Prove that if p is a polynomial with real coefficients, and if is a (complex) solution of P(E)z = 0, then the conjugate of z, the real part of z, and the imaginary part of z are also solutions.
If z is a complex number, and z its conjugate, first you need to prove that conjugation distributes over multiplication and addition:
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(z1 * z2) = z1 * z2
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(z1 + z2) = (z1) + (z2)
If z= x+iy, then the complex number x-iy is called the conjugate of the complext number z and is
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written as z . It is easily seen that numbers conjugate to z1+z2 and z1z2 are (z1) + (z2) and
z1 z2 respectively.
Let P(x) = a x^n + a x^n-1 + a x^n-2 +...+ a x + a
Polynomials, complex solutions, conjugates and shift operators are investigated.