IIT JEE , AIEEE Forum - Mathematics , Algebra - any one with clear concept should try this

animal

Find all the complex numbers z such that

(3z+1)(4z+1)(6z+1)(12z+1)=2.

shreyasnivas

is it to do with mod?

hold on im trying..

animal

I dont think so man!

shreyasnivas

ok i give up.. pls give soln..

spideyunlimited

wait yaar itni kya jaldi hai

shreyasnivas

jab sawaal nahin banta to obvious hai ki soln ke piche main padha rahuunga.

mukundmadhav

Multiply first and last terms, and middle two terms
(36z^2 + 15z + 1)( 24z^2 + 10z +1) = 2
Substitute 12z^2 + 5z = p and solve quadratic..
From the two solutions you get two more quadratics, then solve it again..
Jeez people, you should've been able to do this three years ago...

riku

yaar,, is this a challenging question,, cause u wrote "any one with clear concept should try this "... Hahaha..

nyways,, just arrange ur question as,,

(3z+1)(12z+1) (6z+1)(4z+1)=2.

which gives : (36z² + 15z + 1)( 24z² + 10z +1) = 2

bas substitute (12z² + 5z) with "x" then equation becomes

(3x+1)(2x+1)=2; this equation on being solved yields x= -1, 1/6.

we obtain four solutions such that 12z² + 5z = -1,1/6

giving $z= \frac{-5 + i \sqrt{23}}{24}$ ;

$z= \frac{-5 - i \sqrt{23}}{24}$ ;

$z= \frac{-30 + \sqrt{1188}}{144}$ ; and

$z= \frac{-30 - \sqrt{1188}}{144}$ ....

shayad mere concepts clear hain..

dont 4get to rate dear!!hehe...

(12)

that was easy

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