Tinkutara Equation Editor: Math Forum Question #: 149142


Question Number 149142 by mathdanisur last updated on 03/Aug/21

	Answered by liberty last updated on 03/Aug/21
	$z=a+bi ∣z∣=(√(a^2 +b^2 )) ⇒a+bi+(√(a^2 +b^2 )) = 1+(√3) i { ((b=(√3))),((a+(√(a^2 +3)) =1 ⇒(√(a^2 +3)) =1−a)) :} ⇒a^2 +3=a^2 −2a+1 ⇒a=−1 then z=−1+(√3) i tan ϕ=((√3)/(−1))=−(√3) ϕ=−(π/3)$
	Commented bymathdanisur last updated on 03/Aug/21