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The complex exponential function needs some definition. There are several ways one could define it: Power series As a solution to a differential equation As the unique holomorphic extension of the real exponential function to the whole plane Via the limit definition $\lim_{n \to \infty} (1+\frac{z}{n})^n$ Via Euler's formula ... One can take any one of ...


No, in fact, any discrete subset $I\subset\mathbb{C}$ must be countable. For any $i\in I$, discreteness of $I$ says that we can find an open set $U_i\subset \mathbb{C}$ such that $U_i\cap I=\{i\}$. Inside $U_i$, we can find an open disk $V_i$ containing $i$ such that the real and imaginary parts of the center of $V_i$ as well as the radius of $V_i$ are ...


The identity you quote as being the textbook's answer - $|\cosh(x+i y)|^2 = \cos(x)^2 + \sinh(y)^2$ - is false in the case $y=0$ for all $x \not = 0$. Your identity and proof are correct. It is the case that $|\cosh(x+i y)|^2 = \sinh(x)^2 + \cos(y)^2$.


To show the point $a$ is a removable singularity of $f/g$ suffices to prove that $$\lim_{z\to a}(z-a) \dfrac{f}{g}(z) = 0.$$ In other words, given $\epsilon > 0$ there is $\delta > 0$ such that if $|z-a|<\delta$ then $$\left|(z-a)\dfrac{f(z)}{g(z)}\right| < \epsilon.$$ Constructing $\delta$ $$\left|(z-a)\dfrac{f(z)}{g(z)}\right| = ...


For real $z$, say, $z := x + 0 i$, we have $|\cosh z|^2 = \cosh^2 x$, which is unbounded, but substituting this real quantity in your book's formula gives $\cos^2 x$, which is bounded, and so the book's formula is wrong. Your derivation gives the correct formula.


If $f:D \to D$ is holomorphic and $f(D) \subset D,$ then $|f''(0)| \le 2$ by Cauchy's estimates. We don't need $f(0)= f'(0)$ for this. We do need $f(0)= f'(0)$ for the estimate $|f(z)|\le |z|^2,$ but this is independent of first part; it follows from the proof idea in the Schwarz Lemma as @A.G. suggested in a comment.


Concerning the first part: You want to prove that there is no real root to the polynomial $p(z)=z^6+9z^4+z^3+2z+4$. Now, the idea is to group terms to prove that $p(z)>0$ for all real $z$. First of all, you can get away that $z^3$ term using $z^6+z^3+\frac{1}{4} \ge 0$ which is equivalent to $(z^3+\frac{1}{2})^2 \ge 0$ which holds for all real $z$. ...


This function is bounded and entire, since is is both 1 and i periodic and continuous. It is therefore constant.


$|w|=1\Rightarrow |az+b|=|cz+d|\Rightarrow|a||z+\frac ba|=|c||z+\frac dc|$ This is a straight line iff $|a|=|c|$, namely the perpendicular bisector of $-\frac ba$ and $-\frac dc$. When $|a|\neq|c|$, the locus of $z$ is a Circle of Apollonius.


If you are looking to simplify the angles, then you can use $\frac 13\arctan(\frac{11}{2})=\arctan \frac 12$ You can find this by using $$\tan3\theta=\frac{3t-t^3}{1-3t^2}=\frac{11}{2},$$ where $t=\tan\theta$ $$\Rightarrow2t^3-33t^2-6t+11=0$$ $$\Rightarrow(2t-1)(t^2-16t-11)=0$$ from which the acute angled solution for $\theta$ which is less than ...

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