consider the equation $z^4=5(z-1)(z^2-z+1).$ Let roots of the equation be $z_{1},z_{2},z_{3},z_{4} .$ If the area of quadrilateral $ABCD$ formed by the points $A(z_{1}),B(z_{2}),C(z_{3}),D(z_{4})$ in the arg-and plane is $2\cos^3\alpha.$ Then $\alpha=$
what i try $$z^4=5(z^3-z^2-z^2+z+z-1)$$
$$z^4=5z^3-10z^2+10z-5$$
$$z^4-5z^3+10z^2-10z+5=0$$
$$z^5-5z^4+10z^3-10z^2+5z-1=-1$$
$$\Longrightarrow (z-1)^5=-1$$
$$(z-1)=(-1)^{\frac{1}{5}}=e^{\frac{i(2n+1)\pi}{5}}$$
for $n=0,1,2,3,4$
How do i solve it Help me
First off: The roots are rendered by the OP thusly:
$z=1-\exp(i(2k\pi/5)), k\in\{0,1,2,3,4\}$
but this is wrong. We have instead only
$z=1-\exp(i(2k\pi/5)), k\in\{1,2,3,4\}$
The root $z=0$ cotresponding to $k=0$ was introduced extraneously between the third and fourth lines where there was a multiplication by $z$.
The roots are shown geometrically in the picture below.
Clearly the four good roots, at the corners of the red trapezoid, are at four vertices of a regular pentagon with center $(1,0)$ and radius $1$ unit from the center to each vertex.
The trapezoid may then be divided into four isosceles triangles whose legs (solid blue lines) measure $1$ unit and whose apex angles are $72°$ (three triangles) or $144°$ (one triangle). From the fact that each triangle has area equal to half the product of any two sides times the sine of the included angle we infer the total area of the trapezoid:
$A=(3/2)\sin(72°)+(1/2)\sin(144°)$
At this point one may be tempted to render $\sin(144°)=2\sin(72°)\cos(72°)$ with the double angle formula, but because this introduces the cosine function it offers no real progress. Instead, render the following:
$A=(3/2)\sin(72°)-(1/2)\sin(216°)$
and use the triple-angle formula:
$\sin(3\theta)=\sin(2\theta+\theta)=\sin(2\theta)\cos\theta+\cos(2\theta)\sin\theta$
$=2\sin\theta\cos^2\theta+(1-2\sin^2\theta)\sin\theta$
$=2\sin\theta(1-\sin^2\theta)+(1-2\sin^2\theta)\sin\theta$
$=3\sin\theta-4\sin^3\theta$
Plugging this into the area expression given above with $\theta=72°$ and $3\theta=216°$ then gives
$A=(3/2)\sin(72°)-(3/2)\sin(72°)+2\sin^3(72°)$
$=2\sin^3(72°)=2\cos^3(18°)$
from which we read off $\alpha=18°=\pi/10\text{ radians}$.