Question

What quadratic equation has root `1/2+1/2i` ?

Answer 1

According to what you mean, it could be either of:

`2x^2-2x+1 = 0`

`2x^2-(2+2i)x+i = 0`

Explanation:

First note that:

`1/(1-i) = (1+i)/((1-i)(1+i)) = (1+i)/(1^2-i^2) = (1+i)/2 = 1/2+1/2i`

I am not sure if you are asking for a quadratic with real coefficients, one of whose roots is `1/2+1/2i` or for a quadratic with just one repeated root.

Let's deal with the real coefficient case first:

If a quadratic equation with real coefficients has one root `1/2+1/2i` then the other root must be the complex conjugate `1/2-1/2i` and we have:

`0 = (x-(1/2+1/2i))(x-(1/2-1/2i))`

`color(white)(0) = ((x-1/2)-1/2i)((x-1/2)+1/2i)`

`color(white)(0) = (x-1/2)^2-(1/2i)^2`

`color(white)(0) = x^2-x+1/4+1/4`

`color(white)(0) = x^2-x+1/2`

It's probably a little nicer to multiply this through by `2` to get:

`2x^2-2x+1 = 0`

Now the repeated root case:

If a quadratic equation has a repeated root `1/2+1/2i` then we have:

`0 = (x-(1/2+1/2i))^2`

`0 = x^2-2(1/2+1/2i)x+(1/2+1/2i)^2`

`0 = x^2-(1+i)x+((1/2)^2+2(1/2)(1/2i)+(1/2i)^2)`

`0 = x^2-(1+i)x+(1/4+2(1/2)(1/2i)-1/4)`

`0 = x^2-(1+i)x+1/2i`

Let's multiply this through by `2` to get:

`2x^2-(2+2i)x+i = 0`

Answer 2

`2x^2-2x+1 = 0`

Explanation:

Another approach that avoids having to do a simplification of the given root is to note that the roots of:

`ax^2+bx+c = 0`

are the reciprocals of the roots of:

`cx^2+bx+a = 0`

So let's find a quadratic with roots `(1-i)` and `(1+i)` then reverse the order of the coefficients...

`0 = (x-(1-i))(x-(1+i))`

`color(white)(0) = ((x-1)-i)((x-1)+i)`

`color(white)(0) = (x-1)^2-i^2`

`color(white)(0) = x^2-2x+1+1`

`color(white)(0) = x^2-2x+2`

So reversing the order of the coefficients, we find the quadratic equation:

`2x^2-2x+1 = 0`