Let $f$ be defined as

$$f(x) = \frac{i}{2\pi} Log(-\exp(2i\pi x)) , \forall x \in \Bbb{R} $$

Prove $f(x)=0.45$ when $x=5.05$.

I know that $Log( a + i b) = ln(|a+ib|) + i Arg(a+ib)$ so

$$Log(-\exp(2i \pi x)) = iArg(-\exp(2i \pi x)) = 2i * arctan(\frac{ sin(2\pi x)}{cos(2\pi x) - 1 })$$

I get stuck on arctan. How to go through ?

  • 1
    $\begingroup$ $$f(x)=\lceil x\rceil - x-\frac12$$ for $x\gt0$ $\endgroup$ – Peter Foreman Jun 24 '19 at 10:23
  • $\begingroup$ @PeterForeman Sure, however I'd like to find the result with arctan. $\endgroup$ – Stephan Jun 24 '19 at 10:29
  • $\begingroup$ I don't understand why you would want to do that. $\endgroup$ – Peter Foreman Jun 24 '19 at 10:30

If you consider for $x\in\mathbb{R}$ the expression $\mathrm{exp}(2\pi i x)$, then every time your $x$ is a multiple of an integer, you are in the same spot as for $x=0$ and then begin again (you are going in circles of radius 1 around the origin z = 0). So, $x=5.05$ gives you the same value as $x = 0.05$. Then, by using $-1 = \mathrm{exp}(-i\pi)$ and setting $x=0.05$ you get $$\dfrac{i}{2\pi}\log(-\mathrm{exp}(2\pi i 0.05))= -\dfrac{1}{2\pi}\mathrm{Arg}(\mathrm{exp}(2\pi i 0.05 - i\pi)) = -\dfrac{2 \pi }{2 \pi} 0.05 + \dfrac{\pi}{2\pi}=0.45.$$

Note that your function has multiple "branches", so when you try to find one value, you usually look for the value found in the "principal branch", which you achieve by evaluating your function in $x\mathrm{mod} 1$.

  • $\begingroup$ every time your x is a multiple of an integer, you are in the same spot as for x=0 this hint clearly leads me to the solution. Thanks. $\endgroup$ – Stephan Jun 25 '19 at 7:53

Here is how I solved the problem. With arctan, I was doing things the wrong way.

$$ \begin{align} Log(-\exp(2i\,\pi\,x)) &=iArg(-\exp(2i\,\pi\,x))\\[6pt] &=i\,[2\pi x + (2k+1)\pi]; k \in \Bbb{Z} \\[6pt] \end{align} $$


$$ \begin{align} f(x) &=\frac{i}{2\,\pi}\,i\,[2\pi x + (2k+1)\pi]; k \in \Bbb{Z}\\[6pt] &=\frac{-1}{2\,\pi}\,[2\pi x + (2k+1)\pi]\\[6pt] &=-[x+k+\frac{1}{2}]\\[6pt] \end{align} $$

We want the principal value of $f(x)$ so the following inequality must hold:

$$-\pi < [2\pi x + (2k+1)\pi] \leqslant \pi$$

It holds for any $k \in \Bbb{Z}$ verifying:

$$-x-1 < k \leqslant -x; k \in \Bbb{Z}$$

If $x=5.05$ then $k=-6$ and $f(x) = -[5.05 -6 + 0.5] = 0.45$


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