Okay, as we all know, the square root
of negative 1 is equal to i, but have we ever thought
about what is the square root of i? Oh well. As you know,
i is the imaginary unit, and when you put the
i inside the square root, do we still end up
with a complex number? Most likely, it isn't? So the way to deal
with this is, well, we're just going to set up this
to be a complex number and we can write it down as--
let's say, √(i) is equal to the standard form of
the complex number, which is a+bi. And you know a and b are the real numbers and the i is the imaginary unit, right? If we can find out what's the value of a and b, then we are done, right? Okay, so we'll just do
this the algebraic way, meaning we can just go ahead
and square both sides, so that you see that the square and square root will cancel
and then on the left hand side, we will just have the i, and on the right hand side, we're just
going to multiply this out, right? Okay, so for the first term,
we will just have a², and then for the second term, we are going to have plus
two times this and that, which is 2abi, and then for the third term,
we are going to take this and square, right? And let me put on the plus first and let's see...
let me do this on the side for you. Take bi, and we square that,
and you know that this is b² times i². And what is i²? We know i² is equal to -1,
so all in all, this is -b², right? So this is not plus, it should be minus and
we have the b² from here, right? So it's just like this. And now we'll clean
things up a little bit. On the left hand side,
we will still have just the i, but on the right hand side, let's put
all the real numbers together first--I mean, the real parts. So first, we'll have a²-b². They don't have the i,
so let's put it together like this first. And then I will put this down next,
so we're adding with 2abi, okay? And, as you can see, on the left hand side
we have i, which is the same as saying 0+1i, isn't it? And now, you can just
kind of make a match, because 0 is the real number part, right?
Which has to be the same as this. And, this right here,
we have the 1 in front of the i. On the right-hand side,
we have 2ab in front of the i. So in other words,
1 has to match with 2ab, And this is how we can come with the
system of equations and stuff with a and b, right? So now, let's
make this happen. So first, notice that
I will have a²-b²-- a²-b² equals to 0, so let's
put that down right here first, right here. a²-b² equals to 0, and next, we have 2ab
is equal to 1, right? So that's the system of equations
that we can use to solve. Ok, so from here, we're just going
to be doing the substitution, right? On the second equation from here, we can
divide both sides by 2a, if you would like, so you can see that b=1/(2a), and then just do the typical algebra. You can plug in this into this b right here. So we see that we have
a² minus--the b becomes this: 1/(2a), and then you square that,
and this is equal to 0, like this. And now, we have a² minus 1², which is 1, 2², which is
4 on the bottom, a² on the bottom as well, and this is equal to 0. And to solve this equation, I will prefer you guys to do it this way--let's
get the common denominator here, so multiply the top and bottom by 4a²,
multiply this by 4a² as well. So you see, this is 4a⁴-1
all over the same denominator, which is 4a², and this is equal to 0. How can we have
a fraction equal to zero? The only chance is that
the top equals to zero, right? So all we have to do is
4a⁴-1 is equal to 0, okay? And of course, the typical thing is to add 1
on both sides that you have 4a⁴=1, divide by 4 on both sides, a⁴=1/4, and then, yeah, it's going to be taking
the 4th root on both sides, and, yeah, let's do this. Take the 4th root on both sides. It's like square root, then square root,
but twice, that kind of thing, all right? But this is the fourth root. This and that will cancel. However, when you take the
even numbers root on both sides, you should technically put a plus-or-minus (±),
just like when you do the square root. So a is equal to--you have plus or minus-- this is the 4th root of 1/4, right? Okay, ∜(1), this is 1, no problem, over--
what's the 4th root of 4? Let's look at this right
here down below, all right? 4th root of 4--just in the denominator. This right here--let's do it this way: this 4 is the same as saying 2² and the 4th root is
the same as saying to the 1/4 power, ok? And when you're looking
at this as the polar form, you know that the base is the same
and we can just multiply the powers, So this is 2*(1/4) and 2*(1/4)--
you know is equal to 1/2 of the power, so this is 2 to the 1/2 power, so 1/2 power is the same as √(2), right? so this is how you do
this little thing right here. Okay, so with that being said,
on the bottom here, we have nothing but just √(2);
that's the value of a, all right? And in fact, you have two possibilities
for a, and both are correct. Yeah, in fact we have two forms
for √(i) all together. Anyway, Keep this in mind because I'm going to
plug in this right here for the b, all right? So, let me just write
this down here for you guys. When a is equal to +1/√(2), and the other
one is when a is equal to -1/√(2), in both cases, right here, you know b
is equal to this, right? 1/(2a), So let me just write this down real quick again. b=1/(2a), In this case the a is 1/√(2) so we have
1 over--this 2 is from this right here, this a is that, so I will just
put down 1/√2 like this, okay? And we can simplify this real quick. Just look at the denominator. You have 2/√(2), which is still √(2), so this right here is √(2) on the bottom
and then the 1 is still on the top like that. And likewise, in this case, you still have b=1/(2a), So 1/2 is this 2 and this a is now that,
which is -1/√(2) in the parentheses. Same idea, but this time, it's just a
negative version of 1/√(2). Okay. So first answer is: a is this, b is that. Second answer: a is this, b is that. So in other words, ladies and gentlemen,
√(i), which is equal to a+bi. First answer: a is this, b is that, so we will have 1/√(2)+(1/√(2))i. That's the first answer. Second answer--when a is this and b is that, so I will put down the words better--or, the second answer is that a is -1/√(2), And we add with the -b. Well, b is negative so what? -(1/√(2))i. Two answers: this or that. And we are done. Cool, huh?
