# How do you find a polynomial with complex coefficients of the smallest possible degree for which i and 1+i are zeros and in which the coefficient of the highest power is 1?

Oct 25, 2015

The polynomial is ${x}^{2} - x \left(2 i + 1\right) - 1 + i$.

#### Explanation:

Since using complex numbers a polynomial of degree $n$ has always exactly $n$ solutions ${x}_{1} , \ldots , {x}_{n}$, and can be written as $\left(x - {x}_{1}\right) \left(x - {x}_{2}\right) \ldots \left(x - {x}_{n}\right)$, if we want two numbers to be solution, the smallest possible degree is two, and the polynomial can be written as

$\left(x - i\right) \left(x - \left(i + 1\right)\right) = \left(x - i\right) \left(x - i - 1\right)$

We can expand it into ${x}^{2} - i x - x - i x + {i}^{2} + i$, and since ${i}^{2} = - 1$, it becomes

${x}^{2} - x \left(2 i + 1\right) - 1 + i$.

I haven't fully understood the second question, but if you want the polynomial to have two solution, you can't go for a polynomial of degree one, because it will have only one solution.