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Q: Given complex conjugates $z = a+bi$ and $\overline z = a-bi$, what would be substitution needed to find $R$, $$R= z\ln z+\overline{z}\ln\overline{z}\tag1 $$ such that $R$ is an expression without imaginary numbers?

This question arose as a special case of this post. Given $x^3-x-1=0$, let $x_1\approx1.3247$ be its real root (the plastic constant) and $x_2,x_3$ its complex roots. Then

$$\begin{aligned}\sum\limits_{n=0}^\infty\frac{n!\,(2n)!}{(3n+2)!} &= x_1\ln(1+x_1)+x_2\ln(1+x_2)+x_3\ln(1+x_3)\\ &= \frac32\,x_1\ln(1+x_1)-\frac{1}{2}\sqrt{\frac{3-x_1}{x_1}}\arccos\Big(\frac{x_1-6}{6x_1+2}\Big)\\[2mm] &= 0.5179778\dots\end{aligned}$$

The first is by yours truly, while the second is a simplified version of a result by Reshetnikov. How do we show the equality of the first two lines? In general, how do we get rid of the imaginary unit in $(1)$, and can we always express it as the sum of $\ln$ and $\arccos$ of real numbers?

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  • $\begingroup$ It's easy enough to show that $R = 2 (a \ln |z| - b \arg(z))$, and $|z|$ and $\arg(z)$ can easily be expressed in terms of $a$ and $b$. But I don't think this fact actually helps you prove the identity on its own; it also seems to rely on some relationship between the roots $x_1$, $x_2$, and $x_3$. $\endgroup$ – Michael Seifert Jan 5 '17 at 16:15
  • $\begingroup$ @MichaelSeifert: I'm face-palming myself. Yes, the polar form. $\endgroup$ – Tito Piezas III Jan 5 '17 at 16:28
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If $f(x)=x^3-x-1$, define $$g(x)=f(x-1)=x^3-3x^2+3x-1 - (x-1)-1=x^3-3x^2+2x-1.$$

So $z_i=1+x_i$ are the roots of $g(x)$ and $z_1z_2z_3=1$ so $\log z_1+\log z_2+\log z_3 = 0$ and thus:

$$\sum x_i\log(1+x_i)=\sum z_i\log z_i=z_1\log z_1 +2(\log|z_2|\mathrm{Re}(z_2)-\arg(z_2)\mathrm{Im}(z_2))$$

(By the arguments in the other answers.)

Now since $z_1z_2z_3=1$, you have that $|z_2|=|z_3|=\frac{1}{\sqrt{1+x_1}}$. Since $z_1+z_2+z_3=3$ you have that $\mathrm{Re}(z_2)=\frac{3-z_1}{2}=\frac{2-x_1}{2}$. So $$2\log|z_2|\mathrm{Re}(z_2) = 2\cdot \left(-\frac{1}{2}\log(1+x_1)\right)\cdot\left(1-\frac{x_1}{2}\right)$$

So $$\begin{align}z_1\log z_1 + 2\log|z_2|\mathrm{Re}(z_2) &= x_1\log(1+x_1)+\log(1+x_1) + \frac{1}{2}x_1\log x_1 - \log(1+x_1)\\ &=\frac{3}{2}x_1\log(1+x_1)\end{align}$$

Finally, since $z_2=\frac{2-x_1}{2}+bi$, and $|z_2|$ is known, we can determine both $b=\mathrm{Im}(z_2)$ and $\arg(z_2)=\arccos\left(\frac{\mathrm{Re}(z_2)}{|z_2|}\right)$.

This give $b=\frac{1}{2}x_1\sqrt{\frac{3-x_1}{1+x_1}}$. But $1+x_1=x_1^3$, so $b=\frac{1}{2}\sqrt{\frac{3-x_1}{x_1}}$.

I'm having a hard time getting your result, I get the last term as:

$$-\sqrt{\frac{3+x_1}{x_1}}\arccos\left(\frac{(2-x_1)\sqrt{1+x_1}}{2}\right)$$

You can use that $2\arccos u = \arccos(2u^2-1)$ when $u\in[0,1]$. But that doesn't seem to give your result, either.

[Wolfram alpha tells me that $2u^2-1=\frac{x_1-6}{6x_1+2}$ when $u=\frac{(2-x_1)\sqrt{1+x_1}}{2}$, but I wouldn't have found that final expression myself.]

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If $z=re^{it}$ then:

  • $\bar z=re^{-it}$
  • $\ln z=it+\ln r$
  • $\ln\bar z=-it+\ln r$

So $R=(a+ib)(\ln r+it)+(a-ib)(\ln r-it)=2(a\ln r-bt)=2(\ln|z|\cdot \Re z-\arg(z)\cdot\Im z)$

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With a notation that should be obvious,

$$R=(x+iy)(\log r+i\theta)+(x-iy)(\log r-i\theta)=2(x\log r-y\theta)=2\left(x\log\sqrt{x^2+y^2}-y\,\arccos\frac xr\right).$$

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