This discussion is now closed.
Scroll to see replies
\exp(z_1)\exp(z_2) &= \displaystyle \sum_{n = 0}^\infty\displaystyle\sum_{m = 0}^\infty \frac{z_1^m}{m!}\frac{z_2^n}{n!}\\ &= \displaystyle\sum_{r = 0}^\infty\displaystyle\sum_{m = 0}^r \frac{z_1^{r - m}}{(r - m)!}\frac{z_2^m}{m!}\\ &= \displaystyle\sum_{r = 0}^\infty\frac{1}{r!} \displaystyle \sum_{m = 0}^r \frac{r!}{(r - m)!m!}z_1^{r - m}z_2^m\\ &= \displaystyle\sum_{r = 0}^\infty\frac{(z_1 + z_2)^r}{r!} = \exp(z_1 + z_2)
\exp(z_1)\exp(z_2) &= \displaystyle \sum_{n = 0}^\infty\displaystyle\sum_{m = 0}^\infty \frac{z_1^m}{m!}\frac{z_2^n}{n!}\\ &= \displaystyle\sum_{r = 0}^\infty\displaystyle\sum_{m = 0}^r \frac{z_1^{r - m}}{(r - m)!}\frac{z_2^m}{m!}\\ &= \displaystyle\sum_{r = 0}^\infty\frac{1}{r!} \displaystyle \sum_{m = 0}^r \frac{r!}{(r - m)!m!}z_1^{r - m}z_2^m\\ &= \displaystyle\sum_{r = 0}^\infty\frac{(z_1 + z_2)^r}{r!} = \exp(z_1 + z_2)