Complex numbers, help

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Q13

>>8458791
>>>/hm/

>>8458791
Let's see, it specifically says we're not here to help you with your homework. If you really can't understand what step 3 wants you to do, maybe you're not ready for further math?

Here's the obvious hint: Set 1=Z^3 = (a+bj)^3 and solve, using the facts from I) and ii).

>>8458799
Yeah, i tried that and got one of the roots - 1
But I couldn't figure how to get the others, and I didn't mean to annoy anyone by asking for help but i was stuck and thought it couldn't hurt

>>8458791
london

>>8458791
also underage confirmed, fp1 was easy scrub

>>8458801
-1 isn't a root
or did you mean 1
also look up the geometric interpretation of complex numbers (and how multiplication adds up the angle) and it will become clear how to get the root of any number

>>8458826
I meant 1, i was just using the hyphen to separate it

>high school

sage


also ib math HL is the only worthwile highschool math option, pleb

>>8458791
The first one was pretty easy:
(a^2+2abj-b^2)(a+bj)
a^3+3a^2bj-3ab^2-b^3j
a^3-3ab^2+b(3a^2-b^2)j
a^3-3ab^2=1
a=1, b=0
a^3-9a^3=1
a^3=-1/8
a=-1/2, b=sqrt(3/4)

The second one I can't get either. Maybe I shouldn't be trying to use the quadratic equation?
14+10j+2a+3aj+bj=0
14+2a+(10+3a+b)j=0
a=-7
b=11
z^2+(-7-j)z+(16+11j)=0
[(7+j)+\-sqrt(-16-30j)]/2

>>8458884
I'm stupid, I need to use the other root.
(z-2-3j)(z-a-bj)=z^2-(7+j)z+(16+11j)
z^2-az-bjz-2z+2a+2bj-3jz+3aj-3b
z^2-(a+2+(b+3)j)z+(2a-3b+(2b+3a)j)
a+2=7, a=5
b+3=1, b=-2
2a-3b=16
2b+3a=11
z-5+2j=0
z=5-2j

>>8458791
Very simple if you just go to polar coords:

z = r exp(it)
z^3 = r^3 exp(3it) = 1 = r^3 (cos(3t) + i*sin(3t))

-> r=1, since max val of cos is 1, and the real part of z^3 must be one, then r must be 1 (note: r = |z| > 0 by definition)

-> cos(3t) = 1 and i*sin(3t) = 0, since imag part is 0

-> cos(3t) = 1 for 3t = 0 + k2pi*i, k in Z
-> t = 2/3 * i * k * pi

-> z = r exp(it) = exp(2pi*i*k/3), k in Z

Fundamental theorem of algebra states that any polynomial of order n has n roots -> vary k from 0 to 2 to find all three values of z

>>8458792
Refer to the fundamental theorem of algebra: any polynomial can be written as go fucking read the theorem again, specifically the part about poly being able to the written as product of roots you lazy fat stupid fuck.

>>8458884
>>8458902
Cheers mate, you are a lifesaver

>>8458791
>q8. iii)

Use Demoivre's theorem to find the cubic roots of 1.

>>8458847
>not taking analysis in high school
kys brainlet

>using [math]j[/math] not [math]i[/math] for complex numbers
feelsbadman

>didnt realize that is what j meant

feelsevenbadderman