I am self studying Apostol book and could not think about doubts in this theorem .

Images of proof are at bottom of this page.

EDIT 1 -> in later part of theorem I have some more doubts which I am writing after doubt 1 .

I have doubt in Theorem 2.4 whose statement is -

If f is modular and not identically 0 , then in closure of fundamental region R $\Gamma $ , number of zeroes of f are equal to number of poles.

In proof curve (2) is defined as curve in fundamental region (which is boundary) from $ \rho $ to i and curve (3) is boundary from i to $ \rho $ + 1 . In case , when fundamental region is cut down by Img( $ \tau $ ) = M, where M is taken so large that all zeroes or poles of f are inside truncated region. Let N and P denote the number of zeroes and poles of f inside R. Then N-P = $\frac {1} {2πi } $ ( $\int $ $\frac{f'( \tau ) } {f(\tau) }$ d$\tau $ )and integral is taken over boundary of fundamental region.

The doubt 1 is -> Integral over (2) and (3) cancels because (2) gets mapped onto (3) with a reversal of direction under u = S( $ \tau $ ) = $\frac { -1} {\tau } $ .

Then Apostol shows how integrand remains unchanged and I completely understand it.

But I am not able to understand how u = $\frac {-1} { \tau }$ maps (2) to ( 3) with reversal of direction.

DOUBT 2 ->

Statement of Apostol ->After that mapping cancels each other only 1 integral (5) is remaining which is $ \tau $ = u +iM . Then integral is transformed into x-plane by x = e^(2πi $ \tau $ ) . So, x varies once around a circle K of radius e^(-2πM ) . around x=0 . Where Fourier expansion of f($ \tau $ ) starts from n= -m.

If I denote $ N_F $ and $ P_F $ by number of zeroes and poles of F inside K then it is easily proved that N-P = $ P_F $ - $ N_F $ .

Then I have doubts in following lines -

Argument of Apostol 1. - > If there is a pole of order m at x=0 then m>0 $ N_F $ =0 , $ P_F $ = m so N= P+m. So, from this relation I deduced that f takes value 0 same or greater number of times than it takes $ \infty $ in fundamental region.

    • If there is a zero of order n at x =0 , then m = -n , so $ P_F $ =0 , $ N_F $ = n , hence N+ n = P.

BY this relation I deduced that f takes value $ \infty $ at least as often it takes value 0 in fundamental region.

But Apostol writes , using above mentioned relation that f takes value 0 in fundamental region as often as it takes value$\infty $ .

But I deduced opposite conclusion.

So, can somebody please tell how to deduce conclusion which Apostol writes in his book.

DOUBT 3 - Even if I assumes both conclusions of Apostol to be true , I am unable to deduce the next line which is--> this proves the theorem if f has no zeroes or poles on finite part of boundary of fundamental region ie number of zeroes are equal to number of poles.

What I can deduce by my self assuming Apostol conclusions to be true ( assuming my 2nd deduction is false) -> f takes value 0 as often as $\infty $ which means f assumes 0 at least as number of times as f assumes value $ \infty $ .

So, no. of zeroes could be greater than no. of poles. I don't know how they become equal.

DOUBT 4 - In the case when f has zero or pole at vertex $ \rho $ or i, then detours are introduced. Now, there are 4 integrals, 1 detour avoiding $\rho $ ( C1) , 1 avoiding i (C 2) and 1 avoiding $ \rho $ +1 and 1 on the upper boundary.

My doubt is --

On the path C1, while calculating 1st integral , when r -> 0 how does $\alpha $ ' = π/2 - $ \alpha $ tends to π/3 .

Can somebody please explain these. I know it has became a lengthy problem but I am really struck on it and I am self studying. Please help.

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  • 1
    $\begingroup$ Please reformat your post and use LaTeX. It's pretty hard to read as is. $\endgroup$
    – bsbb4
    Dec 6, 2019 at 13:14
  • 3
    $\begingroup$ This question is in desperate need of some focus. Can you please edit it down to a single question, and ask that question in less than four paragraphs? Provide references to the important definitions, not screenshots, please. $\endgroup$
    – Xander Henderson
    Dec 8, 2019 at 23:29
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    $\begingroup$ @Xander Henderson I wrote it in 2-3 parts just because the idea used in them are not depentent on each other although they are part of same theorem $\endgroup$
    – user775699
    Dec 9, 2019 at 6:55
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    $\begingroup$ @ Xander Henderson I have added screenshots of proof so that any person who doesn't understand any part of what I have written can refer to screenshots of proof for his clarity. $\endgroup$
    – user775699
    Dec 9, 2019 at 7:01
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    $\begingroup$ @Lukas Kofler should this question be moved to math overflow ? $\endgroup$
    – user775699
    Dec 15, 2019 at 6:42

1 Answer 1


1) First a geometric description: For $\lvert \tau\rvert = 1$ we have $1/\tau = \overline{\tau}$ (off the unit circle there's also some scaling involved). Thus when $\tau$ travels along the unit circle from $\rho$ to $i$, then $1/\tau$ travels along the unit circle from $\overline{\rho} = \rho^2$ to $\overline{i} = -i$, and therefore $-1/\tau$ travels along the unit circle from $-\overline{\rho} = -\rho^2 = \rho + 1$ to $-\overline{i} = -(-i) = i$.

Let's do it with a parametrisation: Since $\rho = \exp \bigl(i\frac{2\pi}{3}\bigr)$ and $i = \exp \bigl(i\frac{\pi}{2}\bigr)$ we can parametrise the arc from $\rho$ to $i$ by $\tau(t) = \exp\bigl(i\bigl(\frac{2\pi}{3} - t\bigr)\bigr)$, $0 \leqslant t \leqslant \frac{\pi}{6}$. Then $$-\frac{1}{\tau(t)} = -\frac{1}{\exp\bigl(i\bigl(\frac{2\pi}{3} - t\bigr)\bigr)} = -\exp\bigl(i\bigl(t - \tfrac{2\pi}{3}\bigr)\bigr) = \exp\bigl(i\bigl(t - \tfrac{2\pi}{3}\bigr) + i\pi\bigr) = \exp\bigl(i\bigl(\frac{\pi}{3} + t\bigr)\bigr)\,.$$ We see that this is an arc on the unit circle too, it starts at $-1/\tau(0) = \exp\bigl(i\frac{\pi}{3}\bigr) = \rho + 1$ and it ends at $-1/\tau(\pi/6) = \exp\bigl(i\bigl(\frac{\pi}{3} + \frac{\pi}{6}\bigr)\bigr) = \exp\bigl(i\frac{\pi}{2}\bigr) = i$. The original arc was traversed in direction of decreasing argument (i.e. clockwise), while the transformed arc is traversed in direction of increasing argument (counterclockwise), that is, the orientation is reversed.

2) Let us split the fundamental region $R_{\Gamma}$ in two parts, $A_M$ shall be the part where $\operatorname{Im} \tau < M$, and $B_M$ the part where $\operatorname{Im} \tau > M$, where $M$ is chosen large enough that $f$ has neither zeros nor poles in $B_M$. ($B_M$ shall not contain $i\infty$.) Let $N_M$ be the number of zeros of $f$ in $A_M$, and $P_M$ the number of poles of $f$ in $A_M$. Then by the argument principle $$N_M - P_M = \frac{1}{2\pi i} \int_{\partial A_M} \frac{f'(\tau)}{f(\tau)}\,d\tau\,.$$ By the preceding discussion the integrals over the two vertical segments of the boundary cancel, and the integrals over the two arcs on the unit circle cancel too, thus only the integral over the horizontal line remains, $$N_M - P_M = \frac{1}{2\pi i} \int_{\frac{1}{2} + iM}^{-\frac{1}{2} + iM} \frac{f'(\tau)}{f(\tau)}\,d\tau\,.$$

Now we express this integral in terms of $x = e^{2\pi i\tau}$. Since the real part $u$ of $\tau = u + iM$ decreases in that integral, the circle $\lvert x\rvert = e^{-2\pi M}$ is traversed clockwise, i.e. in the negative direction (hence the minus sign in the next formula). Thus $$N_M - P_M = -\frac{1}{2\pi i} \int_{\lvert x\rvert = e^{-2\pi M}} \frac{F'(x)}{F(x)}\,dx\,.$$ By the argument principle, and taking the sign into account, this is $P_F - N_F$ and we obtain $$N_M - P_M = P_F - N_F$$ or after rearranging $$N_M + N_F = P_M + P_F\,.$$ But $N_M + N_F$ is the total number of zeros of $f$ in $R_{\Gamma}$ (including a possible zero at $i\infty$) and $P_M + P_F$ is the total number of poles of $f$ in $R_{\Gamma}$ (including a possible pole at $i\infty$). So overall $f$ has the same number of zeros and poles in $R_{\Gamma}$, when we include $i\infty$.

Apostol's $N$ and $P$ — my $N_M$ and $P_M$ — are the number of zeros and poles in $R_{\Gamma}$ excepting $i\infty$, whereas the total number of zeros and poles must include $i\infty$. I think that's what threw you off.

3) I hope that's clarified under 2), otherwise I don't understand the problem here. (If not, please explain the problem further, I'll be happy to elaborate.)

4) The circular arc $C_1$ goes from the vertical line $\operatorname{Re} \tau =-\frac{1}{2}$ to the unit circle. Its centre is $\rho$. When the radius $r$ is small, the endpoint of $C_1$ on the unit circle is very close to the intersection of $C_1$ with the tangent to the unit circle at $\rho$. The tangent has the parametrisation $\rho - t\cdot i\rho$, $t \in \mathbb{R}$. The angle it forms with the real axis is the argument of its direction $-i\rho = \exp(-i\pi/2)\exp(2\pi i/3) = \exp(\pi i/6)$. Hence the angle between the vertical line and the tangent is $\frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}$. Since the unit circle lies completely below the tangent (excepting the point of contact $\rho$), the angle subtended by $C_1$ is always larger than $\pi/3$. But rotating the tangent clockwise by an angle $\varepsilon > 0$ produces a secant of the circle, and for all sufficiently small $r$, such that the endpoint of $C_1$ lies between $\rho$ and the other intersction of the secant and the unit circle, the angle subtended by $C_1$ lies betwen $\pi/3$ and $\pi/3 + \varepsilon$. Since $\varepsilon$ can be arbitrarily small it follows that the limit as $r \to 0$ of the angle subtended by $C_1$ is $\pi/3$.

  • $\begingroup$ in (2) Why $B_M$ = img$\tau $ >M shall not contain i$\infty$ ? . I think as fundamental region is cut by Img($\tau$) = M, M>0 , so, I think $B_M$ must contain i$\infty$. Can you please explain? $\endgroup$
    – user775699
    Dec 16, 2019 at 13:28
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    $\begingroup$ I just exclude $i\infty$ so that I can say that "$f$ has neither zeros nor poles in $B_M$", regardless of how $f$ behaves at $i\infty$. It's like when I split the unit disk at the circle $\lvert z\rvert = r$ and then might call the outer annulus $A_r$ and then call $B_r$ the punctured disk $0 < \lvert z\rvert < r$ to be able to say that some function has no singularity in $B_r$ even if it has one at $0$. $\endgroup$ Dec 16, 2019 at 13:33
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    $\begingroup$ So $A_M \cup B_M \cup (-1/2 + iM, 1/2 + iM) = R_{\Gamma} \setminus \{i\infty\}$. $\endgroup$ Dec 16, 2019 at 13:34
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    $\begingroup$ Generally, the parametrisation of a line is $a + tb$, $t \in \mathbb{R}$, where $a$ is a point on the line, and $b \neq 0$ is the direction (a direction, $b$ is only determined up to nonzero real multiples, with $b$ also $-b$, $2b$ etc. work) of the line. Here, we can identify the point $\rho$ with the vector from $0$ to $\rho$. In that sense $\rho$ is the radius vector (also, the direction of the radius vector). To get a line orthogonal to it, we must rotate the direction by $\pm \frac{\pi}{2}$ (a right angle, but we can choose whether to rotate clockwise or counterclockwise). $\endgroup$ Dec 16, 2019 at 14:16
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    $\begingroup$ Rotation by $\varphi$ is in $\mathbb{C}$ achieved by multiplying with $e^{i\varphi}$. Thus rotating the direction $\rho$ by $-\frac{\pi}{2}$ yields the direction $e^{-i\pi/2}\rho = (-i)\rho$. $\endgroup$ Dec 16, 2019 at 14:16

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