What is Root of i ?

That is, indeed the principal square root of i.
We have:
[tex](e^{\frac{i\pi}{2}})^{\frac{1}{2}}=e^{\frac{i\pi}{4}}[/tex]
which, multplied with itself yields:
[tex]e^{\frac{i\pi}{4}}*e^{\frac{i\pi}{4}}=e^{\frac{i\pi}{2}}=i[/tex]
which was what to be shown.

The other square root of i is [tex]e^{\frac{i5\pi}{4}}[/tex]

 

As arildno pointed out, i, like any number (except 0) has two square roots.

If you don't want to use the exponential form, you can use the trigonometric form: [itex]x+ iy= r (cos(\theta)+ i sin(\theta))[/itex] with [itex]r= \sqrt{x^2+ y^2}[/itex] and [itex]\theta= arctan(y/x)[/itex]. By "DeMoivre's formula"- the nth roots are given by
[tex]r^{1/n}(cos(\theta/n)+ i sin(\theta/n))[/tex]
with the n different roots gotten by adding [itex]2\pi[/itex] repeatedly to [itex]\theta[/itex].

In this particular case, [itex]i= 0+ 1i= 1(cos(\pi/2)+ i sin(\pi/2))[/itex].
It's square roots are [itex]1(cos(\pi/4)+ i sin(\pi/4))= \sqrt{2}/2+ i\sqrt{2}/[/itex] and [itex]1(cos((\pi+ 2\pi)/4)+ i sin((\pi+ 2\pi)/4))[/itex][itex]= 1(cos(3\pi/4)+ i sin(3\pi/4))[/itex][itex]=-\sqrt{2}/2+ i\sqrt{2}/2[/itex].

Obviously, I took too long in typing this!

 

here is another proof without using trig at all, and also without using the equivalent form of an img number using e.

let sqrt i =a+bi, then from here only i=(a+bi)^2, if we use the binominal rule here we get:
i=a^2+2abi-b^2, and we know that two img numbers are equal if their reals and imgs are equal, so using this we get
a^2-b^2=0 from here we get a=b, a=-b, and from
2abi=1, if we substitute the value of a we get respectively:

2a^2=1, and -2a^2=1, and from here we get, a=sqrt2/2, and a=-(sqrt2/2), if we substitute these values in
sqrt i =a+bi you will bet what you were looking for

 

-2a² is not equal to 1, for real a.

 

yeah you are righti, i did it too fast so i guess i did not notice this. However i guess it still works for the first one.

 

Yes stupidmaths method is generally the one i show people who don't know anything about complex numbers but just hear that its about the square root of negative 1. They expect that the square roots of these new imaginary numbers must be an even higher level of imaginary-ness! Quite fun to easily show them how they are wrong.

And arildno, I learned the Trig form method shown by Halls because in Australia for some reason they don't teach exponential form, though i've learned it. Just saying, I knew the form but know you've taught me how to use it in a new, time saving way. I love it, thanks lol i sound like a groupie.

 

Hey, be careful, it's not "stupidmath", it's "sutupid math"- completely different meaning! (I guess, actually, I have no idea what it means!)

 

O sorri my bad, when you just read it quickly it looks like stupid math. Im sure the extra u acts a some sort of a retarded prefix which makes it the opposite :D

 

how do we know that the sqrt of i can be denoted as a+bi?

 

a + bi is the general form of any complex number. This said, stupidmath made the assumption that the square root can be expressed as complex number. He didn't know whether or not this is the case first, but lucky him, it is!

 

From a slightly advanced standpoint, because the complex numbers are an algebraically closed field hence the equation x²=i has a solution in the complex numbers.

 

because complex numbers can also be thought of as "vector" in an argand diagram and a is like the "x"/Re coord while b is the "y"/Im coord. so a + ib is just any complex number. as long as sqrt of i is a complex number then it can be expressed as a +ib with a, b real

 

Well, perhaps he knew that the complex numbers are algebraically closed and so did know that the square root of i is a complex number.

 

yeah of course, although i don't know much about complex numbers, i know that they are algebraically closed, otherwise it would be wrong if i would have made such an assumption, at the very beginning. So, it is not about being lucky at all.

 

If you did not know it, it would have made the assumption unsure, not necessarily false.

 

yeah i guess you are right, couse then i would not know what the outcome would be, but this does not necessarely mean that i would not get a complex number, like in this case in particular.