a z4 1 -250 i 6 The mumber of solutions of the equation z2 z 0 where z is a complex number, is (a 1 a 4 z In the quadratic equatian x +(p iq)x +3i 0, p ad ade real. If the sum of the saames of the roots is 8 then 3, q a p 3, q --1 t3, q

In the quadratic equation x^2+(p+i q)x+3i=0,p&q are real. If the sum of the squares of the roots is 8 then: p=3,q=-1 b. p=3,q=1 c. p=-3,q=-1 d. p=-3,q=1

In the quadratic equation x^(2)+(p + iq)x+3i=0 ,where p and q are real If the sum of the squares of the roots is 8 ,then the ordered pair (p, q) is given by : (A) (3,1) (B) (-3,-1) (C) (-3,1) (D) (3,-1)

For a complex number z, the equation z^(2)+(p+iq)z+ r+" is "=0 has a real root (where p, q, r, s are non - zero real numbers and i^(2)-1 ), then

For a complex number z, the equation z^(2)+(p+iq)z r+" is "=0 has a real root (where p, q, r, s are non - zero real numbers and i^(2) = -1 ), then

if the roots of the equation z^(2) + ( p +iq) z + r + is =0 are real wher p,q,r,s, in ,R , then determine s^(2) + q^(2)r .

if the roots of the equation z^(2) + ( p +iq) z + r + is =0 are real wher p,q,r,s, in ,R , then determine s^(2) + q^(2)r .

Let z be a complex number satisfying equation z^p-z^(-q)=0,\ where\ p ,q in N ,\t h e n (A) if p=q , then number of solutions of equation will be infinite. (B) if p=q , then number of solutions of equation will be finite. (C) if p!=q , then number of solutions of equation will be p+q+1. (D) if p!=q , then number of solutions of equation will be p+qdot