Show that the polynomial `f (x )=(x+i)^100+(x-i)^100` has only real zeros. Use modulus
First, suppose that `f(z)=0` for some complex number `z=a+bi.` This implies that
`(z+i)^100=-(z-i)^100,` and if we take the absolute value of each side, we see that
`|z+i|^100=|z-i|^100.` Taking the 100th root of each side, this means that either `|z+i|=|z-i|,` or `|z+i|=-|z-i|.` In the second case, the left side is always non-negative, so the right side must be zero, implying `z=i.` However, `|i+i|!=0,` so the second case is impossible, and thus we have `|z+i|=|z-i|.` Replacing `z` with `a+bi,` we get
`|a+(b+1)i|=|a+(b-1)i|,` which implies `a^2+(b+1)^2=a^2+(b-1)^2,` or after cancelling the `a^2` terms,
`(b+1)^2=(b-1)^2.` Since the squares are the same, we again have two possible cases `b+1=b-1` and `b+1=-(b-1).` The first case is obviously impossible, and in the second case we solve to get `b=0.`
Since `b=0,` we've proved that `z=a+0i=a\in RR,` so all zeros have to be real.