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awn786
 2 years ago
What is the square root of minus 1.
When i do it in a scientific calculator, it says 'i' (imaginary unit) and i have no idea what that means
awn786
 2 years ago
What is the square root of minus 1. When i do it in a scientific calculator, it says 'i' (imaginary unit) and i have no idea what that means

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awn786
 2 years ago
Best ResponseYou've already chosen the best response.1\[\sqrt{1} = x \] what is the value of x

zaynahf
 2 years ago
Best ResponseYou've already chosen the best response.0Imaginary numbers result when you are taking the square root of any negative number: For an example, \[\sqrt{1}\] = i \[\sqrt{25} = 5i\] The 'i' is kind of like the negative sign

awn786
 2 years ago
Best ResponseYou've already chosen the best response.1so what is an imaginary unit

zaynahf
 2 years ago
Best ResponseYou've already chosen the best response.0I know of a video you can watch if you want to

awn786
 2 years ago
Best ResponseYou've already chosen the best response.1yes what is the vid @zaynahf

hba
 2 years ago
Best ResponseYou've already chosen the best response.0did you see the link @awn786 It has everything.

awn786
 2 years ago
Best ResponseYou've already chosen the best response.1yes i did but what is an imaganary unit

zaynahf
 2 years ago
Best ResponseYou've already chosen the best response.01) http://www.youtube.com/watch?v=Iv_uzy4_gec 2)https://www.khanacademy.org/math/algebra/complexnumbers/complex_numbers/v/algebraiiimaginaryandcomplexnumbers?v=C2Ln0pK3kY

awn786
 2 years ago
Best ResponseYou've already chosen the best response.1WOW OMG complicated @hba

zaynahf
 2 years ago
Best ResponseYou've already chosen the best response.0The first link is really straight forward, try that

hba
 2 years ago
Best ResponseYou've already chosen the best response.0@awn786 Well,I tried to simplify and put everything :)

awn786
 2 years ago
Best ResponseYou've already chosen the best response.1listen and can u give a simple i word answer: \[\sqrt{1}=x\] what is the value of x in numbers.

hba
 2 years ago
Best ResponseYou've already chosen the best response.0I can help you understand what imaginary numbers are. So,the base of imaginary numbers is actually laid upon some of the things like \[\huge \ i=\sqrt{1}\] The other thing is \[\huge\ \omega=\frac{ 1+i \sqrt{3} }{ 2 }\]

hba
 2 years ago
Best ResponseYou've already chosen the best response.0We call this omega,It's derivations is given in my link :)

hba
 2 years ago
Best ResponseYou've already chosen the best response.0Now as per your question. x=root{1} x=i Here i is actually iota and it is an imaginary number.

hba
 2 years ago
Best ResponseYou've already chosen the best response.0@awn786 Please check the link i have provided. It has everything Please give it a read.

awn786
 2 years ago
Best ResponseYou've already chosen the best response.1i said in numbers. your answer is basicly my question

hba
 2 years ago
Best ResponseYou've already chosen the best response.0Well,In numbers it is sqrt{1}

awn786
 2 years ago
Best ResponseYou've already chosen the best response.1it should be easy shouldent it.

awn786
 2 years ago
Best ResponseYou've already chosen the best response.1i dont understand ur link

hba
 2 years ago
Best ResponseYou've already chosen the best response.0I have already told you what omega is.

hba
 2 years ago
Best ResponseYou've already chosen the best response.0I said it is 1+i sqrt{3}/2

awn786
 2 years ago
Best ResponseYou've already chosen the best response.1ok (i dont get it but ok) i' only in year 9 doing my gcse's

GCR92
 2 years ago
Best ResponseYou've already chosen the best response.2\[\sqrt{1}\] doesn't exist as a the REAL number you are asking for, because there's no number that you can square (because of the SQUARE root) and gives you 1

GCR92
 2 years ago
Best ResponseYou've already chosen the best response.2you just need to know what a square root is

awn786
 2 years ago
Best ResponseYou've already chosen the best response.1i know what a square root is ofc

GCR92
 2 years ago
Best ResponseYou've already chosen the best response.2that's why there's no sqrt of ANY negative numbers, because everthing squared is BIGGER than 0

awn786
 2 years ago
Best ResponseYou've already chosen the best response.1oh i get it so its an imaganary number because it cant exist?

awn786
 2 years ago
Best ResponseYou've already chosen the best response.1you have the best responce

GCR92
 2 years ago
Best ResponseYou've already chosen the best response.2EXACTLY, it CAN'T EXIST as a "normal number"

GCR92
 2 years ago
Best ResponseYou've already chosen the best response.2so mathematicians around XVII century had to figure out another type of numbers, called COMPLEX numbers, and the first number of complex numbers is \[\sqrt{1}\]

awn786
 2 years ago
Best ResponseYou've already chosen the best response.1you can bring the most complex thing and make others understand. are u sure ur not related to prfsr. brian cox

GCR92
 2 years ago
Best ResponseYou've already chosen the best response.2hahaha you amuse me, nothing is hard when you have an amazing teacher, and I have it right now.

GCR92
 2 years ago
Best ResponseYou've already chosen the best response.2you understand what I drew?

awn786
 2 years ago
Best ResponseYou've already chosen the best response.1yeah u drew a number line. who is ur teacher

GCR92
 2 years ago
Best ResponseYou've already chosen the best response.2My uncle, who's been teaching maths since he was 15 and he's now 57, anyway, let's go back to our imagination

GCR92
 2 years ago
Best ResponseYou've already chosen the best response.2you put the axis pararell and it's perpendicular cutting in 0

GCR92
 2 years ago
Best ResponseYou've already chosen the best response.2that's why i said almost, because you figured out ANOTHER line, and that would make a 300 year old mathematician proud

GCR92
 2 years ago
Best ResponseYou've already chosen the best response.2because they CAN be negative in their new axis, the imaginary axis

GCR92
 2 years ago
Best ResponseYou've already chosen the best response.2it is not allowed to be a negative square root in the REAL line, but it is in the imaginary one

awn786
 2 years ago
Best ResponseYou've already chosen the best response.1so in imaginative can they ever be 3i or always do they have to be 3i

GCR92
 2 years ago
Best ResponseYou've already chosen the best response.2so we just call it i, like we call 1 one

GCR92
 2 years ago
Best ResponseYou've already chosen the best response.2there are imaginary postive numbers and imaginary negative numbers, under 0

GCR92
 2 years ago
Best ResponseYou've already chosen the best response.2real negatives are at the left of 0, so why can't there be negative imaginary ones?

GCR92
 2 years ago
Best ResponseYou've already chosen the best response.2if you delete the minus inside of the root, it would be a real number then!

GCR92
 2 years ago
Best ResponseYou've already chosen the best response.2Yes that's what i was writting

GCR92
 2 years ago
Best ResponseYou've already chosen the best response.2and under 0 negative, i forgot to draw it

GCR92
 2 years ago
Best ResponseYou've already chosen the best response.2just like the real line but perpendicular

GCR92
 2 years ago
Best ResponseYou've already chosen the best response.2above 0 goes posstive until inifinity and under 0 negative until infinity

awn786
 2 years ago
Best ResponseYou've already chosen the best response.1and on the real line is it same?

GCR92
 2 years ago
Best ResponseYou've already chosen the best response.2they are two separated lines.

awn786
 2 years ago
Best ResponseYou've already chosen the best response.1bye i need to go. if i see u tommorw here i will carry on the conversation.

GCR92
 2 years ago
Best ResponseYou've already chosen the best response.2okay, sure! bye! you are smart btw, you figured the other line very well.

AravindG
 2 years ago
Best ResponseYou've already chosen the best response.0i is actually an imaginary number used to give value to \[\sqrt{1}\]

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0\(i\) is defined as the imaginary number such that \(i^2 = 1\), not \(i= \sqrt{1}\).

awn786
 2 years ago
Best ResponseYou've already chosen the best response.1and also its just common knowlige

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0Yes, \(i\) is one of the square roots of \(1\). But \(i \ne \sqrt{1}\).

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0I mean, \(\sqrt{1} =i,i\)

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0So you must define \(i^2 = 1\) if you don't want things to mess up

awn786
 2 years ago
Best ResponseYou've already chosen the best response.1wow gcr92 was WAY better at explaining things

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0What would you like to learn?

awn786
 2 years ago
Best ResponseYou've already chosen the best response.1apparantly his teacher is hgis uncle who started teaching him maths when he was 15 an now is 50

awn786
 2 years ago
Best ResponseYou've already chosen the best response.1look above^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

nikhilchabe
 2 years ago
Best ResponseYou've already chosen the best response.0Here you go with the link for complete solution! http://www.learnalberta.ca/content/memg/Division03/Square%20Root/index.html

awn786
 2 years ago
Best ResponseYou've already chosen the best response.1ermmm it needs to be simple
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