of a function $f(z)$ is a special case of the Taylor series...

$$T_f^{z_0}(z) = \sum_{k=0}^{\infty}{\frac{f^{(k)}(z_0)}{n!}(z - z_0)^k}$$

Where $z_0 = 0$...

$$T_f^{0}(z) = \sum_{k=0}^{\infty}{\frac{f^{(k)}(0)}{k!}z^k}$$

It is important to make the correction that $f(z)$ does not equal, it approximates the finite sequence...

$$f(z) \approx a_0 + a_1z + a_2z^2 + \ldots + a_nz^n$$

It is equal to the infinite sum...

$$f(z) = a_0 + a_1z + a_2z^2 + \ldots + a_nz^n + \ldots$$

So, the absolute value of $f(z)$ would be...

$$\left|T_f^{0}(z)\right| = \left|\sum_{k=0}^{\infty}{\frac{f^{(k)}(0)}{k!}z^k}\right|$$


What is the Maclaurin series for $g(z) = e^z$?

$$ T_g^{0}(z) = \sum_{k=0}^{\infty}{\frac{g^{(k)}(0)}{k!}z^k} \\\implies e^z = \sum_{k=0}^{\infty}{\frac{z^k}{k!}} $$

What is the Maclaurin series for $f(z) = \left|g(z)\right|$?

$$ T_f^{0}(z) = \left|\sum_{k=0}^{\infty}{\frac{g^{(k)}(0)}{k!}z^k}\right| \\\implies\left|e^z\right| = \left|e^{\operatorname{Re}(z) + i\operatorname{Im}(z)}\right| = \left|e^{\operatorname{Re}(z)}\right|\left|e^{i\operatorname{Im}(z)}\right| = e^{\operatorname{Re}(z)} \\\implies\left|e^z\right| = \sum_{k=0}^{\infty}{\frac{\operatorname{Re}(z)^k}{k!}} $$

  • $\begingroup$ Sorry to ask naive questions. But can you please let me know why $|e^{iIm(z)}|$ = 1? $\endgroup$
    – NIT_GUP
    Commented Nov 9, 2019 at 21:17
  • $\begingroup$ @NIT_GUP The answer is made clear by using Euler's formula (math.stackexchange.com/q/721784/490122) $\endgroup$
    – Landon
    Commented Nov 9, 2019 at 21:28

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