 Let's have a look at how Julia deals with complex numbers. Now the complex number in Julia is IM. So I can simply construct 2 plus 2 IM. So that's 2 plus 2 I. And if we execute it, it's 2 plus 2 I. So that IM is an imaginary number, the square root of negative 1. I can also construct it with a function called complex. So I can say 2.2 and that's going to give me the same complex number. Now I can ask what the real part is of a complex number 2 plus 2 IM. Of course, the real part of that is 2 and the imaginary part I can ask for that as well plus 2 IM. And I see the imaginary part is this 2. Just to show you if I made that into a 4, of course the imaginary part is 4. I can ask for the complex conjugate. CONJ, complex conjugate of 2 plus 2 IM. Now remember the complex conjugate. We just take the imaginary part and we multiply it by negative 1. So that's going to become 2 minus 2 IM. If this was a minus, of course, it's very simple. Let's just do that as a minus. It's going to become 2 plus 2 IM. So I take the imaginary part and multiply it by negative 1 and that is called the complex conjugate. Now the absolute value of a complex number. I can do that. What's the absolute value of the complex number 2 plus 2 IM? And there you see. I'll give you a little clue if you don't know what the absolute value is. Let's put a sneakily 3 and a 4 and that's going to return a 5. Because remember we construct the complex numbers or we can represent it on what is called the argant diagram. That is just an x and a y axis where the y axis becomes the imaginary part. So if I were to draw a line from 0.0 to the 0.3,4 you can well imagine I can make a little triangle there and that would be the hypotenuse. So it's the square root of the real part squared plus the imaginary part squared. So it's the square root of 3 squared plus 4 squared and indeed that is just going to be 5. And I can also just ask for it to be squared automatically. That means it removes the square root sign of the hypotenuse by just using this. This 2 3 plus 4 IM ABS 2 and that's going to return the square of that 25. It can also do the angle for me in radians. Now that's called the argument. So once again imagine it is on a plane the 0.3,4. If I draw that line up from 0.0 up to 3,4 it's going to make an angle with the x axis and that's called the argument. So I can say angle of 3 plus 4 IM and that's going to return in radians for me that angle between the x axis and this line. We can calculate the square root of negative 1 doing this. So let's do that the square root of let's do that the 1 IM. 1 IM so that is just I and remember what is I squared. I squared is negative 1 and if I take the square root of negative 1 I get I back 0 plus 1 I. So that was just how to calculate the imaginary number. You cannot say the square root of negative 1. If I would put in just negative 1 and they are going to return an error. It won't work but if I cheat a little bit and I take the imaginary number squared which is negative 1. So I'm asking what is the square root of negative 1 but I am classifying it as an imaginary number. It's going to evaluate the square root of that for me and now give me back the fact that the square root of negative 1 is just I the imaginary number.