MHB Roots of a polynomial with non-real coefficients.

[juantheron]

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 EquationNow 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$

 

[vacuum]

So Option $ \boxed{\boxed{c}} $ is Correct
I would like to provide a more detailed explanation for why the answer is option c) -c.

Firstly, let's define the four roots of the given equation as $\alpha, \beta, \gamma, \delta$. Since the equation has all complex roots, we know that they will exist in conjugate pairs, i.e. $\alpha$ and $\beta$ will be conjugates, and $\gamma$ and $\delta$ will be conjugates.

Now, we know that the roots of the equation lie on the circle $|z| = 1$ in the complex plane. This means that all the roots will have a magnitude of 1, and therefore can be represented in polar form as $e^{i\theta}$, where $\theta$ is the argument of the root.

Since the roots exist in conjugate pairs, we can write them as $e^{i\theta}$ and $e^{-i\theta}$. Now, using Euler's formula, we can expand these roots as $\cos\theta + i\sin\theta$ and $\cos\theta - i\sin\theta$, respectively.

Next, let's look at the sum of the reciprocals of the roots. This can be written as $\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} + \frac{1}{\delta}$. Using the expanded forms of the roots, we can simplify this expression as $\frac{\cos\theta - i\sin\theta}{\cos\theta + i\sin\theta} + \frac{\cos\theta + i\sin\theta}{\cos\theta - i\sin\theta} + \frac{\cos\theta - i\sin\theta}{\cos\theta + i\sin\theta} + \frac{\cos\theta + i\sin\theta}{\cos\theta - i\sin\theta}$.

Now, using the formula for complex conjugates, we can simplify this expression even further to $\frac{\cos^2\theta + \sin^2\theta}{\cos^2\theta + \sin^2\theta} = 1$. Therefore, the sum of the reciprocals of the roots is simply 1.

Finally, using Vieta's formulas, we