The complex exponential satisfies the Cauchy–Riemann equations in Cartesian and polar form

For $h\mathrm{\ne }0$, the addition law and defining series give $\frac{e^{z+h}-e^z}{h}=e^z\frac{e^h-1}{h}=e^z(1+{\sum }_{n\ge 2}\frac{h^{n-1}}{n!})\mathrm{.}$

From [L3], $u=e^x\cos y$ and $v=e^x\sin y$. Therefore $u_x=e^x\cos y=v_y$ and $u_y=-e^x\sin y=-v_x$, and $u_x+i{v}_x=e^{x+iy}$.

Put $c:=\cos \theta$, $s:=\sin \theta$, $A:=e^{rc}$, and $\varphi :=rs$. Then $U=A\cos \varphi$ and $V=A\sin \varphi$, so $U_r=A(c\cos \varphi -s\sin \varphi ),V_r=A(c\sin \varphi +s\cos \varphi ),$ $U_{\theta }=-r{V}_r,V_{\theta }=r{U}_r\mathrm{.}$

When $0<\mathrm{\mid }h\mathrm{\mid }\le 1$, absolute convergence at $1$ gives the finite constant $C:={\sum }_{n\ge 2}1\mathrm{/}n!$, and $\mid {\sum }_{n\ge 2}\frac{h^{n-1}}{n!}\mid \le \mathrm{\mid }h\mathrm{\mid }{\sum }_{n\ge 2}\frac{1}{n!}=C\mathrm{\mid }h\mathrm{\mid }⟶0.$

Thus the quotient tends to $e^z$, directly confirming ${\exp }^{\mathrm{\prime }}(z)=\exp z$ and [L4].

Since $r>0$, step 1.3 is equivalent to $U_r=r^{-1}{V}_{\theta }$ and $V_r=-r^{-1}{U}_{\theta }$, exactly [L5]. At $r=0$ these polar identities are not asserted; the Cartesian calculation in step 1.2 covers the origin.