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proove-e-ipi-1-0-




Question Number 206024 by MaruMaru last updated on 05/Apr/24
proove  e^(iπ) +1=0
$${proove} \\ $$$${e}^{{i}\pi} +\mathrm{1}=\mathrm{0} \\ $$
Answered by Tinku Tara last updated on 05/Apr/24
e^(iπ) =cosπ+isinπ=−1
$${e}^{{i}\pi} ={cos}\pi+{isin}\pi=−\mathrm{1} \\ $$
Answered by Frix last updated on 06/Apr/24
1.  e^(ix) =cos x +i sin x (?)    Let f(x)=((cos x +i sin x)/e^(ix) )    (df/dx)=     [Rule ((u/v))′=((u′v−v′u)/v^2 )]  =((e^(ix) (d/dx)[cos x +i sin x]−(cos x +i sin x)(d/dx)[e^(ix) ])/((e^(ix) )^2 ))=  =((e^(ix) (−sin x +i cos x)−(cos x +i sin x)e^(ix) i)/((e^(ix) )^2 ))=  =((e^(ix) ((−sin x +i cos x)−(i cos x −sin x)))/((e^(ix) )^2 ))=  =0  ⇒  ((cos x +i sin x)/e^(ix) ) is a constant function  f(0)=((1+0i)/1)=1 ⇒ f(x)=1  ⇔ ((cos x +i sin x)/e^(ix) )=1 ⇔ e^(ix) =cos x +i sin x  2.  e^(iπ) =cos π +i sin π =−1+0i=−1  e^(iπ) =−1  e^(iπ) +1=0  q.e.d.
$$\mathrm{1}. \\ $$$$\mathrm{e}^{\mathrm{i}{x}} =\mathrm{cos}\:{x}\:+\mathrm{i}\:\mathrm{sin}\:{x}\:\left(?\right) \\ $$$$ \\ $$$$\mathrm{Let}\:{f}\left({x}\right)=\frac{\mathrm{cos}\:{x}\:+\mathrm{i}\:\mathrm{sin}\:{x}}{\mathrm{e}^{\mathrm{i}{x}} } \\ $$$$ \\ $$$$\frac{{df}}{{dx}}=\:\:\:\:\:\left[\mathrm{Rule}\:\left(\frac{{u}}{{v}}\right)'=\frac{{u}'{v}−{v}'{u}}{{v}^{\mathrm{2}} }\right] \\ $$$$=\frac{\mathrm{e}^{\mathrm{i}{x}} \frac{{d}}{{dx}}\left[\mathrm{cos}\:{x}\:+\mathrm{i}\:\mathrm{sin}\:{x}\right]−\left(\mathrm{cos}\:{x}\:+\mathrm{i}\:\mathrm{sin}\:{x}\right)\frac{{d}}{{dx}}\left[\mathrm{e}^{\mathrm{i}{x}} \right]}{\left(\mathrm{e}^{\mathrm{i}{x}} \right)^{\mathrm{2}} }= \\ $$$$=\frac{\mathrm{e}^{\mathrm{i}{x}} \left(−\mathrm{sin}\:{x}\:+\mathrm{i}\:\mathrm{cos}\:{x}\right)−\left(\mathrm{cos}\:{x}\:+\mathrm{i}\:\mathrm{sin}\:{x}\right)\mathrm{e}^{\mathrm{i}{x}} \mathrm{i}}{\left(\mathrm{e}^{\mathrm{i}{x}} \right)^{\mathrm{2}} }= \\ $$$$=\frac{\mathrm{e}^{\mathrm{i}{x}} \left(\left(−\mathrm{sin}\:{x}\:+\mathrm{i}\:\mathrm{cos}\:{x}\right)−\left(\mathrm{i}\:\mathrm{cos}\:{x}\:−\mathrm{sin}\:{x}\right)\right)}{\left(\mathrm{e}^{\mathrm{i}{x}} \right)^{\mathrm{2}} }= \\ $$$$=\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\frac{\mathrm{cos}\:{x}\:+\mathrm{i}\:\mathrm{sin}\:{x}}{\mathrm{e}^{\mathrm{i}{x}} }\:\mathrm{is}\:\mathrm{a}\:\mathrm{constant}\:\mathrm{function} \\ $$$${f}\left(\mathrm{0}\right)=\frac{\mathrm{1}+\mathrm{0i}}{\mathrm{1}}=\mathrm{1}\:\Rightarrow\:{f}\left({x}\right)=\mathrm{1} \\ $$$$\Leftrightarrow\:\frac{\mathrm{cos}\:{x}\:+\mathrm{i}\:\mathrm{sin}\:{x}}{\mathrm{e}^{\mathrm{i}{x}} }=\mathrm{1}\:\Leftrightarrow\:\mathrm{e}^{\mathrm{i}{x}} =\mathrm{cos}\:{x}\:+\mathrm{i}\:\mathrm{sin}\:{x} \\ $$$$\mathrm{2}. \\ $$$$\mathrm{e}^{\mathrm{i}\pi} =\mathrm{cos}\:\pi\:+\mathrm{i}\:\mathrm{sin}\:\pi\:=−\mathrm{1}+\mathrm{0i}=−\mathrm{1} \\ $$$$\mathrm{e}^{\mathrm{i}\pi} =−\mathrm{1} \\ $$$$\mathrm{e}^{\mathrm{i}\pi} +\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{q}.\mathrm{e}.\mathrm{d}. \\ $$

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