Define
\[ E(z)=e^{x}(\cos y+i \sin y) . \]
- Show that \(E(z)\) is the unique function analytic on \(\mathbb{C}\) that satisfies
\[ E^{\prime}(z)=E(z), \quad E(0)=1 . \]
- Conclude from the first part that \[ E(z)=\sum_{n=0}^{\infty} \frac{z^{n}}{n !} .\]
Solution
Write \(z=x+iy\) and \[ E(z)=u(x,y)+iv(x,y), \qquad u=e^x\cos y,\quad v=e^x\sin y. \] Then \[ u_x=e^x\cos y=v_y, \qquad u_y=-e^x\sin y=-v_x, \] so the Cauchy–Riemann equations hold everywhere and \(E\) is entire. Moreover \[ E'(z)=u_x+iv_x=e^x(\cos y+i\sin y)=E(z), \qquad E(0)=1. \]
For uniqueness, suppose \(F\) is entire with \(F'=F\) and \(F(0)=1\). Since \[ E(z)E(-z)=1, \] \(E\) has no zeros. Hence \[ \left(\frac{F}{E}\right)' =\frac{F'E-FE'}{E^2}=0. \] Thus \(F/E\) is constant, and its value at \(0\) is \(1\). Therefore \(F=E\).
Now let \[ S(z)=\sum_{n=0}^{\infty}\frac{z^n}{n!}. \] The ratio test shows that this series has infinite radius of convergence, so \(S\) is entire and may be differentiated termwise: \[ S'(z)=\sum_{n=1}^{\infty}\frac{n z^{n-1}}{n!} =\sum_{m=0}^{\infty}\frac{z^m}{m!}=S(z), \qquad S(0)=1. \] By the uniqueness just proved, \[ \boxed{E(z)=\sum_{n=0}^{\infty}\frac{z^n}{n!}}. \]