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Question Number 170110 by mathlove last updated on 16/May/22
prove that e^(iθ) =cosθ+isinθ
$${prove}\:{that} \\ $$$${e}^{{i}\theta} ={cos}\theta+{isin}\theta \\ $$
Commented by ajfour last updated on 16/May/22
1×e^(iθ) =z is 1 rotated by θ anticlockwise . hence z=cos θ+isin θ=e^(iθ) draw a diagram n u know.
$$\mathrm{1}×{e}^{{i}\theta} ={z}\:{is} \\ $$$$\:\mathrm{1}\:{rotated}\:{by}\:\theta\:{anticlockwise}\:. \\ $$$${hence}\:{z}=\mathrm{cos}\:\theta+{i}\mathrm{sin}\:\theta={e}^{{i}\theta} \\ $$$${draw}\:{a}\:{diagram}\:{n}\:{u}\:{know}. \\ $$
Answered by floor(10²Eta[1]) last updated on 16/May/22
using the mclaurin series: f(x)=Σ_(n=0) ^∞ ((f^((n)) (0))/(n!))x^n for e^x , sinx, cosx, we have: e^x =1+x+(x^2 /(2!))+(x^3 /(3!))+... sinx=x−(x^3 /(3!))+(x^5 /(5!))−(x^7 /(7!))+... cosx=1−(x^2 /(2!))+(x^4 /(4!))−(x^6 /(6!))+... ⇒e^(iθ) =1+iθ+(((iθ)^2 )/(2!))+(((iθ)^3 )/(3!))+... =1+iθ−(θ^2 /(2!))−i(θ^3 /(3!))+(θ^4 /(4!))+((iθ^5 )/(5!))−(θ^6 /(6!))−((iθ^7 )/(7!))+... =(1−(θ^2 /(2!))+(θ^4 /(4!))−(θ^6 /(6!))+...)+i(θ−(θ^3 /(3!))+(θ^5 /(5!))−(θ^7 /(7!))+...) =cosθ+isinθ
$$\mathrm{using}\:\mathrm{the}\:\mathrm{mclaurin}\:\mathrm{series}: \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{0}\right)}{\mathrm{n}!}\mathrm{x}^{\mathrm{n}} \\ $$$$\mathrm{for}\:\mathrm{e}^{\mathrm{x}} ,\:\mathrm{sinx},\:\mathrm{cosx},\:\mathrm{we}\:\mathrm{have}: \\ $$$$\mathrm{e}^{\mathrm{x}} =\mathrm{1}+\mathrm{x}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}!}+… \\ $$$$\mathrm{sinx}=\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}!}+\frac{\mathrm{x}^{\mathrm{5}} }{\mathrm{5}!}−\frac{\mathrm{x}^{\mathrm{7}} }{\mathrm{7}!}+… \\ $$$$\mathrm{cosx}=\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{4}!}−\frac{\mathrm{x}^{\mathrm{6}} }{\mathrm{6}!}+… \\ $$$$\Rightarrow\mathrm{e}^{\mathrm{i}\theta} =\mathrm{1}+\mathrm{i}\theta+\frac{\left(\mathrm{i}\theta\right)^{\mathrm{2}} }{\mathrm{2}!}+\frac{\left(\mathrm{i}\theta\right)^{\mathrm{3}} }{\mathrm{3}!}+… \\ $$$$=\mathrm{1}+\mathrm{i}\theta−\frac{\theta^{\mathrm{2}} }{\mathrm{2}!}−\mathrm{i}\frac{\theta^{\mathrm{3}} }{\mathrm{3}!}+\frac{\theta^{\mathrm{4}} }{\mathrm{4}!}+\frac{\mathrm{i}\theta^{\mathrm{5}} }{\mathrm{5}!}−\frac{\theta^{\mathrm{6}} }{\mathrm{6}!}−\frac{\mathrm{i}\theta^{\mathrm{7}} }{\mathrm{7}!}+… \\ $$$$=\left(\mathrm{1}−\frac{\theta^{\mathrm{2}} }{\mathrm{2}!}+\frac{\theta^{\mathrm{4}} }{\mathrm{4}!}−\frac{\theta^{\mathrm{6}} }{\mathrm{6}!}+…\right)+\mathrm{i}\left(\theta−\frac{\theta^{\mathrm{3}} }{\mathrm{3}!}+\frac{\theta^{\mathrm{5}} }{\mathrm{5}!}−\frac{\theta^{\mathrm{7}} }{\mathrm{7}!}+…\right) \\ $$$$=\mathrm{cos}\theta+\mathrm{isin}\theta \\ $$$$ \\ $$
Commented by peter frank last updated on 16/May/22
thank you
$$\mathrm{thank}\:\mathrm{you} \\ $$
Commented by mathlove last updated on 16/May/22
thanks
$${thanks} \\ $$
Answered by MJS_new last updated on 18/May/22
cos θ =((e^(iθ) +e^(−iθ) )/2) sin θ =((e^(iθ) −e^(−iθ) )/(2i)) cos θ +i sin θ =((e^(iθ) +e^(−iθ) )/2)+i((e^(iθ) −e^(−iθ) )/(2i))=e^(iθ)
$$\mathrm{cos}\:\theta\:=\frac{\mathrm{e}^{\mathrm{i}\theta} +\mathrm{e}^{−\mathrm{i}\theta} }{\mathrm{2}} \\ $$$$\mathrm{sin}\:\theta\:=\frac{\mathrm{e}^{\mathrm{i}\theta} −\mathrm{e}^{−\mathrm{i}\theta} }{\mathrm{2i}} \\ $$$$\mathrm{cos}\:\theta\:+\mathrm{i}\:\mathrm{sin}\:\theta\:=\frac{\mathrm{e}^{\mathrm{i}\theta} +\mathrm{e}^{−\mathrm{i}\theta} }{\mathrm{2}}+\mathrm{i}\frac{\mathrm{e}^{\mathrm{i}\theta} −\mathrm{e}^{−\mathrm{i}\theta} }{\mathrm{2i}}=\mathrm{e}^{\mathrm{i}\theta} \\ $$

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