# Maths Help; Complex Numbers (smb and others...)

Alright, this question isn't actually for me, it's for a friend.

z^4 = -i

Find all possible values for z.

z = -i^(1/4)

There's one of the 4.
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wat

The answer is 16.
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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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Alright, she doesn't know how to do it completely, so any information involving to work them out step by step explaining why would be awesome.
its impossible. u cant raise a number to an even value and get a negative answer
Quote by Ur all \$h1t
Use De Moivre's theorem.

I hate typing out maths and I've no paper, otherwise I'd do it. Sorry.

What's that?
sureeeeeeeee its for your friend
Quote by RileyB
sureeeeeeeee its for your friend

I'm doing A2 Maths this year which doesn't even cover complex numbers
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."
Quote by Sol9989
fixed.

i'm trying to remember. i'll dig through my uni work from last year.

Cheers, great help
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
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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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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Dude, I thought you said you HAD engineering mathematics?
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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
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."
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?
Oh my love, though our bodies may be parted,
Though our skin may not touch skin
Look for me with the sun-bright swallow,
I will come on the breath of the wind

I'M UNCONDITIONALLY HERS
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
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.
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.
Oh my love, though our bodies may be parted,
Though our skin may not touch skin
Look for me with the sun-bright swallow,
I will come on the breath of the wind

I'M UNCONDITIONALLY HERS
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."
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
Oh my love, though our bodies may be parted,
Though our skin may not touch skin
Look for me with the sun-bright swallow,
I will come on the breath of the wind

I'M UNCONDITIONALLY HERS
Egads. I remember doing that five years ago. It wasn't fun.
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