# sqrt(i)

sqrt(0)
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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 isnt? So the way to deal with this is, well, were just going to set up this to be a complex number and we can write it down as– lets 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 whats the value of a and b, then we are done, right? Okay, so well 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, were 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 lets 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 its just like this. And now well clean things up a little bit. On the left hand side, we will still have just the i, but on the right hand side, lets put all the real numbers together first–I mean, the real parts. So first, well have a²-b². They dont have the i, so lets put it together like this first. And then I will put this down next, so were 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, isnt 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, lets make this happen. So first, notice that I will have a²-b²– a²-b² equals to 0, so lets put that down right here first, right here. a²-b² equals to 0, and next, we have 2ab is equal to 1, right? So thats the system of equations that we can use to solve. Ok, so from here, were 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–lets 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, its going to be taking the 4th root on both sides, and, yeah, lets do this. Take the 4th root on both sides. Its 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– whats the 4th root of 4? Lets look at this right here down below, all right? 4th root of 4–just in the denominator. This right here–lets 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 youre 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); thats 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 Im going to plug in this right 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, its 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. Thats 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? tags:
sqrt(i), sqrt(i) as a complex number, i^(1/2), www.blackpenredpen.com, blackpenredpen, math for fun
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