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Another important result in the global theory of analytic functions is the maximum modulus principle, which states that an analytic function attains its maximum modulus on the boundary of its domain. This result has many applications in complex analysis, including the proof of the fundamental theorem of algebra, which states that every non-constant polynomial with complex coefficients has at least one complex root.
In summary, the theory of analytic functions of one variable is a rich subject with many important results and applications. Its local and global aspects provide powerful tools for understanding the behavior of complex functions, and it has many connections to other areas of mathematics and physics.
For $z=1+\imath$ find $w$ such that the real parts of the following numbers are equal to zero
a) $z+w_i$
b) $z \cdot w$
c) $\frac{z}{w} ;$
d) $\frac{w}{z}$.
Solution. Let $z=1+\imath=(1,1)$ and $w=x+i y=(x, y)$. Then we have
a)
$$
\operatorname{Re}(z+w)=0 \Longleftrightarrow 1+x=0
$$
Hence $x=-1$ and $y \in \mathbb{R}$ is arbitrary.
b)
$$
\operatorname{Re}(z \cdot w)=0 \Longleftrightarrow \operatorname{Re}(x-y+i(x+y))=0 .
$$
Hence $x-y=0$. Therefore $w$ is given by $w=x+\imath x=x(1+\imath)=x \cdot z$, for arbitrary $x \in \mathbb{R}$
c) We have $\frac{z}{w}=0$. Hence
$$
\frac{z}{w}=\frac{1+\imath}{x+2 y} \cdot \frac{x-v y}{x-v y}=\frac{x+y+\imath(x-y)}{x^2+y^2}
$$
Therefore
$$
\operatorname{Re}\left(\frac{z}{w}\right)=0 \Longleftrightarrow \frac{x+y}{x^2+y^2}=0 \Rightarrow x=-y, x \neq 0 .
$$
Finally we have $w=x(1-2)$ for $x \in \mathbb{R}$ and $x \neq 0$.
d) From $\operatorname{Re}\left(\frac{w}{z}\right)=0$ it follows that for every $x \in \mathbb{R}$ :
$$
\frac{w}{z}=\frac{x+\imath y}{1+\imath} \cdot \frac{1-\imath}{1-\imath}=\frac{x+y+\imath(x-y)}{2} .
$$
Hence $w=x(1-\imath)$ for every $x \in \mathbb{R}$.
Let $\mathbb{C}^$ be the set of all complex numbers different from zero. a) Prove that the set $T$ of all complex numbers with modulus 1 is a multiplicative subgroup of the group $\left(\mathbb{C}^, \cdot\right)$.
b) The multiplicative group $\mathbb{C}^*$ is isomorphic with $\mathbb{R}^{+} \times T$.
Solution. a) We have $T={z|| z \mid=1} \subset \mathbb{C}^* . T$ under the multiplication by the equality
$$
\left|z_1 \cdot z_2\right|=\left|z_1\right| \cdot\left|z_2\right| \text {. }
$$
The associativity follows by the associativity of multiplication in $\mathbb{C}$. The neutral element is 1. The inverse of $z=(a, b) \in T$ is
$$
\left(\frac{a}{a^2+b^2}, \frac{-b}{a^2+b^2}\right) \in T \text {. }
$$
b) It is easy to check that an isomorphism is given by
$$
f(z)=(|z|, \cos \theta+i \sin \theta)
$$
where $\theta=\arg z$. nopagebreak c) We define an equivalence relation $\sim$ on $\mathbb{R}$ by $r_1 \sim r_2 \Longleftrightarrow r_1-r_2=2 k \pi, \quad k$ is an integer. Let $\mathbb{R}$ be the corresponding quotient set. Prove that $\mathbb{R}^{+} \times T$ isomorphic with $\mathbb{R}^{+} \times \hat{\mathbb{R}}$. The group $\mathbb{C}^*$ is isomorphic with $\mathbb{R}^{+} \times \tilde{\mathbb{R}}$ (check). Therefore by b) and transitivity of the isomorphisms of groups it follows c).
Starting with a complex number $z \neq 0$, find where the complex numbers $2 z, 3 z, \ldots, n z$ are ?
Solution. For $z=\rho(\cos \theta+\imath \sin \theta)$ we have
$$
n z=n \rho(\cos \theta+i \sin \theta)
$$
the complex numbers $n x$, for $n=1,2, \ldots$ are on the half straight line $y=\tan \theta \cdot x$ with modulus $n \rho$.
