# Tag Info

1

You have to be very carefull with the notation $x^a$ if $x$ is allowed to be an arbitrary element of $\mathbb{C}$. It's natural to define $x^a$ via $$x^a = \exp(a\cdot\ln x)$$ but that leaves the natural logarithm $\ln x$ to be defined for arbitrary $x \in \mathbb{C}$. Now, since the exponential function is not one-to-one over the complex numbers ...

1

For real $x\gt0$, we usually take $\log(x)$ to be real as well. Since $a$ is also real, we have $\log(x^{ia})=ia\log(x)$ is pure imaginary. That means that $$\left|x^{ia}\right|=\left|1\cdot e^{ia\log(x)}\right|=1$$ Therefore, $r=1$.

2

As $$x=e^{ln(x)}$$ Therefore $$x^{ia}=e^{ln(x)\cdot ia}=e^{i\cdot ln(x)a}$$ So $$r=1$$ and your phase is different from what you thought.

2

You have solved the problem yourself. Just check this. $$\lim_{n \to \infty} (1+\frac{1}{n+2})^{2n+3}$$ $$=\lim_{n \to \infty} \left(\left(1+\frac{1}{n+2}\right)^{n+2}\right)^{\frac{2n+3}{n+2}}$$ $$= \left(\lim_{(n+2) \to \infty}\left(1+\frac{1}{n+2}\right)^{n+2}\right)^{\lim_{n \to \infty}\frac{2n+3}{n+2}}$$ =e^{\lim_{n \to \infty} \frac{2n+3}{n+2}} ...

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