#2
z = -i^(1/4)

There's one of the 4.
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#4
The answer is 16.
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#6
Use De Moivre's theorem.

I hate typing out maths and I've no paper, otherwise I'd do it. Sorry.
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#8
its impossible. u cant raise a number to an even value and get a negative answer
#12
Quote by cameronc
its impossible. u cant raise a REAL number to an even value and get a negative answer


fixed.

i'm trying to remember. i'll dig through my uni work from last year.
"And after all of this, I am amazed...

...that I am cursed far more than I am praised."
#15
I was just messing with you if you didn't get it....

when people start off saying I NEED HELP FOR MY "FRIEND".. ahh nvm
#16
Well, we did this today in class kinda. i^4=1 because i^2=-1. Make it so z= -i^1/4 I guess that is a start, we didn't get to in depth that is what I would do though and maybe you could cancel out the i and get -1 somehow?
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#17
Like Ur all $h1t said, use De Moivre's theorem to turn it into polar (I think that's what it's called, haven't done this in a while) form, ie. with sin's and cos's, and solve from there.
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#19
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Dude, I thought you said you HAD engineering mathematics?

I didn't say that. I said it was an awesome book. I always go through it and University Physics as mentioned when with this friend
#20
ok, you put it into polar by getting the argument (θ and modulus (r) and putting it into the polar form: r ( cos θ+ i sin θ.



as you can see -i has a modulus of 1 and an argument of -π/2 with the x-axis (2π being a full rotation) so the equation can be expressed as: z^4 = 1 ( cos [-π/2] + i sin [-π/2])

at this point we can just forgot the 1.

you then express z^4 as more things by adding 2π to every arguement to bring the circle back round again.

so z^4 = ( cos [-π/2] + i sin [-π/2])
=( cos [3π/2] + i sin [3π/2])
=( cos [7π/2] + i sin [7π/2])
=(cos [11π/2] + i sin [11π/2])

so then you can find z by 4th rooting r (in this case 1 so it's easy) and dividing all theta by 4.

so z = ( cos [-π/8] + i sin [-π/8])
=( cos [3π/8] + i sin [3π/8])
=( cos [7π/8] + i sin [7π/8])
=(cos [11π/8] + i sin [11π/8])

all in polar form obviously, and you can plot them on an argand diagram quite happily and convert them back to normal form using: x = r cos θ, y = r sin θ but the numbers aren't very nice.
"And after all of this, I am amazed...

...that I am cursed far more than I am praised."
#21
Looks right to me.

But what the hell do I know? I'm only a devastatingly handsome stripper.




EDIT: Do we have to censor nips on dewds as well?
#22
Quote by LordBishek
Looks right to me.

But what the hell do I know? I'm only a devastatingly handsome stripper.




EDIT: Do we have to censor nips on dewds as well?
Hellloooo! Can I borrow him while I answer this question?

Edit: looks like Sol answered you pretty well. I'll just take the man-goodness
#23
Quote by Sol9989
ok, you put it into polar by getting the argument (θ and modulus (r) and putting it into the polar form: r ( cos θ+ i sin θ.



as you can see -i has a modulus of 1 and an argument of -π/2 with the x-axis (2π being a full rotation) so the equation can be expressed as: z^4 = 1 ( cos [-π/2] + i sin [-π/2])

at this point we can just forgot the 1.

you then express z^4 as more things by adding 2π to every arguement to bring the circle back round again.

so z^4 = ( cos [-π/2] + i sin [-π/2])
=( cos [3π/2] + i sin [3π/2])
=( cos [7π/2] + i sin [7π/2])
=(cos [11π/2] + i sin [11π/2])

so then you can find z by 4th rooting r (in this case 1 so it's easy) and dividing all theta by 4.

so z = ( cos [-π/8] + i sin [-π/8])
=( cos [3π/8] + i sin [3π/8])
=( cos [7π/8] + i sin [7π/8])
=(cos [11π/8] + i sin [11π/8])

all in polar form obviously, and you can plot them on an argand diagram quite happily and convert them back to normal form using: x = r cos θ, y = r sin θ but the numbers aren't very nice.

This looks like it...

And yes, isn't that the DeMoivre's theorm.

Damn its been like 3 years since i've done high school maths. But i do remember it uses a graph and equations like the ones towards the end.
#24
Quote by smb
Hellloooo! Can I borrow him while I answer this question?

Edit: looks like Sol answered you pretty well. I'll just take the man-goodness


All yours big boy.
#25
damn it, my maths drowned out by male porn.

it MUST be a nightmare.
"And after all of this, I am amazed...

...that I am cursed far more than I am praised."
#26
Quote by Sol9989
damn it, my maths drowned out by male porn.

it MUST be a nightmare.


Must be quite the anticlimax.

You're my hero
#27
Egads. I remember doing that five years ago. It wasn't fun.
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