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7

\begin{align} |e^{it}-1|^2&=(e^{it}-1)(\overline{e^{it}-1})\\ &=(e^{it}-1)(e^{-it}-1)\\ &=2-(e^{it}+e^{-it})\\ &=2-2 \cos t\\ &=4\sin^2\left(\frac{t}2\right) \end{align}

3

Use that $$e^{i\theta}-1=e^{i\theta/2}(e^{i\theta/2}-e^{-i\theta/2})$$ Then what is $|e^{i\theta/2}|$ and how can we express $e^{i\theta/2}-e^{-i\theta/2}$ as a $\sin$?

3

The principal branch of $\log$ is analytic on $\Bbb{C}-(-\infty,0]$ $\frac{f(z)}{g(z)}$ is analytic and $\in \Bbb{C}-(-\infty,0]$ for $|z| \in (1-\epsilon,1+\epsilon)$ thus $h(z)=\log \frac{f(z)}{g(z)}$ is analytic for $|z| \in (1-\epsilon,1+\epsilon)$ so that $$0=\int_{|z|=1} h'(z)dz=\int_{|z|=1} (\frac{f'(z)}{f(z)}-\frac{g'(z)}{g(z)})dz$$ The proof ...

1

You first use binomial expansion and re-ordering, using absolute convergence inside the radius of convergence to justify the re-ordering, to get $$\sum_{k\ge 0} c_kz^k=\sum_k c_k(a+(z-a))^k=\sum_{0\le j\le k}c_k\binom{k}{j}a^{k-j}(z-a)^j =\sum_{j\ge 0}\left(\sum_{k\ge j}c_k\binom{k}{j}a^{k-j}\right)(z-a)^j$$ Then use that by Cauchy-Hadamard (in the ...

1

$$|e^{it}-1|=|e^{it/2}||e^{it/2}-e^{-it/2}|=|2i|\left|\sin\left(\frac t2\right)\right|=2\left|\sin\left(\frac t2\right)\right|$$

1

$|\cos(t)-1+i\sin(t)|=\sqrt{(\cos(t)-1)^2+(\sin(t))^2}=\sqrt{\cos^2(t)-2\cos(t)+1+\sin^2(t)}=\sqrt{2-2\cos(t)}=\sqrt{2(2\sin^2(\frac{t}{2})}=\sqrt{4\sin^2(\frac{t}{2})}=|2\sin(\frac{t}{2})|$

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