Prove Coefficients of Complex Polynomials are Real

Click For Summary
To prove that the coefficients of a complex polynomial P(z) are real if and only if P(x) is real for all real x, one can utilize the properties of complex conjugates. By expressing the polynomial as a Taylor series and comparing P(x) with its conjugate P*(x) when x is real, it becomes evident that both must equal each other. This leads to the conclusion that the differences between the coefficients and their conjugates must be zero for all powers of x. Therefore, each coefficient must be equal to its conjugate, confirming that they are real. The argument effectively demonstrates the relationship between the reality of polynomial coefficients and the polynomial's behavior over real inputs.
de1irious
Messages
20
Reaction score
0
I need to show that the coefficients of a complex polynomial P(z) are real iff P(x) is real for all real x. Thanks!
 
Physics news on Phys.org
Try a Taylor series about 0.
 
try to look at P(x) and p(x)* where x is real (what do you know about z and z* if they are real). And write

a + b x + c x^2 + ... d x^n

and

(a + b x + c x^2 + ... d x^n)*

and compare them exploiting that x and p(x) is real. If this isn't enough i have made a spoiler in the bottom of the post, just follow the dots.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.




a + b x + c x^2 + ... d x^n = real = real* = (a + b x + c x^2 + ... + d x^n)* = a* + b* x* + c* x*^2 + ... + d* x*^n = a* + b* x + c* x^2 + ... + d* x^n

so

0 = (a-a*) + (b-b*)x + (c-c*)x^2 + ... + (d-d*)x^n for all x

so

a=a*, b=b*, c=c* ... d=d*
 
Last edited:

Similar threads

  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 10 ·
Replies
10
Views
3K
Replies
6
Views
1K
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
48
Views
4K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K