#1

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

z^4 = -i

Find all possible values for z.

Cheers in advance!

z^4 = -i

Find all possible values for z.

Cheers in advance!

#2

z = -i^(1/4)

There's one of the 4.

There's one of the 4.

#3

Doesn't summoning smb require some kind of sacrifice?

#4

The answer is 16.

#5

Doesn't summoning smb require some kind of sacrifice?

Meh.

#6

Use De Moivre's theorem.

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

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

#7

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.

#8

its impossible. u cant raise a number to an even value and get a negative answer

#9

Use De Moivre's theorem.

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

What's that?

#10

sureeeeeeeee its for your friend

#11

sureeeeeeeee its for your friend

I'm doing A2 Maths this year which doesn't even cover complex numbers

#12

its impossible. u cant raise aREALnumber 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.

#13

fixed.

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

Cheers, great help

#14

Keeping this thread alive.

#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

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?

#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.

#18

Dude, I thought you said you HAD engineering mathematics?

#19

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.

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.

#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?

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

Hellloooo! Can I borrow him while I answer this question?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?

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

#23

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

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.

it MUST be a nightmare.

#26

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.