# Show that the polynomial f (x )=(x+i)^100+(x-i)^100 has only real zeros.  Use modulus

Asked on by danedraft

### 1 Answer |Add Yours

degeneratecircle | High School Teacher | (Level 2) Associate Educator

Posted on

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.

We’ve answered 318,989 questions. We can answer yours, too.