"Find mininal polynomial of i32+12" was answered by our experts

Alan Zavala

Answered question

2022-04-12

Find mininal polynomial of

i \frac{\sqrt{3}}{2} + \frac{1}{2}

uvredio0of6

Beginner2022-04-13Added 17 answers

x = i \frac{\sqrt{3}}{2} + \frac{1}{2} \Rightarrow 2 x - 1 = i \sqrt{3} \Rightarrow x^{2} - x + 1 = 0

is the minimal polynomial with integral coefficients.
Another way to address this is to put

i \frac{\sqrt{3}}{2} + \frac{1}{2} = R (\cos y + i \in y) = R e^{i y}

where R is a positive real number.
Squaring and adding we get,

R^{2} = 1 \Rightarrow R = 1

\cos y = \frac{1}{2}

and

\sin y = \frac{\sqrt{3}}{2} \Rightarrow \tan y = \sqrt{3}

As the cosine and sine ratios of y are positive, so

y = \frac{π}{3}

So,

x = i \frac{\sqrt{3}}{2} + \frac{1}{2} = e^{\frac{i π}{3}}

To rationalize

e^{\frac{t p x}{q}}

where p,q are integers with (p,q)=1, we need to take the q-th power as

e^{i p π} = {(- 1)}^{p}

x^{3} = - 1 \Rightarrow x^{3} + 1 = 0

, but clearly,

x \neq - 1

So,

\frac{x^{3} + 1}{x + 1} = 0 \Rightarrow x^{2} - x + 1 = 0

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