- #1

- 1,462

- 44

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- I
- Thread starter Mr Davis 97
- Start date

- #1

- 1,462

- 44

- #2

- 3,987

- 1,538

Expressed in formal logic, we have

$$x^3=1 \Leftrightarrow (x=1\vee x=\omega\vee x=\omega^2)$$

Given the properties of ##\vee## (OR), this allows us to deduce that

$$x=1\Rightarrow x^3=1$$

but not that

$$x^3=1 \Rightarrow x=1$$

- #3

- 1,462

- 44

I think I see. So with square roots, ##+ \sqrt[2]{x}## is the principal square root which defines a function, while ##\pm \sqrt[2]{x}## is the value of both roots. So analogously ##\sqrt[3]{x}## is the principal cube root, while... What's the analogous way of retrieving the two complex roots easily?threeseparate equations, not one: ##x=1,\ x=\omega## and ##x=\omega^2##, where ##\omega=e^{2\pi/3}##.

Expressed in formal logic, we have

$$x^3=1 \Leftrightarrow (x=1\vee x=\omega\vee x=\omega^2)$$

Given the properties of ##\vee## (OR), this allows us to deduce that

$$x=1\Rightarrow x^3=1$$

but not that

$$x^3=1 \Rightarrow x=1$$

- #4

lurflurf

Homework Helper

- 2,445

- 140

##exp(2k \pi i /n)\sqrt[n]{y}##

for

k=0,1,...,n-2,n-1

thus the complex cube roots of a positive real y are

##exp(2 \pi i /n)\sqrt[n]{y}=\left(-\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)\sqrt[n]{y}##

and

##exp(4 \pi i /n)\sqrt[n]{y}=\left(-\frac{1}{2}-\frac{i \sqrt{3}}{2}\right)\sqrt[n]{y}##

- #5

- 16,451

- 15,516

Polynomials split over the complex numbers. So we have to look at complex multiplication. If we represent complex numbers as points in the plane, then multiplication means to multiply the real length of two numbers (= distance to the origin), and add there angles (= towards the real axis). This means in return, that numbers of length one are all on the unit circle, and multiplication of those is adding angles, i.e. in order to solve ##x^n=x \cdot \ldots \cdot x = 1##, we have to divide the unit circle in ##n## equal angles and all possible ##x## are those points on the circle which @lurflurf listed.What's the analogous way of retrieving the two complex roots easily?

Share: