How do I find these two limits? I've tried everything! I just don't know how to do them.Tried De Moivres formulas , putting the complex number as $x+ yi$
$$1.)\lim_{n\to \infty}(1+\frac{z}{n})^n,z=x+yi  \\ 2.)\lim_{n\to \infty}n (\sqrt[n]{z}-1), $$
 A: HINT: (first question answer, you can solve the second one)
$$\lim_{n\to\infty}\left(1+\frac{z}{n}\right)^n=$$
$$\lim_{n\to\infty}\exp\left(\ln\left(\left(1+\frac{z}{n}\right)^n\right)\right)=$$
$$\lim_{n\to\infty}\exp\left(n\ln\left(1+\frac{z}{n}\right)\right)=$$
$$\exp\left(\lim_{n\to\infty}n\ln\left(1+\frac{z}{n}\right)\right)=$$
$$\exp\left(\lim_{n\to\infty}\frac{\ln\left(1+\frac{z}{n}\right)}{\frac{1}{n}}\right)=$$
$$\exp\left(\lim_{n\to\infty}\frac{\frac{\text{d}}{\text{d}n}\left(\ln\left(1+\frac{z}{n}\right)\right)}{\frac{\text{d}}{\text{d}n}\left(\frac{1}{n}\right)}\right)=$$
$$\exp\left(\lim_{n\to\infty}\frac{-\frac{z}{n^2\left(1+\frac{z}{n}\right)}}{-\frac{1}{n^2}}\right)=$$
$$\exp\left(\lim_{n\to\infty}\frac{nz}{n+z}\right)=$$
$$\exp\left(\lim_{n\to\infty}\frac{z}{1+\frac{z}{n}}\right)=\exp\left(\frac{z}{1+0}\right)=\exp\left(\frac{z}{1}\right)=\exp\left(z\right)=e^z$$
A: Firstly, notice that complex number $x+iy$ can be written as
$$\sqrt{x^2+y^2}\left(\cos\left(\tan^{-1}\frac yx\right)+i\sin\left(\tan^{-1}\frac yx\right)\right)$$
For the first limit use facts that
$$(1+x)^{\frac1x}=e$$
at $x\to0$ and
$$e^{ix}=\cos x+i\sin x$$
Apply these facts
$$\begin{align}\lim_{n\to\infty}\left(1+\frac zn\right)^n&=\lim_{n\to\infty}e^{\frac znn}\\&=e^{x+iy}\\&=e^x\cos y+ie^x\sin y\end{align}$$
For the second one use fact that
$$\frac{a^x-1}{x}=\ln a$$
at $x\to0$.
Write your limit as
$$\begin{align}\lim_{n\to\infty}n\left(\sqrt[n]z-1\right)&=\lim_{n\to0}\frac{z^n-1}{n}\\&=\ln z\\&=\ln\sqrt{x^2+y^2}+i\tan^{-1}\frac yx\end{align}$$
