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Question Number 206024 by MaruMaru last updated on 05/Apr/24

$$proove$$$$e^{i\pi }+1=0$$

Answered by Tinku Tara last updated on 05/Apr/24

$$e^{i\pi }=cos\pi +isin\pi =-1$$

Answered by Frix last updated on 06/Apr/24

$$1.$$$${\mathrm{e}}^{\mathrm{i}x}=\cos x+\mathrm{i}\sin x(?)$$$$$$$$\mathrm{Let}f\left(x\right)=\frac{\cos x+\mathrm{i}\sin x}{{\mathrm{e}}^{\mathrm{i}x}}$$$$$$$$\frac{df}{dx}=[\mathrm{Rule}{\left(\frac{u}{v}\right)}^{\prime }=\frac{u^{\prime }v-v^{\prime }u}{v^2}]$$$$=\frac{{\mathrm{e}}^{\mathrm{i}x}\frac{d}{dx}[\cos x+\mathrm{i}\sin x]-(\cos x+\mathrm{i}\sin x)\frac{d}{dx}\left[{\mathrm{e}}^{\mathrm{i}x}\right]}{{\left({\mathrm{e}}^{\mathrm{i}x}\right)}^2}=$$$$=\frac{{\mathrm{e}}^{\mathrm{i}x}(-\sin x+\mathrm{i}\cos x)-(\cos x+\mathrm{i}\sin x){\mathrm{e}}^{\mathrm{i}x}\mathrm{i}}{{\left({\mathrm{e}}^{\mathrm{i}x}\right)}^2}=$$$$=\frac{{\mathrm{e}}^{\mathrm{i}x}((-\sin x+\mathrm{i}\cos x)-(\mathrm{i}\cos x-\sin x))}{{\left({\mathrm{e}}^{\mathrm{i}x}\right)}^2}=$$$$=0$$$$\Rightarrow$$$$\frac{\cos x+\mathrm{i}\sin x}{{\mathrm{e}}^{\mathrm{i}x}}\mathrm{is}\mathrm{a}\mathrm{constant}\mathrm{function}$$$$f\left(0\right)=\frac{1+0\mathrm{i}}{1}=1\Rightarrow f\left(x\right)=1$$$$\iff \frac{\cos x+\mathrm{i}\sin x}{{\mathrm{e}}^{\mathrm{i}x}}=1\iff {\mathrm{e}}^{\mathrm{i}x}=\cos x+\mathrm{i}\sin x$$$$2.$$$${\mathrm{e}}^{\mathrm{i}\pi }=\cos \pi +\mathrm{i}\sin \pi =-1+0\mathrm{i}=-1$$$${\mathrm{e}}^{\mathrm{i}\pi }=-1$$$${\mathrm{e}}^{\mathrm{i}\pi }+1=0$$$$\mathrm{q}.\mathrm{e}.\mathrm{d}.$$

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