# Show that all roots of polynomial (x+i)^10+((x-i)^10 are real.

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You need to make the assumption that `alpha` is a root of polynomial and `alpha in C` , hence `alpha = x+i*y` .

Since `alpha` is a root of polynomial, hence `f(alpha) = 0` such that:

`{(f(alpha) = (alpha+i)^10 + (alpha-i)^10),(f(alpha)=0):}` `=gt(alpha+i)^10 + (alpha-i)^10 = 0 =gt (alpha+i)^10=- (alpha-i)^10=gt |alpha + i| = |alpha - i|`

You need to substitute `x+iy` for `alpha` such that:

`|x+iy+i| = |x+iy-i| => |x+i(y+1)| = |x+i(y-1)|`

You need to evaluate the absolute values above such that:

`sqrt(x^2 + (y+1)^2) = sqrt(x^2 + (y-1)^2)`

Raising to square both sides yields:

`x^2 + (y+1)^2 = x^2 + (y-1)^2`

Expanding the squares yields:

`x^2 + y^2 + 2y + 1 = x^2 + y^2 - 2y + 1`

Reducing like terms yields:

`2y = -2y => 4y = 0 => y = 0`

Notice that y represents the imaginary part of the complex number `alpha = x + i*y` .

**Since the imaginary part of the root alpha is zero, hence all the roots of the polynomial are real numbers: `alpha = x` .**

**Sources:**

yes all roots will be real of the equation.