At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

\[\sqrt{-1} = x \] what is the value of x

please dont put links

so what is an imaginary unit

I know of a video you can watch if you want to

yes i did but what is an imaganary unit

The first link is really straight forward, try that

You are welcome :)

listen and can u give a simple i word answer: \[\sqrt{-1}=x\] what is the value of x in numbers.

1 word answer*

not i word

whats w

We call this omega,It's derivations is given in my link :)

Now as per your question.
x=root{-1}
x=i
Here i is actually iota and it is an imaginary number.

whats omega

i said in numbers. your answer is basicly my question

Well,In numbers it is sqrt{-1}

it should be easy shouldent it.

i dont understand ur link

what is omega

I have already told you what omega is.

I said it is -1+i sqrt{3}/2

ok (i dont get it but ok) i' only in year 9 doing my gcse's

you just need to know what a square root is

i know what a square root is ofc

that's why there's no sqrt of ANY negative numbers, because everthing squared is BIGGER than 0

(possitive)

oh i get it so its an imaganary number because it cant exist?

you have the best responce

EXACTLY, it CAN'T EXIST as a "normal number"

thank u so much.

they INVENTED it

you are welcome

not really, but thanks!

ill draw you something

|dw:1358026644045:dw|

hahaha you amuse me, nothing is hard when you have an amazing teacher, and I have it right now.

you understand what I drew?

yeah u drew a number line. who is ur teacher

|dw:1358026974868:dw|

ALMOST

|dw:1358027044754:dw|

|dw:1358027175723:dw|

|dw:1358027034198:dw|

you put the axis pararell and it's perpendicular cutting in 0

|dw:1358027255837:dw|

yepeee

because they CAN be negative in their new axis, the imaginary axis

the imaginary line

it is not allowed to be a negative square root in the REAL line, but it is in the imaginary one

so in imaginative can they ever be 3i or always do they have to be -3i

so we just call it i, like we call 1 one

there are imaginary postive numbers and imaginary negative numbers, under 0

real negatives are at the left of 0, so why can't there be negative imaginary ones?

|dw:1358027454145:dw|

above 0 is positive?

if you delete the minus inside of the root, it would be a real number then!

Yes that's what i was writting

and under 0 negative, i forgot to draw it

just like the real line but perpendicular

above 0 goes posstive until inifinity and under 0 negative until infinity

and on the real line is it same?

|dw:1358027632303:dw|

they are two separated lines.

bye i need to go. if i see u tommorw here i will carry on the conversation.

okay, sure! bye! you are smart btw, you figured the other line very well.

figured out*

thanks

complex numbers

i is actually an imaginary number used to give value to \[\sqrt{-1}\]

yes i get that.

Look above^

Wait, no!

what?

\(i\) is defined as the imaginary number such that \(i^2 = -1\), not \(i=
\sqrt{-1}\).

no ur wrong

wait

How so...?

watch

wait

http://imgur.com/kwevvqi

and also its just common knowlige

Knowledge*

Yes, \(i\) is one of the square roots of \(-1\). But \(i \ne \sqrt{-1}\).

I mean, \(\sqrt{-1} =i,-i\)

So you must define \(i^2 = -1\) if you don't want things to mess up

wow gcr92 was WAY better at explaining things

What would you like to learn?

apparantly his teacher is hgis uncle who started teaching him maths when he was 15 an now is 50

wait

carry on from:

|dw:1358591619083:dw|

look above^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

Re and Co, yes.

ermmm it needs to be simple

nope its not

sorry

where is GCR92

Here. What's up?