$z^4+2 z^2 \cos \alpha+1=0

Substitute y = z 2 to transform the equation into a quadratic equation in y .
Solve for y using the quadratic formula: y = − cos α ± i sin α .
Express y in polar form: y = e i ( π − α ) or y = e i ( π + α ) .
Solve for z by taking the square root: z = ± e i ( π − α ) /2 and z = ± e i ( π + α ) /2 .
The solutions are: z = ± e i ( π − α ) /2 , ± e i ( π + α ) /2 ​ .

Explanation

Understanding the Problem We are given the equation z 4 + 2 z 2 cos α + 1 = 0 , where z is a complex number and α is a real number. Our goal is to find the solutions for z . Notice that this equation is a quadratic equation in terms of z 2 .

Applying the Quadratic Formula Let y = z 2 . Then the equation becomes y 2 + 2 y cos α + 1 = 0 . We can solve for y using the quadratic formula: y = 2 ( 1 ) − 2 cos α ± ( 2 cos α ) 2 − 4 ( 1 ) ( 1 ) ​ ​ .

Simplifying the Expression Simplifying the expression for y , we get: y = 2 − 2 cos α ± 4 cos 2 α − 4 ​ ​ = 2 − 2 cos α ± 2 cos 2 α − 1 ​ ​ = − cos α ± cos 2 α − 1 ​ . Since cos 2 α − 1 = − sin 2 α , we have y = − cos α ± − sin 2 α ​ = − cos α ± i sin α .

Expressing in Polar Form Now we need to solve for z . Since y = z 2 , we have z 2 = − cos α ± i sin α . We can express − cos α ± i sin α in polar form. Recall that − cos α = cos ( π − α ) and sin α = sin ( π − α ) . Thus, − cos α + i sin α = cos ( π − α ) + i sin ( π − α ) = e i ( π − α ) . Similarly, − cos α − i sin α = cos ( π + α ) + i sin ( π + α ) = e i ( π + α ) . Therefore, z 2 = e i ( π − α ) or z 2 = e i ( π + α ) .

Solving for z To solve for z , we take the square root of both sides. For z 2 = e i ( π − α ) , we have z = ± e i ( π − α ) ​ = ± e i ( π − α ) /2 . For z 2 = e i ( π + α ) , we have z = ± e i ( π + α ) ​ = ± e i ( π + α ) /2 .

Final Solutions Thus, the four solutions for z are z = e i ( π − α ) /2 , − e i ( π − α ) /2 , e i ( π + α ) /2 , − e i ( π + α ) /2 . We can express these solutions in terms of cosine and sine using Euler's formula: z = cos ( 2 π − α ​ ) + i sin ( 2 π − α ​ ) , z = − cos ( 2 π − α ​ ) − i sin ( 2 π − α ​ ) , z = cos ( 2 π + α ​ ) + i sin ( 2 π + α ​ ) , z = − cos ( 2 π + α ​ ) − i sin ( 2 π + α ​ ) .


Examples
Understanding the roots of complex polynomials like this is crucial in fields like signal processing and quantum mechanics. For example, in signal processing, the roots of the characteristic equation of a system determine its stability. Similarly, in quantum mechanics, solving polynomial equations involving complex numbers helps determine the energy levels of a quantum system. This problem provides a foundation for analyzing such systems by finding the values of z that satisfy the given equation, which can represent physical states or system behaviors.

Answered by GinnyAnswer | 2025-07-08