 I am Zor. Welcome to Unizor education. Today we will solve a problem, problem number one for complex numbers. You remember that complex numbers were introduced to be able to solve certain equations which we could not solve within the real numbers. And one of the equations was something like x square equals to minus one. And we have introduced i as a pure artificial new number. The only property of which is that this is a root of this equation, basically. And now with generalized complex numbers which are expressed as a times i plus b, we can say that where a and b are real numbers, now this represents a much broader universe of numbers. Actually the broadest people are mathematicians are dealing with nowadays. Maybe they will invent something else. But right now these are the numbers, the broadest universe of numbers we are dealing with. So the question is, can we do anything we want with these numbers? Can we solve any equation or do some operation which we know about within the set of these numbers? Well, here is an exercise or problem if you wish. For instance, we know what square root of minus one now is. This is i among the complex numbers. So my problem is what is a square root of i? Well, I said everything is possible. So this is also supposed to be some kind of rational number which has its own imaginary and real part. We have to know these coefficients a and b. So this is supposed to be something of this type and the only property which we want is that this b squared is equal to i. So basically if you wish, this is an equation. Alright, so that's the problem. Find a and b which would satisfy this particular equation which means that you will know how to extract the square root of i. Now press pause button, think about this problem and I will continue with the solution. Okay, so we have to know what square root of i is which means we have to solve this equation. How can it be solved? Well, very easily. Since a i plus b square is equal to i, first of all, let's write down what this actually is. This is the multiplication obviously, a i plus b times a i plus b. And we know this is i. Alright, now this thing, let's open all the parenthesis and what should we have? As an imaginary part we will have a times b plus b times a i. And as a real part we will have b times b which is b square a times a times i square which is minus a square because i square is minus 1. So this is what we are looking for, right? We have actually an equation with two different variables among the complex numbers. Well, how it can be solved? Well, obviously if some complex number is equal to zero then its real time, real part is supposed to be equal to zero and imaginary part would be equal to zero. So basically from one equation we actually have two different equations. This part is a real part, we have a real part on the left and no real part zero on the right which means we have this, that's one equation which we have from here. Another is imaginary parts should be equal, imaginary part of imaginary coefficient is 1, so a b plus b a which is 2 a b is equal to 1. Now we have two equations with two variables which obviously is solvable and easily actually solvable because from the first equation we get that a square is equal to b square which means a can be either plus or minus b and if we can substitute the a's expression into the second equation that we will have is the following. If we will substitute a equals b then it will be 2 b square equals 1 which means b square is equal to 1 half and b is equal to plus or minus 1 over square root of 2. So we have two different solutions for b and a as I said was b. If we substitute minus b into this equation we will have minus 2 b square equals 2 1 and this does not have any roots among real numbers, don't forget that b is a real number. So there is no such thing as a real number which we have to go actually invent the imaginary numbers to be able to solve these equations because from here you get this and obviously this has no solutions. Alright so basically the only solution we have is a equals to b is equal to either plus or minus 1 over square root of 2 which means that first solution is this, another solution is this. So these are two solutions both of them represent square root of i. Well what should we do after we solve the equation or prove some formula or whatever? We should verify, we should check it out. So it's extremely important after you have found the solution of the equation check it, it's really true. Well we have to check that i is equal to a times i plus b square. That's what we have to check. Alright let's do it for this case for instance 1 over square root of 2 i plus 1 over square root of 2 square root of 2 will be imaginary pot will be, let me expand it so it will be easier to, imaginary pot this times this, 1 over square root of 2 times 1 over square root of 2 will be 1 over 2 and this will also give us imaginary pot which is also 1 over 2. How about the real pot is 1 over square root of 2 times 1 over square root of 2 which is 1 2, 1 over square root of 2 times 1 over square root of 2 and minus 1, minus 1. This is 0, this is 1, this is 1. So as a result of this multiplication we have, we have 1 i plus 0 which is i which is supposed to be 0. Now if i will use the second solution which is negative here, here, here, here and here it will be negative here, sorry it will be, this is still 1 over 2, it will be minus 1 over 2 times minus 1. So it will be still plus 1 over 2 as it should be, same thing here, minus 1 over square root of 2 times 1 minus over square root of 2 will be 1 over 2. So this is exactly the same, here, again the same thing because we have two negatives which will give us the positive here. So it's exactly the same here and we get exactly the same result here. So this is also the solution with negative minus over square root of 2. Now what we have here is a very interesting case when we have two different solutions actually, we have two different square roots of i, but if you remember this is not in the usual situation because if I were to ask you what is square root of 4, what should you answer? Not 2, it's plus or minus 2, right? So we have two solutions to this as well. So there is no wonder that square root of 2 also has plus or minus 1 over square root of 2. So you have two solutions. I would rather put it not this way, but this way. So as you see, square root of something always has two different solutions in this case, which is obvious, that's how it's supposed to be, right? Okay, that's it for this particular problem. If you want to continue, you can do the root of the third degree or fourth degree root or whatever from i just as an exercise and you will see different answers and you will see different numbers of solutions, something like the fourth degree root of i probably should have four different solutions. Well, try to find out. That's it for this test. Thank you very much.