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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 $f(z)=0$?

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    $\begingroup$ What is the only possible $z$ that makes $z^2=0\,?$ $\endgroup$
    – Kurt G.
    Commented Feb 24 at 8:55
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    $\begingroup$ I am afraid that calculating the roots of the Riemann Zeta function is slightly more complicated than determining the roots of $z^2 = 0$ ... $\endgroup$
    – Martin R
    Commented Feb 24 at 8:56
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    $\begingroup$ 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 $z\mapsto z^2\,.$ $\endgroup$
    – Kurt G.
    Commented Feb 24 at 9:19
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    $\begingroup$ 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. $\endgroup$
    – GameTime With Aryan
    Commented Feb 24 at 9:33
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    $\begingroup$ 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. $\endgroup$
    – Zubin Mukerjee
    Commented Feb 24 at 12:46

1 Answer 1

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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.

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  • $\begingroup$ 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: mathematica.stackexchange.com/questions/298498/… Why can we plot $f(z)=z^2$ in a similar way to see the zeros geometrically? $\endgroup$
    – zeynel
    Commented Feb 26 at 7:31
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    $\begingroup$ 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. $\endgroup$
    – GameTime With Aryan
    Commented Feb 26 at 14:20

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