Find all complex number z such that (3z+1)(4z+1)(6z+1)(12z+1)=2


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Question Number 152757 by EDWIN88 last updated on 01/Sep/21

$F i n d a l l c o m p l e x n u m b e r z s u c h$ $t h a t (3 z + 1) (4 z + 1) (6 z + 1) (12 z + 1) = 2$
	Answered by john_santu last updated on 01/Sep/21
	$note that 8(3z+1)6(4z+1)4(6z+1)2(12z+1)=768 (24z+8)(24z+6)(24z+4)(24z+2)=768 let u=24z+5 ⇒(u+3)(u+1)(u−1)(u−3)=768 ⇒(u^2 −1)(u^2 −9)=768 i.e w^2 −10w−759=0 , w=u^2 → { ((w_1 =33⇒z=((±(√(33))−5)/(24)))),((w_2 =−23⇒z=((±(√(23i))−5)/(24)))) :}$
	$n o t e t h a t 8 (3 z + 1) 6 (4 z + 1) 4 (6 z + 1) 2 (12 z + 1) = 768$ $(24 z + 8) (24 z + 6) (24 z + 4) (24 z + 2) = 768$ $l e t u = 24 z + 5$ $\Rightarrow (u + 3) (u + 1) (u - 1) (u - 3) = 768$ $\Rightarrow (u^{2} - 1) (u^{2} - 9) = 768$ $i . e w^{2} - 10 w - 759 = 0, w = u^{2}$ $\to {\begin{cases} w_{1} = 33 \Rightarrow z = \frac{\pm \sqrt{33} - 5}{24} \\ w_{2} = - 23 \Rightarrow z = \frac{\pm \sqrt{23 i} - 5}{24} \end{cases}$
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