How to prove that if a quartic equation ( with real coefficients ) has 4 imaginary roots they all will be in conjugate pairs?I proved this fact for qudratic equation in the following way , let a qudratic equation have a imaginary root p+iq(q is not 0), and let other root be (a+ib). Now here sum of roots will be a real number lets say R, ⇒(p+iq)(a−iq)=R′⇒(pa+q2)+i(aq−pq)=0⇒aq−pq=0⇒aq=pq⇒a=p. So finally the roots are, p+iq and p−iq, hence proved. I tried to prove this fact for quartic equation in a similar way but could not reach to the conclusion. Please guide me by answering by my method or by suggesting any other simple method to prove that if a quartic equation has all imaginary roots then they will occur in conjugate pairs.

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How to prove that if a quartic equation ( with real coefficients ) has 4 imaginary roots they all will be in conjugate pairs?I proved this fact for qudratic equation in the following way , let a qudratic equation have a imaginary root p+iq(q is not 0), and let other root be (a+ib). Now here sum of roots will be a real number lets say R, ⇒(p+iq)(a−iq)=R′⇒(pa+q2)+i(aq−pq)=0⇒aq−pq=0⇒aq=pq⇒a=p. So finally the roots are, p+iq and p−iq, hence proved. I tried to prove this fact for quartic equation in a similar way but could not reach to the conclusion. Please guide me by answering by my method or by suggesting any other simple method to prove that if a quartic equation has all imaginary roots then they will occur in conjugate pairs.

prangijahnot · Accepted Answer

If P(z)=∑i=0naiziis a polynomial having real coefficients with z0 as a root, then taking conjugate on both sides we see that z0― is also a root.Now, if a+ib is a root of a quartic with real coefficients, then by this reasoning, a−ib is also a root. Hence, given polynomial is divisible by the quadratic (x−a−bi)(x−a+bi)=(x−a)2+b2. Divide it by this quadratic, get a quadratic and apply the same procedure.

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