- Thread starter
- #1

**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: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

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$

Last edited: