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Question Number 170110 by mathlove last updated on 16/May/22

$$provethat$$$$e^{i\theta }=cos\theta +isin\theta$$

Commented by ajfour last updated on 16/May/22

$$1\times e^{i\theta }=zis$$$$1rotatedby\theta anticlockwise.$$$$hencez=\cos \theta +i\sin \theta =e^{i\theta }$$$$drawadiagramnuknow.$$

Answered by floor(10²Eta[1]) last updated on 16/May/22

$$\mathrm{using}\mathrm{the}\mathrm{mclaurin}\mathrm{series}:$$$$\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{f}}^{\left(\mathrm{n}\right)}\left(0\right)}{\mathrm{n}!}{\mathrm{x}}^{\mathrm{n}}$$$$\mathrm{for}{\mathrm{e}}^{\mathrm{x}},\mathrm{sinx},\mathrm{cosx},\mathrm{we}\mathrm{have}:$$$${\mathrm{e}}^{\mathrm{x}}=1+\mathrm{x}+\frac{{\mathrm{x}}^2}{2!}+\frac{{\mathrm{x}}^3}{3!}+...$$$$\mathrm{sinx}=\mathrm{x}-\frac{{\mathrm{x}}^3}{3!}+\frac{{\mathrm{x}}^5}{5!}-\frac{{\mathrm{x}}^7}{7!}+...$$$$\mathrm{cosx}=1-\frac{{\mathrm{x}}^2}{2!}+\frac{{\mathrm{x}}^4}{4!}-\frac{{\mathrm{x}}^6}{6!}+...$$$$\Rightarrow {\mathrm{e}}^{\mathrm{i}\theta }=1+\mathrm{i}\theta +\frac{{\left(\mathrm{i}\theta \right)}^2}{2!}+\frac{{\left(\mathrm{i}\theta \right)}^3}{3!}+...$$$$=1+\mathrm{i}\theta -\frac{{\theta }^2}{2!}-\mathrm{i}\frac{{\theta }^3}{3!}+\frac{{\theta }^4}{4!}+\frac{\mathrm{i}{\theta }^5}{5!}-\frac{{\theta }^6}{6!}-\frac{\mathrm{i}{\theta }^7}{7!}+...$$$$=(1-\frac{{\theta }^2}{2!}+\frac{{\theta }^4}{4!}-\frac{{\theta }^6}{6!}+...)+\mathrm{i}(\theta -\frac{{\theta }^3}{3!}+\frac{{\theta }^5}{5!}-\frac{{\theta }^7}{7!}+...)$$$$=\cos \theta +\mathrm{isin}\theta$$$$$$

Commented by peter frank last updated on 16/May/22

$$\mathrm{thank}\mathrm{you}$$

Commented by mathlove last updated on 16/May/22

$$thanks$$

Answered by MJS_new last updated on 18/May/22

$$\cos \theta =\frac{{\mathrm{e}}^{\mathrm{i}\theta }+{\mathrm{e}}^{-\mathrm{i}\theta }}{2}$$$$\sin \theta =\frac{{\mathrm{e}}^{\mathrm{i}\theta }-{\mathrm{e}}^{-\mathrm{i}\theta }}{2\mathrm{i}}$$$$\cos \theta +\mathrm{i}\sin \theta =\frac{{\mathrm{e}}^{\mathrm{i}\theta }+{\mathrm{e}}^{-\mathrm{i}\theta }}{2}+\mathrm{i}\frac{{\mathrm{e}}^{\mathrm{i}\theta }-{\mathrm{e}}^{-\mathrm{i}\theta }}{2\mathrm{i}}={\mathrm{e}}^{\mathrm{i}\theta }$$

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