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Question Number 152757 by EDWIN88 last updated on 01/Sep/21
Find all complex number z such  that (3z+1)(4z+1)(6z+1)(12z+1)=2
Findallcomplexnumberzsuchthat(3z+1)(4z+1)(6z+1)(12z+1)=2
Answered by john_santu last updated on 01/Sep/21
note that 8(3z+1)6(4z+1)4(6z+1)2(12z+1)=768  (24z+8)(24z+6)(24z+4)(24z+2)=768  let u=24z+5  ⇒(u+3)(u+1)(u−1)(u−3)=768  ⇒(u^2 −1)(u^2 −9)=768  i.e w^2 −10w−759=0 , w=u^2   → { ((w_1 =33⇒z=((±(√(33))−5)/(24)))),((w_2 =−23⇒z=((±(√(23i))−5)/(24)))) :}
notethat8(3z+1)6(4z+1)4(6z+1)2(12z+1)=768(24z+8)(24z+6)(24z+4)(24z+2)=768letu=24z+5(u+3)(u+1)(u1)(u3)=768(u21)(u29)=768i.ew210w759=0,w=u2{w1=33z=±33524w2=23z=±23i524

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