1Find all those roots of z12- 56z6 -512 =0 whose
imaginary part is +ve.
Q.If Z1 & Z2 are two complex no such that Z2≠0 and whole mod of Z1-1/Z1+Z2=1.Then prove dat i*Z1/Z2=K,where K is a real no.Find the angle between the lines frm the original to the points Z1+Z2 & Z1-Z2 in terms of K???????????
plzzz xplain elaboratly................
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9 Answers
1pls post the question again clearly particularly the"whole mod"
1If |z| =1 and omega = (z-1)/(z+1) (where z not equal to -1) thn
Re(omega) is
1Given z=cos (2pi/2n+1) + i sin(2pi/2n+1),n is a +ve integer,find the eq
wose roots are
alpha= z+z3+........+ z 2n+1
and beta = z2+ z4+............z2n
1
62first question.. prove that (1-aplha1)(1-aplha2)...........(1-alphan-1) = n what are alpha1, alpha2.. etc? non real nth roots of unity?
IF that is the case, which i think it is, then
(x-a1)(x-a2)........(x-an-1) (x-1)=xn-1 = (x-1)(1+x+x^2....+x^n-1))
now divide both sides by x-1
thus you get
(x-a1)(x-a2)........(x-an-1)=xn-1 =(1+x+x^2....+x^n-1))
NOw substitute x=1 to get the result :)
62If |z| =1 and omega = (z-1)/(z+1) (where z not equal to -1) thn
Re(omega) is
take z=cos@+i sin @ and try... that should help you easily!
62Find all those roots of z12- 56z6 -512 =0 whose
imaginary part is +ve.
take z6=t
I think you should get z6=-8 or 64 as the roots... check that ...
1|z1+z2| = |z1-z2|
or, |z1/z2 + 1|=|z1/z2 - 1|
it means that the distance of the point representing z1/z2 from the points(1 , 0) and (-1 , 0)is the same.
this only means that z1/z2 lies in the pependicular bisector of the line joining (1 , 0) and (-1 , 0).
for any z1/z2, this perpendicular bisector is the y-axis,which means z1/z2 is purely imaginary.
so Re(z1/z2) = 0 and correspondingly, i x (z1/z2) is a real number k.
let k = cosθ + i sinθ
(suppose the angle in quesion is θ)
so θ = amp(z1 + z2z1 - z2)
=amp(z1/z2 + 1z1/z2 - 1 )
i(z1z2) = k,
or, z1/z2 = k/i = ki/i2 = -ki,
substituting,
θ = amp (-ki + 1 / -ki - 1)
=amp{(-ki + 1)2 / (-ki + 1)(-ki - 1)}
=amp{k2 + i2 + 2ki / -(k2 i2 - 1)}
=amp(k2 - 1 + 2ki / k2 + 1)
=amp{(k2 - 1 / k2 + 1) + i (2k / k2 + 1)}
but k = cosθ + i sinθ,
therefore, cosθ = k2 - 1 / k2 + 1, --------1
i sinθ = i (2k / k2 + 1)
or, sinθ = 2k / k2 + 1 ---------2
dividing 2 by 1;
tanθ = 2k / k2 - 1;
or θ = tan-1 ( 2k / k2 - 1 )
brain storming, it was. :)