1Find the mistake
z2z1+ z2z + z3=0 is quadratic equation in z.Let its roots be 1,μ
μ+1=-(z2z1) and μ=z3/z1
Substituting value of μ ,z3z1+1=-z2z1 giving z1+z2+z3=0
plz solve
A(z1) , B(z2) and C(z3) are 3 points lying on the curve
|z|=√3
1. If z2z1+ z2z + z3=0 has one root of modulus unity.
z1,z2,z3are in:
a)AP
b)Gp
c)HP
D)all of the above
2.If z2z1+ z2z + z3=0 and
z2z2+ z3z + z1=0 each have a common root unity then
triangle ABC is:
a)isosceles
b)equilateral
c)right
d)scalene
ans
1.b
2.b
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6 Answers
341z^2z_1+zz_2+z_3 = 0 \Rightarrow \overline{z^2z_1+zz_2+z_3}= \overline{z^2}\overline{z_1} + \overline{z}\overline{z_2} + \overline{z_3}=0
Now |z| =1 \Rightarrow \overline{z} = \frac{1}{z} and |z_1| = \sqrt 3 \Rightarrow \overline{z_1} = \frac{3}{z_1} etc.
So we have after dividing by 3 throughout, \frac{1}{z^2z_1} + \frac{1}{zz_2}+\frac{1}{ z_3}=0
So if p=z^2z_1; q=zz_2; r=z_3 we have
p+q+r=0; \frac{1}{p} + \frac{1}{q} + \frac{1}{r} = 0 \Rightarrow pq+qr+rp=0
pq+qr+rp=0 \Rightarrow q(p+r) + rp = 0 \Rightarrow q^2=pr.
This means (zz_2)^2 = z^2z_1z_3 or z_2^2=z_1z_3
Hence z_1, z_2, z_3 are in GP
1
1@prophet sir plz help, I can't understand
"p+q+r=0 therefore
1/p + 1/q + 1/r =0 .............."
how??
