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[SOLVED] Roots of a polynomial with non-real coefficients.

jacks

Well-known member
Apr 5, 2012
226
Let a,b,c,d be real numbers. Sauppose that all the roots of the equation $z^4 + az^3 + bz^2 + cz + d = 0$ are complex numbers

lying on the circle $\mid z\mid = 1$ in the complex plane. The sum of the reciprocals of the roots is necessarily:


options

a) a
b) b
c) -c
d) d


---------- Post added at 22:03 ---------- Previous post was at 21:31 ----------

Thanks Friends I have Got it

Let $\alpha\;,\beta\;,\gamma\;,\delta$ be the roots of Given Equation


Now all Roots are Imaginary and lie on $\mid z\mid = 1$


and Imaginary Roots are always occur in pair


so Let $\alpha = x_{1}+iy_{1}\;\;\beta = x_{1}-iy_{1}$ and $ \alpha.\beta = x^2_{1}+y^2_{1} = 1$


Similarly $ \gamma = x_{2}+iy_{2}\;\;\delta = x_{2}-iy_{2}$ ] and $ \gamma.\delta = x^2_{2}+y^2_{2} = 1$


Now Using Vieta, s Formula


$ \alpha+\beta+\gamma+\delta = -a$


$ \alpha.\beta.\gamma+\beta.\gamma.\delta+\gamma \delta.\alpha+\delta.\alpha.\beta = -c$


$ \alpha.\beta.\gamma.\delta = d$


So $\displaystyle \frac{\alpha.\beta.\gamma+\beta.\gamma \delta+\gamma.\delta.\alpha+\delta.\alpha.\beta}{\alpha \beta.\gamma.\delta} = -\frac{c}{d} = - c$


So $ \displaystyle \frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}+\frac{1}{\delta} = -c$


because $ \alpha.\beta.\gamma.\delta = d=1$
 
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