Roots of complex numbers (1 Viewer)

[abdooooo!!!]

what? you're a cheat ND... that is if you know how to use it. what kind of calculator is it? what model? i need one...

 

[Grey Council]

lol, yup. So you wanna cheat too abdooooo!!!

humph, can i have your name, number, address, student ID, school?

 

[ND]

Umm it's a Casio fx-100s or something like that. I don't know if it finds roots or anything though, which would be the only useful function really.

 

[Grey Council]

ND, why have you got a 'silent' account? It'd be nice to know when your online.

 

[ND]

Cos i leave the page open when i'm doing other stuff alot, and people might think i'm purposely not replying to their PM's or posts or something...

edit: 1000 posts! ... again ...

 

[Grey Council]

oo k, fair enough.

1000 posts again? mm, i remember something about your account getting deleted, or you being unable to log on or something. Is that what the 'again' means? and its 1001, a palindromic post. Don't post EVER again, the post count looks mad.

 

[abdooooo!!!]

Originally posted by GuardiaN
lol, yup. So you wanna cheat too abdooooo!!!

humph, can i have your name, number, address, student ID, school?

my name is... nah im telling you.

there is only gonna be 1 or 2 questions involving roots of complex no. so its not that much of cheat to be using a calculator... its just good to conform your results fast.

yeah don't post ND... even if we needed you're help. LOL

 

[Xayma]

Originally posted by GuardiaN
oo k, fair enough.

1000 posts again? mm, i remember something about your account getting deleted, or you being unable to log on or something. Is that what the 'again' means? and its 1001, a palindromic post. Don't post EVER again, the post count looks mad.

What about if ND posts 110 posts really quickly for 1111 posts, or all questions for them could be posted in Non-school

 

[ND]

Oops, too late...

1000 posts again? mm, i remember something about your account getting deleted, or you being unable to log on or something. Is that what the 'again' means?

Yeh, my account was lost or something... Though Laz kindly edited my new account to look like the other one (minus about 400 posts, but i don't mind).

 

[Grey Council]

rofl

Lazarus added 400 posts to your account ND. lol, you shoulda said 'minus 3000 posts, but I don't mind', maybe he would have given you 3000 posts extra.

lol, talk about a waste of a wish. You ask 'god' to give you 300 posts, and just at that time he happens to be listening. tehehehe

 

[ND]

Hahahahah, thanks Laz! Did i say 400? I meant 4000 heheh

 

[KeypadSDM]

Originally posted by GuardiaN
x^2 - y^2 = 0
x^2 + y^2 = 1
lol, didn't think of doing x^2=y^2, the way wogboy did it. Thats quite neat, actually. Let me try it here, but my solution will be identical to Wogboy's now.
x^2 = y^2
2*x^2 = 1
x^2 = 1/2
x = 1/(sqrt 2)

Day-YUM! That is neat. very elegant. ffs, why couldn't I have thought of that.

It's also very long ...
Just take the first 2 equations and add, you skip a WHOLE LINE of working

 

[Grey Council]

One day i'm going to become rich and famous for inventing a way to stab people in the face over the net.

*takes out sword and lops of Keypad's head*

Bleh, I was so proud of that working out (even though I did nothing), and then Keypad comes along and ruins it all. gngngngn

 

[mazza_728]

Hey guys,
I actualy successfully tried to answer one of these questions:
z³+8i=0
Let z=r(cos@+isin@)=rcis@
and -8i=0-8i=8(0-i)=8(cos(3pi/2) + isin (3pi/2))
=8 cis (3pi/2)
Hence z³+8i becomes r³3@=8cis270
Therefore r³=8
r=2
cis3@=cis(3pi/2)
cos3@+isin3@=cos270+isin270
3@=270+2k pi
@=90+(2kpi/3)
Hence z=2cis(90+(2kpi/3)
k=0 z=2i
k=-1 z= (sqrt)3 -1
k=1 z=-(sqrt)3-i

Its the right answer but how was my working? did i miss anything? should i have done anything diff? is there another method? Also does it matter what order u do for the final answers, because the actual answers say k=-1, k=0, k=1 should i do it in numerical order or doesnt it matter?

 

[Constip8edSkunk]

you dont really need convert z^3 to mod arg form

z^3 = - 8i
=8cis[(4k-1)pi/2]

z = 2 cis[(4k-1)pi/6]

= 2 cis[ -pi/6 ] if k = 0
= sqrt3 - i

= 2 cis[ pi/2 ] if k = 1
= 2i

= 2 cis[ -5pi/6 ] if k = 2
= -sqrt3 - i

edit:
it doesnt matter what order u do it in or even what number, you getthe same answers with -1, 0, 1 as 0, 1, 2, or 123, 124, 125

 

[mazza_728]

I understand ur way too, i think but where did the 4k-1 come from??

I was on a roll thought i was understanding everything when i came across
z⁴=8((sqrt)3+i)
Let z=r(cos@+isin@)=rcis@
8((sqrt)3+i) -- because i cant find a common theta with this can i change it to 8(sqrt3/2+i/2)?? or is this messing it up??
can someone take a look and give me an idea, thankyou xoxo

 

[Constip8edSkunk]

-8i = 8cis[ -pi/2 + 2kpi ]
= 8cis[ (4k-1)pi ]

 

[mazza_728]

understood.. thankyou

 

[mazza_728]

can some one please attempt
z⁴=8(SQRT3+i)
Anyone???
thanks xoxo

 

[KeypadSDM]

Originally posted by mazza_728
can some one please attempt
z⁴=8(SQRT3+i)
Anyone???
thanks xoxo

|z⁴|=16
Angles are easy enough ...