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I'm having difficulty grasping how the zeros of Riemann zeta function is calculated. I thought studying a simpler function such as $f(z)=z^2$ may help.

I'm reading this LibreText document (PDF). I understand how the function $f(z)=z^2$ maps the lines $x=1$ and $y=1$ in the z-plane into parabolas in the w-plane. (Fig. 8.3.2)

Can you give any clues about how to calculate $f(z)=0$?

zeynel
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    What is the only possible $z$ that makes $z^2=0,?$ – Kurt G. Feb 24 '24 at 08:55
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    I am afraid that calculating the roots of the Riemann Zeta function is slightly more complicated than determining the roots of $z^2 = 0$ ... – Martin R Feb 24 '24 at 08:56
  • @KurtG. Thanks. That should be $z=(0+0i)$. So, in Fig. 8.3.2, the plot on the right hand side, the dotted line and the normal line cross each other at -2 and 2. Are these the zeros? I assume the dotted line is the Re(z) and the other line is the Im(z). Correct? If this is true, the definition of "zeros" in complex valued functions must be different than real valued functions. Are the two points where the parabolas cross the horizontal axis zeros too? – zeynel Feb 24 '24 at 09:15
  • @MartinR Yes, this is true. But as I replied to @Kurt G I'm trying to understand the basic idea. – zeynel Feb 24 '24 at 09:18
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    We have just determined the zeroe(s) of $z^2=0$ and you ask if $z=2$ and $z=-2$ are the zeroes. Think about this. On top of that: your understanding of what those lines are is completely wrong. They are the images of two orthogonal straight lines under the mapping $z\mapsto z^2,.$ – Kurt G. Feb 24 '24 at 09:19
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    You should not be looking up Riemann-zeta function at this point. Your understanding of functions seem to have gaps, so you should review your concepts on functions, polynomials, graphs etc. – GameTime With Aryan Feb 24 '24 at 09:33
  • @KurtG. I can see that the real valued $x^2$ is zero at $x=0$ https://www.wolframalpha.com/input?i=x%5E2 but I don't see where to look for the zeros of the complex valued function. Do I look at the first plot on Fig. 8.3.2 or the second (mapping plot)? And, in the mapping plot parabolas cross at $u=0$. This is confusing to me that's why I asked. – zeynel Feb 24 '24 at 09:34
  • As I see it you found the correct unique zero of the complex $z^2$ right after my first comment. Stop using Wolfram Alpha for such simple stuff and looking at graphs that have nothing to do with that problem. – Kurt G. Feb 24 '24 at 10:56
  • @KurtG. Then, maybe you can refer me to a plot of $f(z)=z^2$ with that zero visible. – zeynel Feb 24 '24 at 11:36
  • @GameTimeWithAryan This is exactly what I'm doing, I'm studying complex functions from LibreText. I'm having difficulty visualizing the zeros of complex functions. If you can refer me to a plot of $f(z)=z^2$ showing the zero, that would be appreciated. – zeynel Feb 24 '24 at 11:46
  • If you absolutely want a plot, hint: write $z\mapsto z^2$ as vector field $\mathbb R^2\to\mathbb R^2,.$ Then plot. – Kurt G. Feb 24 '24 at 11:50
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    A useful graph might be $f(x) = (x-2)(x-3)$. Here, $f$ is $\mathbb{R} \to \mathbb{R}$, but this will demonstrate some ideas that extend to $\mathbb{C}\to \mathbb{C}$ functions. Note that the zeroes of this function are the points at which the graph intersects the $x$-axis. Move the 2 and 3 toward 0 and observe how the graph changes. You will understand why $g(x) = x^2$ has only one root, but with multiplicity 2. This idea of roots having multiplicities extends to roots of complex functions. – Zubin Mukerjee Feb 24 '24 at 12:46
  • @ZubinMukerjee Thanks. I watched this video https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:poly-graphs/x2ec2f6f830c9fb89:poly-intervals/v/polynomial-zero-multiplicity#:~:text=The%20polynomial%20p(x)%3D(,a%20zero%20of%20multiplicity%201. to understand what multiplicity of 2 means. Fine. But, I don't see the relation to plotting the zeros of complex valued functions. – zeynel Feb 26 '24 at 06:31
  • For anybody who comes here in the future with same problem visualizing complex functions the key seems to be to work in polar coordinates. This great video by Michael Penn https://youtu.be/El1pwxq87Ws?si=kOP6JPFfNxEoeu3U clarified many of the problems for me. It's hard to understand why in a mathematical forum like this a legitimate mathematical question is closed by overzealous moderatiors. I think the question should be opened and a proper answer be given. – zeynel Feb 26 '24 at 09:29

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$$f(z) = z^2 = (x+iy)^2$$

Remember that $a$ is a zero of the function if $f(a) = 0$, now

$$a= x_a + iy_a$$

Where $x_a$ and $y_a$ are real numbers

$$(x_a+iy_a)^2 = 0$$

$$x_a + iy_a = 0$$

Which is only possible when $(x_a,y_a)=(0,0)$ as $x$ and $y$ are real numbers, hence $a=0 + i0 = 0$ is the only zero of $f(z)=z^2$

In LibreText document (PDF), in fig 8.3.3, the point (x=0,y=0) in the z-plane gets mapped to (u=0, v=0) in the w-plane. The point from the z-plane which maps to (u=0, v=0) will be the zero of the mapping function.

Notice that the axis of the graphs in Fig 8.3.2 and 8.3.3 do not cross at the origin, which may be confusing you. Zeros are not defined by the fact that they touch an axis (the axis can be shifted as in your case), but by the fact that they make the function equal to $0$

Plotting

In the comments, you also asked for a plot. The plot is given in your LibreText document in Fig 8.3.3 for a set of grid lines. You have to understand that a plot for mapping all the set of points in the z-plane into the w-plane will be meaningless as almost all points in the w-plane will be plotted out, so it would look like a dark mess. The standard way to plot complex functions is by taking a set of grid lines in the z-plane and mapping them to the w-plane using the complex function.

You must realize that plotting works differently for complex-valued functions than real ones.

These plots are different from your plots of real-valued functions. The input and output are each 2 dimensional, hence you need 1 plot for input and 1 plot for output. Hence, you cannot determine the zeroes of the function just by looking at either plot.

The y-axis on each plot represents the imaginary axis and x-axis represents the real number line.

GameTime With Aryan
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  • Thanks, good answer but I still have to work on this. Also you wrote, "you cannot determine the zeroes of the function just by looking at either plot." But for the zeta function, they have this famous plot showing the zeros graphically: https://mathematica.stackexchange.com/questions/298498/what-is-the-exact-formula-mathematica-uses-for-riemann-zeta-function Why can we plot $f(z)=z^2$ in a similar way to see the zeros geometrically? – zeynel Feb 26 '24 at 07:31
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    I am also a student hence I have no idea about the linked graph. Your question asked about "calculating" the zeros, but if you want to understand more about graphing then maybe you can ask a separate question specifically for that. I apologize for not being able to help. – GameTime With Aryan Feb 26 '24 at 14:20