Show that limlimits_(n→∞)(1+dfrac zn)^n=e^x(cos y+isin…

Show that $\ \lim\limits_{n\to\infty}\left(1+\dfrac zn\right)^n=e^x(\cos y+i\sin y)=e^z$, where $z=x+iy$.

Step 1 · Rewrite as an exponential

$$ \left(1+\frac zn\right)^n=e^{n\ln(1+z/n)} $$

Step 2 · Expand the logarithm

Using $\ln(1+w)=w-\frac{w^2}2+\frac{w^3}3-\cdots$ with $w=\frac zn$:

$$ n\ln\left(1+\frac zn\right)=n\left(\frac zn-\frac{z^2}{2n^2}+\cdots\right)=z-\frac{z^2}{2n}+\frac{z^3}{3n^2}-\cdots $$

Step 3 · Take the limit of the exponent

Every term beyond the first carries a positive power of $\frac1n$ in the denominator, so:

$$ \lim_{n\to\infty}n\ln\left(1+\frac zn\right)=z $$

Step 4 · Conclude

Since $w\mapsto e^w$ is continuous:

$$ \boxed{\lim_{n\to\infty}\left(1+\frac zn\right)^n=e^z} $$

Alternative · separate modulus and argument

Writing $1+\frac zn=1+\frac xn+i\frac yn=r_ne^{i\theta_n}$: one finds $\ln r_n\approx\frac xn$ (so $r_n^n\to e^x$) and $\theta_n\approx\frac yn$ (so $n\theta_n\to y$), giving $\left(1+\frac zn\right)^n=r_n^ne^{in\theta_n}\to e^xe^{iy}=e^z$ — the same conclusion via the polar route.

Special cases

$$ z=0:\ \left(1+0\right)^n=1=e^0 \quad\checkmark; \qquad z=1:\ \left(1+\frac1n\right)^n\to e \quad(\text{the classical definition}); \qquad z=i:\ \left(1+\frac in\right)^n\to e^i=\cos1+i\sin1 $$