Complex numbers

[Matt No.1265235]

Hi /sci/
any anons with good understanding of complex numbers able to tell me the other 2 solutions to 2^3
(i know the basic answer = 8)
pic unrelated

[Anonymous No.1265244]

>that pic
Oh lord, do you think it hurts for it to live?

[Anonymous No.1265252]

>>1265244
Probably not. Plants are actually kind of cool that way. They'll grow together and just be all "sup".

[Anonymous No.1265255]

Ask the solution to a quadratic equation which doesn't have a real solution if you want an anon to give yo complex numbers. Your question shows you don't understand what a complex number is.

[Matt No.1265256]

very much so, it suffers, yet if there is a god, he does nothing

[Matt No.1265266]

sorry, i meant

z^3 = 8

so simple solution is z=2

[Anonymous No.1265281]

<span class="math">2e^{i\pi/3}, 2e^{i2\pi/3}[/spoiler]

[Anonymous No.1265317]

There are no other solutions

[Anonymous No.1265321]

Demoivre's formula.. use it..

[Anonymous No.1265329]

>>1265266
z = 2, -2, -2

[Anonymous No.1265350]

>>1265255

You know damn well what he means.

[Anonymous No.1265348]

http://www.wolframalpha.com/input/?i=z^3+%3D+8+

[Anonymous No.1265346]

Draw three lines of length two equally spaced around 360 degrees, like the Mercedes symbol. Set one line on the real axis.

There are your solutions.

[Anonymous No.1265622]

once you know one solution you can easily find the others because of symmetry:
3 roots means they must lie at 120 degrees from each other.

you know that z^3 =8 has the solution z=2, so then you must add 120 degrees and 240 degrees to get your answer.

multiplication by i rotates a vector counterclockwise by 90degrees, so to rotate it by 120degrees we multiply by 4/3i .
so (2+0i)*4/3i = -4/3 + 8/3i
and the other root is
(-4/3+8/3i)*4/3i = -32/9 -16/9i
so the 3 roots are 2, -4/3 + 8/3i, -32/9 -16/9i

[Anonymous No.1265625]

Solve <span class="math"> 2^3 [/spoiler]

Answer: 8

There are no non trivial answers

/thread

[Anonymous No.1265930]

>>1265625
this

[Matt No.1265978]

it's ok, i have just finished my lesson, the answer was -1 +/- 3i^(1/2) for those interested

[Anonymous No.1265990]

leaving the answer in i^(1/2) makes me vomit

[Anonymous No.1266027]

>>1265622

Astigmatic?
Try holding down shift or ctrrl (one of those) and it'll keep your circle circular.

[Anonymous No.1266259]

>>1266027
try shutting your mouth and leave the math to the big kids

[Anonymous No.1266302]

8e^3ix=8(cos 3x +isin 3x)

x= 0, 2/3pi, -2pi/3

z=2e^ix

I think, it's been a while.

[Anonymous No.1266309]

Working:

z^3=8 <=> z^3-8=0

z-2 is a factor so

(z-2)/(z^3-8) = z^2+2z+4
by long division

so z={-2+/-(-12^1/2)} /2 when you put it into the quadratic formula

{-2 +/- (12^1/2)(-1^1/2)}/2
{-2 +/- (4^1/2)(3^1/2)(-1^1/2)}/2
i=(-1^1/2)
hence z= -1+/- (3^1/2)(-1^1/2)

the answers are conjugates ;)

that is all

[Anonymous No.1266322]

>>1265346
a winrar is you

[Anonymous No.1266341]

>>1265346
>>1266322
a samefag is you