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Question Number 208398 by mokys last updated on 14/Jun/24
writez=13+iineiθ
Answered by A5T last updated on 14/Jun/24
z=3−i(3)2−(i)2=3−i4,∣z∣=(34)2+(−14)2=14tanθ=−1434=−13⇒θ=−π6⇒z=e−iπ62
Answered by mathzup last updated on 15/Jun/24
z=3−i4=34−14i(=x+iy)∣z∣=(34)2+(−14)2=316+116=12wehavez=∣z∣eiarctan(yx)=12eiarctan(−1.3)=12e−iπ6
Answered by lepuissantcedricjunior last updated on 15/Jun/24
z=13+i=1232+i12=12eiπ6=12e−π6=>eiθ=12e−iπ6=>θ=−π6
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