need help with equations

[hollow sapphire]

How did they get complex numbers?? I don't understand how those trigs suddenly appeared. Can someone explain?

[fading seal]

1. Wait patiently for a helper to come along.
2. Once someone helps you, say thank you and close the thread with:

+close

3. Feel free to nominate the person for helper of the week in #helper-nominations
4. Do not ping the mods, unless someone is breaking the rules.
5. If you're happy with the help you got here, and the server overall, you can contribute financially as well:

[hollow sapphire]

need help with equations

[rapid thunder]

oh this

ok

basically

to find the complex roots of a number

all complex roots of $x^{\frac{1}{n}}$ where n is natural are $|Re(x^{\frac{1}{n}})|\times e^{\frac{2ik\pi}{n}}$ for all $k \in {m: m \in \mathbb{N}, m \leq n}$

[frosty edge]

Inverse Cupid

[rapid thunder]

Re denotes the real root of x^(1/n)

and $e^{ix} = \cos{x} + i\sin{x}$

[frosty edge]

Inverse Cupid

[rapid thunder]

@hollow sapphire

[hollow sapphire]

Where can i learn this?

[rapid thunder]

uh, is someth not clear here?

I can tell you

[hollow sapphire]

Uhm yeah, I never knew about this stuff...

[rapid thunder]

what part do you not understand

I can write the thing in a little less techincal way

[hollow sapphire]

Sure, I just don't get how those trigs suddenly appeared.

[rapid thunder]

oh just that

[hollow sapphire]

And i'm confused why they put 0, pi/2, etc in theta

[rapid thunder]

use this

[rapid thunder]

try it out with the formula that I gave

root of whatever's in the question

[rapid thunder]

this one

it returns all the roots

[hollow sapphire]

Solve for x?

I just don't get it💀

[rapid thunder]

oof

alr

what the formula basically says is

suppose you want to find the complex roots of $1^{\frac{1}{4}}$

[frosty edge]

Inverse Cupid

[rapid thunder]

now, the positive real root of that is simply 1

so

all complex roots of $x^{\frac{1}{n}}$ where n is natural are $|Re(x^{\frac{1}{n}})|\times e^{\frac{2ik\pi}{n}}$ for all $k \in {m: m \in \mathbb{N}, m \leq n}$

[frosty edge]

Inverse Cupid

[rapid thunder]

so, the formula becomes

$1 \times e^{\frac{2ik\pi}{n}}$ for all $k \in {m: m \in \mathbb{N}, m \leq n}$

[frosty edge]

Inverse Cupid

[rapid thunder]

now, let's define the set ${m: m \in \mathbb{N}, m \leq n}$

[frosty edge]

Inverse Cupid

[rapid thunder]

this set becomes ${1,2,3,4}$

[frosty edge]

Inverse Cupid

[rapid thunder]

now, the formula is

$1 \times e^{\frac{2ik\pi}{4}}$ for all $k \in {1,2,3,4}$

[frosty edge]

Inverse Cupid

[rapid thunder]

now, let's plug in all the values of k and see what we get

$e^{\frac{i\pi}{2}}, k=1$

$e^{i\pi}, k=2$

$e^{\frac{i3\pi}{2}}, k=3$

$e^{2i\pi}, k=4$

[frosty edge]

Inverse Cupid

[rapid thunder]

now let's use this handy formula

$\cos{(\frac{\pi}{2})} + i\sin{(\frac{\pi}{2})}, k=1$

$\cos{(\pi)} + i\sin{(\pi)}, k=2$

$\cos{(\frac{3\pi}{2})} + i\sin{(\frac{3\pi}{2})}, k=3$

$\cos{(2pi)} + i\sin{(2\pi)}, k=4$

[frosty edge]

Inverse Cupid

[rapid thunder]

you get all the possible values of $1^{\frac{1}{4}}$ as

$i, -1, -i, 1$

[frosty edge]

Inverse Cupid

[rapid thunder]

the complex roots are $i, -i$

[frosty edge]

Inverse Cupid

[rapid thunder]

@hollow sapphire read from here

[hollow sapphire]

Is this in calculus or something? What topic is this?

[rapid thunder]

er it's just actually euler's identity

e^ix = cosx + isinx

it seems a bit complex, no doubt