数学代写|MTH829 Complex Analysis

Proposition 0.1 (Exercise IV.8.1). The solution set to $\cos z=2$ is
$$
{x+i y \in \mathbb{C}: x=2 \pi n, y=-\log (2 \pm \sqrt{3}), n \in \mathbb{Z}}
$$
Proof. First we reduce to a quadratic in $e^{i z}$.
$$
2=\cos z=\frac{e^{i z}+e^{-i z}}{2} \Longrightarrow e^{i z}+e^{-i z}-4=0 \Longrightarrow\left(e^{i z}\right)^2-4 e^{i z}+1=0
$$
Then applying the quadratic formula,
$$
e^{i z}=\frac{4 \pm \sqrt{12}}{2}=2 \pm \sqrt{3}
$$

Let $z=x+i y$, then
$$
e^{i z}=e^{i(x+i y)}=e^{-y+i x}=e^{-y}(\cos x+i \sin x)
$$
Since $e^{i z}$ is real, $\sin x=0$, so $x$ is an integer multiple of $\pi$. Thus $\cos x= \pm 1$. Since $2+\sqrt{3}$ and $2-\sqrt{3}$ are positive, we must have $\cos x=+1$. Thus $x$ is an integer multiple of $2 \pi$.
$$
\begin{aligned}
e^{i z} & =2+\sqrt{3} \Longrightarrow e^{-y}=2+\sqrt{3} \Longrightarrow-y=\log (2+\sqrt{3}) \
e^{i z} & =2-\sqrt{3} \Longrightarrow e^{-y}=2-\sqrt{3} \Longrightarrow-y=\log (2-\sqrt{3})
\end{aligned}
$$
Thus the solution set is
$$
{x+i y: x=2 \pi n, y=-\log (2 \pm \sqrt{3}), n \in \mathbb{Z}}
$$

Lemma 0.2 (Exercise IV.8.2, not assigned). Let $z=x+i y$. Then
$$
\begin{aligned}
\cos z & =\cos x \cosh y-i \sin x \sinh y \
\sin z & =\sin x \cosh y+i \cos x \sinh y
\end{aligned}
$$

Proposition 0.3 (Exercise IV.8.3). Let $\phi$ be the map $z \mapsto \cos z$. Let $c \in \mathbb{C}$. The image of a line $\operatorname{Re} z=c$ is a hyperbola with transverse axis in the real direction, except for degenerate cases when $\sin c=0$ or $\cos c=0$. The image of a line $\operatorname{Im} z=c$ is an ellipse centered at the origin, except for the degenerate case $c=0$.

Proof. Fix $c \in \mathbb{R}$. First we determine the image of the line $\operatorname{Im} z=c$ under $\cos z$. When $c=0$, we take the image of the embedded real axis, which is the line segment $[-1,1]$. When $c \neq 0$, note that $\sinh c$ and $\cosh c$ are nonzero, so using the formula
$$
\cos (x+i c)=u+i v=\cos x \cosh c-i \sin x \sinh c
$$
We notice that the real and imaginary parts $u, v$ satisfy
$$
\frac{u^2}{\cosh ^2 c}+\frac{v^2}{\sinh ^2 c}=\left(\frac{\cos x \cosh c}{\cosh c}\right)^2+\left(\frac{-\sin x \sinh c}{\sinh c}\right)^2=\cos ^2 x+\sin ^2 x=1
$$

so the image is an ellipse centered at the origin. It intersects the real axis at $\pm \cosh c$, and intersects the imaginary axis at $\pm \sinh c$. Now we determine the image of the line $\operatorname{Re} z=c$. Applying the formula again,
$$
\cos (c+i y)=u+i v=\cos c \cosh y-i \sin c \sinh y
$$
When $\sin c=0$, the image is contained in the real axis, and when $\cos c=0$, the image is contained in the imaginary axis. When $\sin c \neq 0$ and $\cos c \neq 0$, we see that
$$
\frac{u^2}{\cos ^2 c}-\frac{v^2}{\sin ^2 c}=\left(\frac{\cos c \cosh y}{\cos c}\right)^2-\left(\frac{\sin c \sinh y}{\sin c}\right)^2=\cosh ^2 y-\sinh ^2 y=1
$$
so the image is a hyperbola centered at the origin with the real axis as the transverse axis. It intersects the real axis at $\pm \cos c$.

Proposition 0.4 (Exercise IV.13.3). Let $G$ be the open set $\mathbb{C} \backslash[-1,1]$. There is a branch of $\sqrt{\frac{z+1}{z-1}}$ in $G$

Proof. Let $\phi: \overline{\mathbb{C}} \rightarrow \overline{\mathbb{C}}$ be given by $z \mapsto \frac{z+1}{z-1}$. Then $\phi(1)=\infty, \phi(0)=-1$, and $\phi(-1)=0$, so $\phi$ maps the extended real line onto itself (because $\phi$ is a fractional linear transformation and preserves clircles). In particular $\phi$ maps the segment $[-1,1]$ onto the negative real axis $(-\infty, 0]$. We know that there is branch of $\log$ on the slit complex plane $\mathbb{C} \backslash(-\infty, 0]$, so there is a branch of $\log$ on $\phi(G)$. Then by the discussion in IV.13, there is a branch of $\phi^{1 / 2}$ in $G$.

For exercises V.6.2 and following, I need the following lemma, which is Exercise V.6.1, which was not assigned. I won’t mentioned the use of this lemma, I’ll just use the phrases “converges uniformly” and “uniformly Cauchy” interchangeably.