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Question Number 208398 by mokys last updated on 14/Jun/24
write z = (1/( (√3)+i)) in e^(iθ)
$${write}\:{z}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}+{i}}\:{in}\:{e}^{{i}\theta} \\ $$
Answered by A5T last updated on 14/Jun/24
z=(((√3)−i)/(((√3))^2 −(i)^2 ))=(((√3)−i)/4), ∣z∣=(√((((√3)/4))^2 +(((−1)/4))^2 ))=(1/4) tanθ=(((−1)/4)/((√3)/4))=((−1)/( (√3)))⇒θ=((−π)/6) ⇒z=(e^((−iπ)/6) /2)
$${z}=\frac{\sqrt{\mathrm{3}}−{i}}{\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}} −\left({i}\right)^{\mathrm{2}} }=\frac{\sqrt{\mathrm{3}}−{i}}{\mathrm{4}},\:\mid{z}\mid=\sqrt{\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\right)^{\mathrm{2}} +\left(\frac{−\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${tan}\theta=\frac{\frac{−\mathrm{1}}{\mathrm{4}}}{\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}}=\frac{−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\Rightarrow\theta=\frac{−\pi}{\mathrm{6}} \\ $$$$\Rightarrow{z}=\frac{{e}^{\frac{−{i}\pi}{\mathrm{6}}} }{\mathrm{2}} \\ $$
Answered by mathzup last updated on 15/Jun/24
z=(((√3)−i)/4)=((√3)/4)−(1/4)i (=x+iy) ∣z∣=(√((((√3)/4))^2 +(−(1/4))^2 ))=(√((3/(16))+(1/(16)))) =(1/2) we have z=∣z∣e^(iarctan((y/x))) =(1/2) e^(iarctan(−(1/(.(√3))))) =(1/2)e^(−((iπ)/6))
$${z}=\frac{\sqrt{\mathrm{3}}−{i}}{\mathrm{4}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{4}}{i}\:\:\left(={x}+{iy}\right) \\ $$$$\mid{z}\mid=\sqrt{\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\right)^{\mathrm{2}} +\left(−\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} }=\sqrt{\frac{\mathrm{3}}{\mathrm{16}}+\frac{\mathrm{1}}{\mathrm{16}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\:{we}\:{have}\:{z}=\mid{z}\mid{e}^{{iarctan}\left(\frac{{y}}{{x}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:{e}^{{iarctan}\left(−\frac{\mathrm{1}}{.\sqrt{\mathrm{3}}}\right)} =\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\frac{{i}\pi}{\mathrm{6}}} \\ $$
Answered by lepuissantcedricjunior last updated on 15/Jun/24
z=(1/( (√3)+i))=((1/2)/(((√3)/2)+i(1/2)))=((1/2)/e^(i(𝛑/6)) )=(1/2)e^(−(𝛑/6)) =>e^(i𝛉) =(1/2)e^(−i(𝛑/6)) =>𝛉=−(𝛑/6)
$$\boldsymbol{{z}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}+\boldsymbol{{i}}}=\frac{\frac{\mathrm{1}}{\mathrm{2}}}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+\boldsymbol{{i}}\frac{\mathrm{1}}{\mathrm{2}}}=\frac{\frac{\mathrm{1}}{\mathrm{2}}}{\boldsymbol{{e}}^{\boldsymbol{{i}}\frac{\boldsymbol{\pi}}{\mathrm{6}}} }=\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{e}}^{−\frac{\boldsymbol{\pi}}{\mathrm{6}}} \\ $$$$=>\boldsymbol{{e}}^{\boldsymbol{{i}\theta}} =\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{e}}^{−\boldsymbol{{i}}\frac{\boldsymbol{\pi}}{\mathrm{6}}} \\ $$$$=>\boldsymbol{\theta}=−\frac{\boldsymbol{\pi}}{\mathrm{6}} \\ $$

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