What is the Root of i? A Surprising Discovery for a 3rd Year Physics Student

  • Thread starter neu
  • Start date
  • Tags
    Root
In summary, the discussion revolved around the square root of i, which is the principal square root of i, and how it can be expressed in different forms, such as the exponential form and the trigonometric form. It was also discussed how the complex numbers are algebraically closed, which explains why the square root of i can be expressed as a complex number in the form of a+bi. The conversation also touched on different methods for finding the square root of i, such as using the binomial rule and the trigonometric form.
  • #1
neu
230
3
I am a 3rd Physics student and this is news to me!

apparently root of i is exp(iPi/2))^1/2

Why?

Of course by i I mean Root of minus 1
 
Mathematics news on Phys.org
  • #2
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]
 
  • #3
worked it out just after i posted, but another way

expi(Pi/2)=CosPi/2 + i Sin Pi/2 = i

As the cos pi/2 =0 & sinPi/2 =1

badboy
 
  • #4
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!
 
  • #5
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
 
  • #6
-2a2 is not equal to 1, for real a.
 
  • #7
neutrino said:
-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.
 
  • #8
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.
 
  • #9
Hey, be careful, it's not "stupidmath", it's "sutupid math"- completely different meaning! (I guess, actually, I have no idea what it means!)
 
  • #10
O sorri my bad, when you just read it quickly it looks like stupid math. I am sure the extra u acts a some sort of a retarded prefix which makes it the opposite :D
 
  • #11
how do we know that the sqrt of i can be denoted as a+bi?
 
  • #12
mathis314 said:
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!
 
  • #13
mathis314 said:
how do we know that the sqrt of i can be denoted as a+bi?

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.
 
  • #14
mathis314 said:
how do we know that the sqrt of i can be denoted as a+bi?

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
 
  • #15
Werg22 said:
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!
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.
 
Last edited by a moderator:
  • #16
HallsofIvy said:
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.
 
  • #17
If you did not know it, it would have made the assumption unsure, not necessarily false.
 
  • #18
Werg22 said:
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.
 

What is Root of i?

The root of i is a mathematical concept that represents the solution to the equation x^2 = i. It is a complex number that can be expressed as ± (√2/2 + √2/2i).

What is the value of Root of i?

The value of root of i is approximately 0.707 + 0.707i or -0.707 - 0.707i. It is a complex number that cannot be simplified any further.

How is Root of i calculated?

The root of i is calculated by solving the equation x^2 = i. This can be done by using the quadratic formula or by using the method of completing the square.

What is the significance of Root of i?

The root of i is significant in mathematics and physics as it is used in various calculations involving complex numbers, such as in electrical engineering and quantum mechanics.

Can Root of i be expressed in simpler terms?

No, the root of i cannot be expressed in simpler terms as it is a complex number that cannot be simplified any further. It is the solution to a specific mathematical equation and has its own unique value and properties.

Similar threads

Replies
10
Views
1K
  • General Math
2
Replies
45
Views
3K
Replies
5
Views
3K
  • Science and Math Textbooks
Replies
1
Views
643
Replies
9
Views
2K
Replies
1
Views
586
Replies
5
Views
1K
  • General Math
2
Replies
44
Views
3K
Replies
1
Views
43
Replies
26
Views
1K
Back
Top