অয়লারের সূত্র

${\displaystyle {\begin{aligned}e^{ix}&{}=1+ix+{\frac {{(ix)}^{2}}{2!}}+{\frac {{(ix)}^{3}}{3!}}+{\frac {{(ix)}^{4}}{4!}}+{\frac {{(ix)}^{5}}{5!}}+{\frac {{(ix)}^{6}}{6!}}+{\frac {{(ix)}^{7}}{7!}}+{\frac {{(ix)}^{8}}{8!}}+\cdots \\[8pt]&{}=1+ix-{\frac {{x}^{2}}{2!}}-{\frac {{ix}^{3}}{3!}}+{\frac {{x}^{4}}{4!}}+{\frac {{ix}^{5}}{5!}}-{\frac {{x}^{6}}{6!}}-{\frac {{ix}^{7}}{7!}}+{\frac {{x}^{8}}{8!}}+\cdots \\[8pt]&{}=(1-{\frac {{x}^{2}}{2!}}+{\frac {{x}^{4}}{4!}}-{\frac {{x}^{6}}{6!}}+{\frac {{x}^{8}}{8!}}+\cdots )+i(x-{\frac {{x}^{3}}{3!}}+{\frac {{x}^{5}}{5!}}-{\frac {{x}^{7}}{7!}}+\cdots )\\[8pt]&{}=\cos {x}+i\sin {x}\\[8pt]\end{aligned}}}$

ধরা যাক,

${\displaystyle f(x)=e^{-ix}(\cos {x}+i\sin {x})}$

${\displaystyle f(x)}$ কে ${\displaystyle x}$ এর সাপেক্ষে অন্তরীকরণ করে, ${\displaystyle f'(x)=e^{-ix}(-\sin {x}+i\cos {x})-ie^{-ix}(\cos {x}+i\sin {x})=0}$

যেহেতু ফাংশন ${\displaystyle f}$ এর অন্তরীকরণ শূন্য সেহেতু এটি একটি ধ্রুব ফাংশন। অর্থাৎ, ${\displaystyle f(x)=e^{-x}(\cos {x}+i\sin {x})=c}$ , যেখানে ${\displaystyle c}$ একটি ধ্রুবক।

${\displaystyle x=0}$ বসিয়ে, ${\displaystyle f(0)=1=c}$

সুতরাং, ${\displaystyle f(x)=e^{-x}(\cos {x}+i\sin {x})=1}$

অর্থাৎ ${\displaystyle e^{ix}=\cos {x}+i\sin {x}}$ (প্রমাণিত)

ধরা যাক,

${\displaystyle {\begin{aligned}z&{}=\cos {x}+i\sin {x}\\[8pt]\therefore dz&{}=(-\sin {x}+i\cos {x})dx\\[8pt]dz&{}=(i^{2}\sin {x}+i\cos {x})dx\\[8pt]dz&{}=i(\cos {x}+i\sin {x})dx\\[8pt]dz&{}=izdx\\[8pt]{\frac {dz}{z}}&{}=idx\\[8pt]\int {\frac {dz}{z}}&{}=i\int {dx}\\[8pt]\ln {z}&{}=ix+c\\\end{aligned}}}$

যেখানে ${\displaystyle c}$ সমাকলন ধ্রুবক। প্রথম লাইনের সমীকরণে ${\displaystyle x=0}$ বসালে ${\displaystyle z=1}$ হয়। শেষ সমীকরণে ${\displaystyle x=0}$ এবং ${\displaystyle z=1}$ বসিয়ে ${\displaystyle c=0}$ পাওয়া যায়।

সুতরাং,

${\displaystyle \ln {z}=ix}$

${\displaystyle e^{ix}=\cos {x}+i\sin {x}}$ (প্রমাণিত)