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Question Number 208398 by mokys last updated on 14/Jun/24

write z = (1/( (√3)+i)) in e^(iθ)

writez=13+iineiθ

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=3i(3)2(i)2=3i4,z∣=(34)2+(14)2=14tanθ=1434=13θ=π6z=eiπ62

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=3i4=3414i(=x+iy)z∣=(34)2+(14)2=316+116=12wehavez=∣zeiarctan(yx)=12eiarctan(1.3)=12eiπ6

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)

z=13+i=1232+i12=12eiπ6=12eπ6=>eiθ=12eiπ6=>θ=π6

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