Shane Macguire
Last Activity: 11 Years ago
i am reframing the solution may be this is helpful for the staudent
we will use the method of contradiction
let us ASSUME there is a complex root p+iq of the given equation a/x-1 + b/x-2 + c/x-3 + d/x-4 =1
so put x = p + i q in the given equation then let us see the first term of the
a/(x-1) = a/(p+iq - 1) = a / {(p-1) + iq}
now multiply the numerator and denominator by (p-1) - iq
we get
a{(p-1) - i q}/{(p-1)^2 + q^2)} -----------1
again we take the first term only
similarly we can write all the other three terms
now we will put these terms in the given equation and write REAL PART and IMAGINARY PART seperately
imaginary part of L.H.S = imaginary part of R.H.S
and imaginary part of R.H.S = 0
so imaginary part of L.H.S is also 0
IGNORE the real part just concentrate on IMAGINARY PART of L.H.S, which will be
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)}]
there would be a term of q common so we took q as common
and we know this expression is equal to 0
so
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
let us take
( a/[(p-1)^2 + q^2)} + b/{(p-2)^2 + q^2)}+c/{(p-3)^2 + q^2)}+d/{(p-4)^2 + q^2)}= Y
now it of the form q * ( Y ) = 0
so either q = 0 or Y = 0
Y can''t be 0, it''s value depend on a,b,c,d, as the denominator is greater than 0 in all the terms ,as a,b,c,d are
either all +ve or -ve value Y can''t be zero as if all are +ve then Y will be +ve if all -ve Y will be -ve
so only one case remaining q = 0
q is the IMAGINARY PART so if imaginary part = 0 for all the cases where a,b,c,d are all +ve or -ve
so the root will be real it can''t be imaginary
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
