What is the square root of i, where i is the imaginary unit thats stands for the square root of negative 1?

Answer 1

There are two square roots of i (of course). One of them is
sqrt(1/2)+i*sqrt(1/2)
and the other is the negative of that one. Just try it to verify!

Answer 2

There are a couple of ways to answer this. Best is if you can figure it out for yourself. Compex numbers are pairs of numbers usually expressed like a+bi (or (a,b)). So lets assume an answer is of the form a+bi, i.e. (a+bi)^2 = i
expand the left side a^2 +2abi +(b^2)(i^2) = a^2+2abi-b^2 = i
looking at that we see that the only way to get a number in the form ci is if a^2 = b^2 (so they cancel)
now its prety easy. look at the middle term 2abi. that has to be the same as 2(c^2)i which must be our answer = i
obviously c=1/sqrt(2) or -1/sqrt(2) since 2/(sqrt(2)*sqrt(2)) =1
hence a=1/sqrt(2) = b (don't forget the i in a+bi)
also dont forget that we can have the negatives so there are two answers, as we would expect for a square root problem.
Also note that there is a way to solve this trigonomically if we put the b on the y axis and use rotation to substitute for multiplication - but thats another (complex) subject.

Answer 3

i = cosπ/2 + i sinπ/2
= e^(iπ/2)

So square root i

= ±√i

= ±i ^(½)

= ±(e^(iπ/2))^½

= ±e^(iπ/4)

= ±cos π/4 + i sinπ/4

= ±½ √2(1 + i)

Note (±½ √2(1 + i))²
= 2/4 (1 + 2i + i²)
= ½(1 + 2i - 1)
= i

So square root (i) = ±½ √2(1 + i)

sahsjing

How can you say that square root i = ±√i is 'obviously wrong'???????????????

If you were asked for the square root of 9 you would have NO hesitation in saying:

the square root of 9 = ±√9 = ±3

After all this is the solution of x² = 9

If x² = 9 then x = square root of 9 = ±√9 = ±3.

So if z² = i then z = ±√i = ±½ √2(1 + i)

Please think carefully before you criticise others for what they say!!!!

falzoon

you have not just the square root of i you have the square root of -i and that is why you have 4 solutions to x^4 = -1

if x^4 = -1 then x^2 = ±i

The solution of square root of i is being sought not square root of ±i

Answer 4

Let x = sqrt(i)

Squaring both sides gives :
x^2 = i = sqrt(-1)

Squaring both sides again, gives :
x^4 = -1

Rearranging gives the 4th degree polynomial :
x^4 + 1 = 0

which shows that there are 4 roots of sqrt(i).

Factorising gives :
(x^2 + x*sqrt(2) + 1)(x^2 - x*sqrt(2) + 1) = 0

Setting each of these terms to zero
and applying the quadratic formula,
gives the 4 solutions :

(-1 - i) / sqrt(2)

(-1 + i) / sqrt(2)

(1 - i) / sqrt(2)

(1 + i) / sqrt(2)

Wal C

Good point. Thank you.

Answer 5

The square root of i is:

±(1 + i)/√2

Square it and you will see. It comes out i.

Answer 6

The square root of (i) is (c). (i) is the ninth letter of the Alphabet. The square root of 9 is 3. Since (c) is the third letter of the Alphabet, there you go.

Answer 7

it's just the square root of "i". There isn't a name for it. The imaginary number is created because of the inability to take the square root of negative one; which is the very first operation you are trying. The next operations cannot be done because the first operation cannot be done...

Answer 8

Damn I wish they had the Internet when I was in high school. This is a nifty way to get your homework done!

Answer 9

i = e^[i (pi/2+2npi)]
√i = e^(i pi/4+npi)
= (√2/2)(1+i), if n=0
=-(√2/2)(1+i), if n=1
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Wal C,

"square root i = ±√i" is obviously wrong even through you have the right answer.

Answer 10

(-1)^(1/4)