Hi, I'm Toby and welcome to the Joy of Mathematics.
All you will need to be able to follow along at home is a pen some paper and a vision in your mind.
Today we're going to be exploring the square root of i.
It's a curious question with a delightful answer. Don't let the mathematics scare you away
I'm here to walk through every step with you.
Talent is just pursued interest.
Anything that you're willing to practice you can do. i is our favorite imaginary friend
But our relationship with them is complicated. It's mostly their fault
In fact, i is a complex friend if we use a smiley face
to refer to all of our complex friends we can write them as
x
plus i y where x and y
Give the address of our friend on the real and imaginary
axis
Real friends can be complex, too
they just have a y-value of 0 and
it's our y on this axis and our x down here. If we draw a straight line between the origin and
Where our friend is we can call that r.
And it actually allows us to set up some triangles in here
Where this angle here is Theta.
Using a bit of trigonometry we can actually rewrite
expressions for x and y using
Sine and cosine. So x will be this distance here and it will be our
cosine of theta
y, this one here will be r sine of theta.
Now, if you've seen my first Joy of Mathematics video you may remember something called Euler's formula
And it was a very beautiful expression. I will remind you of it here. It is e to the i theta
is equal to
cosine theta plus i
sine theta.
Now you may see that we're dealing with this already.
We have a cosine theta plus i sine theta in here.
So actually we can use Euler's formula to rewrite this expression up here.
I'll get rid of it since we now know what we can do.
Go and just beat the Euler out of it. Friend is going to be r e
to the i theta. If we just want to talk about our friend
i for a second we can get rid of all this general case.
Well, i is going to be up here at a distance of one from the origin.
That in fact places i on the unit circle which are the points that are all a distance of one away from the origin
Don't know how good I'm doing a circle today, but there we go. Good enough to write down. I like this
We know that r is going to be equal to one but what is theta going to be?
One way to find theta would be going like this up from the real axis
To our i to our imaginary axis. Now if two pi radians is the
rotation it takes to get all the way around the circle, then this quarter of a rotation right here will be
PI over
two.
Now there's another way to write i and that's by rotating around this way.
Now likewise, this one is going to be e to the minus this time because we're rotating the other way, i
And it's going to be three pi over two.
It's time for a friendship test here and we want to test to see that
If the square root of our complex friend is written like this. What's the solution going to be? Is he complex?
Can we trust him?
I don't know. We can say that the square root of any complex friend is some other number
Z. now if we make both of these guys a little bit more square a little bit more uptight we can write
complex friend is
equal to Z
squared.
Now if we subtract this from this arrange everyone to equal zero we get Z squared minus
our complex friend
is
equal to 0.
Now this might not look like but it's actually kind of important for us. What we have here is a polynomial
It's a polynomial with a degree two. It's a quadratic and that means that we can use something called the
FTA.
Now I know what you're thinking but this is not friendly trees anonymous instead this is the fundamental theorem of algebra.
The fundamental theorem of algebra means that any polynomial with complex coefficients
Will have complex solutions now. Do we have complex coefficients here?
well our coefficient in front of the Z squared term is one
we don't have as it term on its own so the B, as it's usually called, is zero and
This is our constant term here and the coefficient and of our constant term is our complex friend.
So we do have complex coefficients. Therefore, we know that the answer to the square root of i
will be a complex number too. In fact, we know that there will be two solutions because this gives us a polynomial of degree two.
This is just useful to know
So we know how to write and treat our answer as we're going forwards
Now back to I you might know that the square root of a number can be written as that number to the power of 1/2
so we could write the square root of i as i to the power of 1/2 and
Then we can just do the same thing to these terms. We would have e to the i pi
over two to the power of 1/2
I'm writing or for these two solutions, e to the minus i three pi over two
It's the power of 1/2.
To work these out just bring this exponent inside of this one and multiply them out. We would get e to the i pi
over 4
and e
to the minus i
three pi over four.
Now we can actually find these solutions on our unit circle over here. So a theta of pi over four
is going to be to do half way up here.
That's going to be our first solution to the square root of i.
Our next solution will be minus
3/4 pi which would place us opposite over here.
That's what our two solutions look like
graphically. I've made a little bit more room for ourselves so we can then take these answers that we've already found graphically and if you
choose to
convert them back into the original form, we just do that
using the reverse of Euler's formula. So putting them back into sine and cosine for our first solution we would have
theta as pi over four, so we would have
cosine
pi over 4 plus
i sine
pi over
four now
Cosine of pi over four actually works out to one over root two and in a similar way
We would have plus i times the sine of that which is also one over root two
so that is actually our first solution to the square root of i.
Now you could do the exact same thing using a negative
3/4 pi as
you know our argument in here or you can look at our visual plot and know that if this was our first solution,
our second solution will be the negative of both of these values because we're in the opposite quadrant here on the graph.
So our second solution
will just be
minus one over root two
minus i one over root two. And those will be our two solutions as
shown there for the square root of i. Now the dislikes on this video might come from people who are upset
I didn't rationalize this denominator. You could rewrite these fractions as
Root two over two, I believe
But they look cute enough as they are to me
These solutions kind of makes sense if you think about it visually for example this rotation from here to here
If you repeat it once again
you get to i and same with this on here you repeat this rotation again you get back to i so
That's how you find the square root
If you wanted to find the cube root you would end up with three equally spaced
solutions where if you repeat them twice more you get back to i
and you can even try finding out the cube root of i it's a very similar technique although you might want to
think twice about how you find the two, I guess, definitions of i
Instead of like this way in this way. You might want to just keep adding on two pi or a full rotation
Every time you're trying to define i to get the number of solutions you need.
Thank you for joining me here today in the garden and I hope you have an absolutely
mathematical day.
This is Henry and he's jealous that Skittles got included in one of my previous videos
But he didn't so now it's his time for a video.