Deficiencies of the exponential and sine

(Exponential: characteristic) By [F1] and [F6], for the entire function $e^z$ one has $m_0(r,e^z)\le T(r,e^z)\le m_0(r,e^z)+\frac{1}{2}\log 2$. For $z=r{e}^{it}$ the modulus formula [F5] gives ${\log }^{+}\mathrm{\mid }{e}^z\mathrm{\mid }=\max (0,r\cos t)=r{\cos }^{+}t$, and ${\int }_0^{2\pi }{\cos }^{+}tdt={\int }_{-\pi \mathrm{/}2}^{\pi \mathrm{/}2}\cos tdt=2$, so $m_0(r,e^z)=\frac{r}{2\pi }\cdot 2=\frac{r}{\pi }$. Hence $T(r,e^z)=\frac{r}{\pi }+O(1)$.

(Exponential: $a$-point counts) By [F5] the kernel of the exponential is $2\pi i\mathbb{Z}$ and every nonzero value is attained, so for fixed $a\mathrm{\ne }0$ and a logarithm $b$ of $a$ the $a$-points of $e^z$ are exactly the points $b+2\pi ik$, $k\in \mathbb{Z}$, and they are simple because $(e^z{)}^{\mathrm{\prime }}=e^z\mathrm{\ne }0$ there. The number of $k\in \mathbb{Z}$ with $\mathrm{\mid }b+2\pi ik\mathrm{\mid }\le t$ is $\frac{t}{\pi }+O(1)$ for large $t$: writing $b=u+iv$, the condition is $\mathrm{\mid }v+2\pi k\mathrm{\mid }\le \sqrt{t^2-u^2}$, a nonempty integer interval of length $\frac{t}{\pi }+O(1)$. Therefore $n(t,a;e^z)=\frac{t}{\pi }+O(1)$ and $N(r,a;e^z)=\bar{N}(r,a;e^z)=\frac{r}{\pi }+O(\log r)$ with $N_1(r,a;e^z)=0$.

(Sine: characteristic) By [F7] sine is entire, so [F6] gives $m_0(r,\sin )\le T(r,\sin )\le m_0(r,\sin )+\frac{1}{2}\log 2$. For $z=r{e}^{it}$ the definition [F7] and the modulus formula [F5] give $\mathrm{\mid }\sin z\mathrm{\mid }\le \frac{1}{2}(e^{\mathrm{\mid }\mathrm{Im}z\mathrm{\mid }}+e^{-\mathrm{\mid }\mathrm{Im}z\mathrm{\mid }})\le e^{\mathrm{\mid }\mathrm{Im}z\mathrm{\mid }}=e^{r\mathrm{\mid }\sin t\mathrm{\mid }}$, so ${\log }^{+}\mathrm{\mid }\sin z\mathrm{\mid }\le r\mathrm{\mid }\sin t\mathrm{\mid }$ and $m_0(r,\sin )\le \frac{r}{2\pi }{\int }_0^{2\pi }\mathrm{\mid }\sin t\mathrm{\mid }dt=\frac{2r}{\pi }$ by [F7]. Conversely, $\mathrm{\mid }\sin z\mathrm{\mid }\ge \frac{1}{2}(e^{\mathrm{\mid }\mathrm{Im}z\mathrm{\mid }}-e^{-\mathrm{\mid }\mathrm{Im}z\mathrm{\mid }})\ge \frac{1}{4}{e}^{r\mathrm{\mid }\sin t\mathrm{\mid }}$ whenever $r\mathrm{\mid }\sin t\mathrm{\mid }\ge 1$, so ${\log }^{+}\mathrm{\mid }\sin z\mathrm{\mid }\ge r\mathrm{\mid }\sin t\mathrm{\mid }-\log 4$ on that set, and since the integrand is nonnegative everywhere, $m_0(r,\sin )\ge \frac{1}{2\pi }{\int }_0^{2\pi }(r\mathrm{\mid }\sin t\mathrm{\mid }-\log 4)dt=\frac{2r}{\pi }-\log 4$. Hence $T(r,\sin )=\frac{2r}{\pi }+O(1)$.

(Sine: reduction to a quadratic) Fix $a\in \mathbb{C}$; putting $w=e^{iz}$, one has $\sin z=a$ if and only if $w^2-2iaw-1=0$, because $e^{iz}\mathrm{\ne }0$ and multiplying $\frac{w-w^{-1}}{2i}=a$ by $2iw$ is reversible. If $q(w)=w^2-2iaw-1$ has a double root $w$, then $w=ia$ and $w^2=-1$, forcing $a^2=1$; so for $a\mathrm{\ne }\pm 1$ the polynomial $q$ has two distinct roots $w_1,w_2$, and $q(0)=-1$ shows both are nonzero. By [F7] a root $w_1\mathrm{\ne }0$ exists, and $w_2:=-1\mathrm{/}{w}_1$ is a second root, since $w_1^2-2ia{w}_1-1=0$ gives $q(-1\mathrm{/}{w}_1)=w_1^{-2}(1+2ia{w}_1-w_1^2)=0$.

(Sine: simplicity away from $\pm 1$) At a solution of $\sin z=a$ with $w=e^{iz}$ one has $\cos z=\frac{w+w^{-1}}{2}=\frac{w^2+1}{2w}$ by [F7], which vanishes exactly when $w=\pm i$ by [F7]; then $\sin z=\frac{w-w^{-1}}{2i}$ equals $1$ for $w=i$ and $-1$ for $w=-i$. Hence for $a\mathrm{\ne }\pm 1$ no $a$-point has $\cos z=0$, and since ${\sin }^{\mathrm{\prime }}=\cos$ by [F7] every $a$-point is simple.

(Exponential: deficiencies) Since $e^z$ is entire, $N(r,\mathrm{\infty };e^z)=0$ and [F3] gives $\delta (\mathrm{\infty },e^z)=1$; since $e^z\mathrm{\ne }0$ for all $z$ the value $0$ is omitted and [F3] gives $\delta (0,e^z)=1$; for $0\mathrm{\ne }a\in \mathbb{C}$ step 1.2 gives $\frac{N(r,a;e^z)}{T(r,e^z)}\to 1$, hence $\delta (a,e^z)=1-{\mathrm{lim\hspace{0.17em}sup}}_r\frac{N(r,a;e^z)}{T(r,e^z)}=0$.

(Exponential: ramification indices) For finite $a\mathrm{\ne }0$ all $a$-points of $e^z$ are simple by step 1.2, so $N_1(r,a;e^z)=0$ and $\epsilon (a,e^z)=0$; for $a=0$ the counting function is identically zero, so $\epsilon (0,e^z)=0$; and $N_1(r,\mathrm{\infty };e^z)=0$ because $e^z$ has no poles, so $\epsilon (\mathrm{\infty },e^z)=0$.

(Sine: the $a$-point progressions) With $w_j\mathrm{\ne }0$ as in step 1.4, fix $b_j$ with $e^{b_j}=w_j$ by [F5]; then $e^{iz}=w_j$ if and only if $iz-b_j\in 2\pi i\mathbb{Z}$, that is, $z\in -i{b}_j+2\pi \mathbb{Z}$ by [F5]. Hence for $a\mathrm{\ne }\pm 1$ the solutions of $\sin z=a$ are exactly the two disjoint arithmetic progressions $-i{b}_1+2\pi \mathbb{Z}$ and $-i{b}_2+2\pi \mathbb{Z}$.

(Sine: the values $\pm 1$) For $a=1$ the quadratic of step 1.4 is $q(w)=(w-i{)}^2$, so $\sin z=1$ if and only if $e^{iz}=i$, that is, $z\in \frac{\pi }{2}+2\pi \mathbb{Z}$; at these points $\cos z=0$ by [F7] and ${\sin }^{\mathrm{\prime }\mathrm{\prime }}=-\sin =-1\mathrm{\ne }0$ by [F7], so each is a double point: $n(t,1;\sin )=\frac{2t}{\pi }+O(1)$ with multiplicity and $\bar{n}(t,1;\sin )=\frac{t}{\pi }+O(1)$, whence $N(r,1;\sin )=\frac{2r}{\pi }+O(\log r)$, $\bar{N}(r,1;\sin )=\frac{r}{\pi }+O(\log r)$ and $N_1(r,1;\sin )=\frac{r}{\pi }+O(\log r)$. The same computation with $-i$ and the progression $-\frac{\pi }{2}+2\pi \mathbb{Z}$ gives the analogous statements for $a=-1$.

(Sine: counting for $a\mathrm{\ne }\pm 1$) For a fixed $c\in \mathbb{C}$, the number of $k\in \mathbb{Z}$ with $\mathrm{\mid }c+2\pi ik\mathrm{\mid }\le t$ is $\frac{t}{\pi }+O(1)$ as $t\to \mathrm{\infty }$: the condition is $\mathrm{\mid }2\pi k+\mathrm{Im}c\mathrm{\mid }\le \sqrt{t^2-(\mathrm{Re}c{)}^2}$, an integer interval of length $\frac{t}{\pi }+O(1)$. The two disjoint progressions of step 2.3 therefore give $n(t,a;\sin )=\frac{2t}{\pi }+O(1)$, and by step 1.5 all $a$-points are simple, so $N(r,a;\sin )=\bar{N}(r,a;\sin )=\frac{2r}{\pi }+O(\log r)$ and $N_1(r,a;\sin )=0$.

(Sine: deficiencies and ramification indices) Since sine is entire, $N(r,\mathrm{\infty };\sin )=0$, so $\delta (\mathrm{\infty },\sin )=1$ by [F3]. For every finite $a$, steps 3.1 and 2.4 give $\frac{N(r,a;\sin )}{T(r,\sin )}\to 1$ because $T(r,\sin )=\frac{2r}{\pi }+O(1)$ by step 1.3, so $\delta (a,\sin )=1-{\mathrm{lim\hspace{0.17em}sup}}_r\frac{N}{T}=0$. For $a\mathrm{\notin }\{1,-1\}$ step 3.1 gives $N_1(r,a;\sin )=0$, so $\epsilon (a,\sin )=0$, including $a=0$; and $\epsilon (\mathrm{\infty },\sin )=0$ because the $N_1$-count at $\mathrm{\infty }$ vanishes for an entire function; while for $a=\pm 1$ step 2.4 gives $\frac{N_1(r,a;\sin )}{T(r,\sin )}\to \frac{r\mathrm{/}\pi }{2r\mathrm{/}\pi }=\frac{1}{2}$, so $\epsilon (1,\sin )=\epsilon (-1,\sin )=\frac{1}{2}$.

(Sine: combined defect sum) By step 4.1 the only nonzero terms of the total deficiency sum of [F3] are $\delta (\mathrm{\infty },\sin )=1$, $\epsilon (1,\sin )=\frac{1}{2}$ and $\epsilon (-1,\sin )=\frac{1}{2}$; the supremum of finite subsums is therefore $1+\frac{1}{2}+\frac{1}{2}=2$, so the combined defect sum of sine equals $2$, in agreement with the general upper bound of [F4].

(Exponential: combined defect sum) For $e^z$ steps 2.1 and 2.2 give the only nonzero terms $\delta (0,e^z)=\delta (\mathrm{\infty },e^z)=1$, so its combined defect sum is $2$ as well, again with equality in the bound of [F4].