Amit Saxena
Last Activity: 11 Years ago
we will use the method of contradiction
let us assume there be a complex root p+iq
after multiplying the conjugate in numerator and denominator we get
a{(p-1) - i q}/{(p-1)^2 + q^2)} similarly we will write all the three terms
there would be a term of q common in all the four terms as coefficient of i
a/x-1 + b/x-2 + c/x-3 + d/x-4 =1
now comparing l.h.s and r.h.s
q( a/[(p-1)^2 + q^2)} + b/{(p-2)^2 + q^2)}+c/{(p-3)^2 + q^2)}+d/{(p-4)^2 + q^2)}]=0
imaginary part has to be 0 as a , b, c, d all are of same sign
the second term will not be zero so q has to be zero
so our assumption was wrong there can''t be a imaginary root when a,b,c,d has
same sign