Question

Polynomial equation with real coefficients that has roots i,\ 1+ii,\ 1+i

Polynomial arithmetic
Polynomial equation with real coefficients that has roots $$\displaystyle{i},\ {1}+{i}{i},\ {1}+{i}$$

2021-08-03
Step 1
Formula Used:
$$\displaystyle{\left({a}+{b}\right)}{\left({a}-{b}\right)}={a}^{{{2}}}-{b}^{{{2}}}{\left({a}+{b}\right)}{\left({a}-{b}\right)}={a}{2}-{b}{2}$$
Calculation:
If the polynomial has real coefficients, then it's imaginary roots occur in conjugate pairs. So, a polynomial with the given root $$\displaystyle{i},\ {1}+{i}{i},\ {1}+{1}{i}$$ must have other roots as $$\displaystyle-{i},\ {1}-{i}-{i},\ {1}-{i}.$$
Since each root of the equation corresponds to a factor of the polynomial also the roots indicate zeros of that polynomial. Hence, below mentioned equation is written.
$$\displaystyle{\left({x}-{i}\right)}{\left[{x}-{\left(-{1}\right)}\right]}{\left[{x}-{\left({1}+{i}\right)}\right]}{\left[{x}-{\left({1}-{i}\right)}\right]}={0}$$
$$\displaystyle{\left({x}-{i}\right)}{\left({x}+{i}\right)}{\left({x}-{1}-{i}\right)}{\left({x}-{1}+{i}\right)}={0}$$
$$\displaystyle{\left({x}-{i}\right)}{\left[{x}-{\left(-{i}\right)}\right]}{\left[{x}-{\left({1}+{i}\right)}\right]}{\left[{x}-{\left({1}-{i}\right)}\right]}={0}{\left({x}-{i}\right)}{\left({x}+{i}\right)}{\left({x}-{1}-{i}\right)}{\left({x}-{1}+{i}\right)}={0}{\left({x}-{i}\right)}{\left({x}+{i}\right)}{\left[{\left({x}-{1}\right)}-{i}\right]}{\left[{\left({x}-{1}\right)}+{i}\right]}={0}$$
$$\displaystyle{\left({x}-{i}\right)}{\left({x}+{i}\right)}{\left[{\left({x}-{1}\right)}-{i}\right]}{\left[{\left({x}-{1}\right)}+{i}\right]}={0}$$
Further use arithmetic rule.
$$\displaystyle{\left({a}+{b}\right)}{\left({a}-{b}\right)}={a}^{{{2}}}-{b}^{{{2}}}{\left({a}+{b}\right)}{\left({a}-{b}\right)}={a}{2}-{b}{2}$$
Now, the polynomial equation is,
$$\displaystyle{\left({x}^{{{2}}}-{\left({i}\right)}^{{{2}}}\right]}{\left[{\left({x}-{1}\right)}^{{{2}}}-{\left({i}\right)}^{{{2}}}\right]}={0}{\left({x}{2}-{\left({i}\right)}{2}\right)}{\left[{\left({x}-{1}\right)}{2}-{\left({i}\right)}{2}\right]}={0}$$
Use arithmetic rule.
$$\displaystyle{\left({a}-{b}\right)}^{{{2}}}={a}^{{{2}}}-{2}{a}{b}+{b}^{{{2}}}{\left({a}-{b}\right)}{2}={a}{2}-{2}{a}{b}+{b}{2}$$
And $$\displaystyle{i}^{{{2}}}=-{1}{i}{2}=-{1}.$$
Now,
$$\displaystyle{\left({x}^{{{2}}}+{1}\right)}{\left({x}^{{{2}}}-{2}{x}+{1}+{1}\right)}={0}$$
$$\displaystyle{\left({x}^{{{2}}}+{1}\right)}{\left({x}^{{{2}}}-{2}{x}+{2}\right)}={0}$$
$$\displaystyle{\left({x}{2}+{1}\right)}{\left({x}{2}-{2}{x}+{1}+{1}\right)}={0}{\left({x}{2}+{1}\right)}{\left({x}{2}-{2}{x}+{2}\right)}={0}{\left({x}{4}+{x}{2}-{2}{x}{3}-{2}{x}+{2}{x}{2}\right)}={0}{x}{4}-{2}{x}{3}-{2}{x}+{2}={0}$$
$$\displaystyle{\left({x}^{{{4}}}+{x}^{{{2}}}-{2}{x}^{{{3}}}-{2}{x}+{2}{x}^{{{2}}}\right)}={0}$$
$$\displaystyle{x}^{{{4}}}-{2}{x}^{{{3}}}+{3}{x}^{{{2}}}-{2}{x}+{2}={0}$$
Hence, the polynomial expression of given roots Hence, the polynomial equation of given roots
$$\displaystyle{i},\ {1}+{i}{i},\ {1}+{i}$$ is $$\displaystyle{x}^{{{4}}}-{2}{x}^{{{3}}}+{3}{x}^{{{2}}}{2}{x}+{2}={0}{x}{4}-{2}{x}{3}+{3}{x}{2}-{2}{x}+{2}={0}$$