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The most cliché motivation of complex numbers is to stare at this quadratic equation, and say <sarcastic> “Oh look, squares are always nonnegative, and if we add 1 to it, it is surely positive, so this equation has no solution, and that’s why we invent a solution, called i.” </sarcastic> But the problem is that having no solution doesn’t mean we have to invent one. If anything, it makes you think that mathematicians actively seek trouble,
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addressing a problem that could be ignored, because having no roots is not a problem. The actual problem arises when we try to solve this cubic equation with a general method. By inspection, 4 is a totally legitimate root, because its cube is 64, and multiplying by 15 gives 60, and indeed 64 minus 60 minus 4 is 0, so 4 is indeed a root. But we might want to work this out in a more systematic way than *inspection*. This can be done using
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Cardano’s formula, and all we need to do is to substitute p and q, and this is what we got. So we really have to come to terms with square root of negative numbers. Not only that, we also need to know how to take cube roots with these weird stuff. This time we can’t just ignore this, because THERE IS A ROOT: 4, and this was really what prompted the study of complex numbers. [CHAPTER 1: Complex plane] This is one way of representing complex numbers. i here denotes the square root of -1, called
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the imaginary unit. a and b are plain old real numbers, and this time a is called the real part, without the attachment of i, and b is the imaginary part, multiplying the imaginary unit. In other words, both real and imaginary parts are real. So we need two real numbers, a and b, to specify a complex number, which could naturally be put on a coordinate plane. So this point, having coordinates (a,b), corresponds to the complex number a + bi. Because of this
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correspondence, we can represent all complex numbers on a plane, called the complex plane. We usually denote a complex number as z, which is why it can also be called the z-plane. Now since a is the real part, b is the imaginary part, the horizontal axis is the real axis, usually labelled as Re(z), and the vertical axis is the imaginary axis, usually labelled as Im(z). However, this is only one way of thinking about this point - the Cartesian way. There
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is another - the polar coordinates. r here specifies the distance from the origin, called the modulus, and denoted as the absolute value of z. Knowing the modulus is not enough, because those on the circle all have the same modulus. So we also need to specify the angle theta made with the horizontal, called the argument, and denoted as arg(z). However, this can spark some arguments (pun very much intended). What if the point is down here? The modulus is
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perfectly fine, but for the argument, would you go clockwise, registering as a negative angle? Or would you still go counterclockwise all the way around, registering as a positive angle? It really depends on your convention. I mean, technically, we could add another 2 pi radian to it whenever we want. As long as you have a clear convention and stick to it, you are fine. In this case, the usual convention is to consider the negative argument. Essentially we want
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the argument to be in the range (-pi, pi]. So on the upper half plane, we get a positive argument, and on the lower half plane, we get a negative argument. The edge case is the negative real axis. In this case, we say that the argument is positive pi. If we plot the argument against the complex plane, this is what we get. So we can see that on the positive real axis, the argument is 0. As we go along the upper half plane, the argument
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increases up to positive pi. On the other hand, if we go to the lower half plane, the argument decreases to negative pi. You can see that there is a massive drop-off on the negative real axis, which is sometimes annoying. We will talk more about these problems in the next video. For now, let’s go back to our normal complex plane. This is the Cartesian way of representing complex numbers, and this is the polar way. Using simple trigonometry on this right-angled
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triangle, it is not difficult to see that a is r cos theta, and b is r sin theta. So apart from writing z as a+bi, the other way is to use the modulus and the argument. As we’ll see later, this bit can be further simplified. Anyway, this white box represents how we know a and b from r and theta, but what about the other way around? Using Pythagorean theorem, r can be quite easily found. And if we focus on this triangle, we can represent
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tan theta as b/a, which seems to suggest that theta is arctan(b/a), or is it? [Vsauce music plays] Remember that the argument is in the range (-pi, pi], but the arctangent function only ranges from -pi/2 to +pi/2, so clearly something’s wrong. But this isn’t *completely* wrong though. If we are in the right half plane, then this is perfectly fine, because in this case, the range of theta matches exactly that of the
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arctangent function. But what’s happening is that the ratio b/a is the same on this line, but still the same even for points on the opposite side. But instead of having a positive argument, we would have a negative one. Since the arguments differ by pi, in this case, it should be the arctangent minus pi. Similarly, the ratio b/a is unchanged even on the other side of the line, and since this formula is perfectly fine in the fourth
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quadrant, and the argument should have been positive rather than negative, in the second quadrant, the argument should be the arctangent plus pi. Coupled with the issue of that massive drop-off, the argument function is not exactly the nicest function to work with, but we need to deal with it. As a very quick recap, for any complex number, we have the usual Cartesian way of representation in terms of its real and imaginary parts, but also the polar way of representation using
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the modulus and the argument. And this box is the interconversion between the two ways of representation. However, one thing before we close this chapter. Instead of thinking about the complex number as a point, we can also think of it as a vector emanating from the origin, and now we move on. [CHAPTER 2: EULER’S FORMULA] Euler’s formula is the gateway to study more about complex numbers. But to actually derive it, it depends on how you define the exponential function. There will be three
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proofs based on three definitions in this video. The most common definition would be the power series. If this is how you define exponentials, this is the most classic way to prove the formula. We basically group every other term together. And based on the definition that i^2 = -1, we rewrite each bracket a little bit. And now we “recognise” that the first series is that of cos theta, and the second series is sin theta, and so we have e^(i theta)
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= cos theta + i sin theta. But this proof isn’t that illuminating, because why do these series correspond to sines and cosines? To address this problem, people usually simply *define* sines and cosines using these series, which, well, rigorously speaking isn’t a problem, but how do these series even relate to the geometric characterisations of sines and cosines? It isn’t impossible, but that’s a story for another time.
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So we turn to the second definition of the exponential, which uses its property related to differentiation, namely, the derivative is simply i times itself, which is the basis of a nice visual proof. Before explaining this proof, we need a little fact, which is how we multiply a complex number by i. When a is multiplied by i, not too surprisingly, we get ai; and when bi is multiplied by i, we get -b, because i^2 = -1. This means that
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the point (a,b) is mapped to the point (-b,a) by multiplication of i. But there is a really simple geometric illustration. This is the point (a,b), and after multiplying by i, we will get here, (-b,a). If we add the vectors and the angle, it might be a bit clearer what multiplication by i does. Yes, it is simply a counterclockwise rotation about the origin by a right angle. And so we get onto the proof. First of all, we need to understand what this derivative means. e^(i theta) can be thought
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of as your position on the complex plane. But the position changes with theta, so the changing positions create a trajectory. In this case, the trajectory isn’t random. It is governed by this derivative equation, where the left hand side denotes the velocity of a particle following the trajectory. The right side is the position vector, rotated anticlockwise about the origin by a right angle. This relation, together with the fact
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that when theta is 0, the exponential is 1, allows us to determine the trajectory. So this is the complex plane. When theta is 0, it is 1, so the particle starts here. Since the velocity is perpendicular to the position vector, the initial velocity points upwards. So after a very small time, the velocity vector no longer points directly upwards, because the position vector is no longer perfectly horizontal. In fact, as time goes on, the
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trajectory is exactly a unit circle. Throughout the trajectory, the position vector has length 1, and so the speed is 1 throughout the journey, because the velocity is simply rotating the position vector with no stretching. So after time theta, the distance travelled is precisely theta, and so by definition of angles in radian, the argument of this point is exactly theta. Because this point is e^(i theta), we finally get that e^(i theta) = cos theta + i sin theta, Euler’s formula!
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This amazing proof was in “Visual complex analysis” mentioned in the previous video, and also adapted by 3Blue1Brown in this video. So this proof uses the second definition of the exponential. There is actually a third proof, using the limit definition, but I am going to put this off until later where we develop an even better intuition of complex numbers. But let’s consider the case where theta is pi, then cos pi = -1, and sin pi
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= 0, and so e^(i pi) = -1. This is usually called Euler’s identity, which is often considered the most beautiful equation. [CHAPTER 3: OPERATIONS OF COMPLEX NUMBERS] How do we manipulate these complex numbers then? Let’s start with addition and subtraction, which is what you will expect. Adding two complex numbers simply means adding the real and imaginary parts separately. Similarly, subtracting two complex numbers means subtracting
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the real and imaginary parts separately. On the complex plane, adding two complex numbers is the same as adding two vectors together, so you either shift one of the vectors, or apply the parallelogram rule to get the sum. If we are considering the difference, then this is simply the relative position of z_1 from z_2. We simply translate so that it is emanating from the origin, and now z_1 - z_2 is the tip of the vector. So addition and
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subtraction of complex numbers are rather easy. Just one more thing: by normal triangle inequality on this triangle, we get the modulus of the sum is smaller than the sum of the moduli, and now let’s get onto multiplication. To multiply two complex numbers, expand normally, treating i as just another variable, but then note that i^2 = -1, and separate the real and imaginary parts. This looks rather complicated.
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Is there a better way to do it? Remember that in the first chapter, we have looked at the polar way of representing complex numbers. Now we know that the bracket is precisely e^(i theta), because of the Euler’s formula presented in the second chapter. So now, if we represent the two complex numbers in this form, multiplication becomes a lot easier: the non-exponential parts multiply, and multiplying exponentials become addition in the exponent.
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In other words, we multiply the modulus, and add the arguments together, well technically, not necessarily just adding, because while theta_1 and theta_2 are in the range for argument, it isn’t necessarily the case that the sum is in the range. This equation is still true, just that the thing inside the exponent is not necessarily the *argument*. The argument could differ from this by multiples of 2pi. [IMPORTANT CAVEAT]
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On the complex plane, how do we multiply two complex numbers? To do that, we identify where 1 is, and construct a triangle with the origin, 1, and z_1 as the vertices. We then rotate the triangle so that the correct side aligns with z_2, and then stretch it so that that side perfectly matches the vector representing z_2, and the tip would represent the product! And I encourage you to think for yourself how this interpretation is related to the
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previous equation we have talked about. And we now consider conjugation. The complex conjugate of any complex number is simply obtained by negating its imaginary part. In this video series, I am going to use a bar above to denote complex conjugation. The geometric intuition is not too difficult to grasp either. For any complex number z, its conjugate is simply its reflection across the real axis, which gives us another way of looking at it.
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Since we can always write a complex number in terms of its modulus and argument, let’s see what the corresponding information for z bar might be. A simple reflection doesn’t change the modulus, but the argument changes from positive to negative, and so z bar is actually r e^(- i theta), so this is another way of thinking about conjugation. With all these, let’s consider the product of z and its conjugate. The algebraic way would be
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to first write z in its Cartesian form. Then the product can be written this way, and through some basic algebraic manipulations, we get that it is equal to a^2 + b^2. But looking back at the form of z, it is simply the modulus squared! This simply came from applying the Pythagorean theorem. The other approach would be to write z in the polar form. Then from what we discussed, the conjugate has its argument negated, and the exponentials cancel each
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other, and we get, again, modulus squared, and this will be the inspiration for division of complex numbers. When it comes to division, the Cartesian form is quite messy. We first multiply both numerator and denominator by the same complex number, namely the conjugate of the original denominator, because now the denominator would be real, which is simply modulus squared. The numerator, however, is a mess, but this is only the product of two complex numbers, which we knew how
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to deal with. To be honest, this expression is quite disgusting, but sometimes we do need to know *how* to divide complex numbers this way. The much better way would be to write everything in polar form. Quite similar to multiplication, the result is rather nice. We divide the moduli, and subtract the arguments, well, again technically, the argument isn’t simply the difference: it could be different up to multiples of 2 pi.
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On the complex plane, how do we divide two complex numbers? We first construct a triangle with z_1, z_2, and the origin as vertices, then rotate it so that the side originally corresponding to z_2 is aligned with the positive real axis. Then stretch or squish it so that the vertex on the real axis goes to 1. And then the tip would represent the quotient! Again, I encourage you to think about how this is related to the formula we described
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earlier. As a fun little exercise, consider these products or quotients, and see how these are actually related to calculation of pi. More details in this video. The final boss is a complex number to the power of another complex number. To do that, we need to know both exponentiation and logarithm, but let’s tackle exponentiation first. Given z in this Cartesian form, what is the value of e^z? Well substitute normally into the
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expression, and simply separate the real and imaginary parts. If you really want, we can further rewrite e^(iy) using Euler’s formula. But let’s focus on this second step. Isn’t this similar to the polar form of a complex number? So the modulus is precisely e^x, and y is the argument, well again, only up to multiples of 2 pi. And that could directly lead to how we define logarithms. If we have e^z to be in polar
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form, how do we find out z? Well, just like what we said on the left, the modulus is e^x, and so x is the usual real natural logarithm of r, but what about y? As we saw just earlier, it should be the argument, right? Well, actually it could be the argument plus any multiple of 2 pi, so does this mean that the logarithm of w, which is z, or x + iy can take a lot of different values? Usually, we only want one of the answers, and we choose the one
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such that the imaginary part is in the range (-pi, pi], so really, it is just the argument of w. So with that, let’s give a very simple example. The logarithm of -1 is then the natural logarithm of its absolute value, 1, plus i times the argument, which is pi in this case, which means it is i pi. And finally, we see how we do complex exponents. This is completely analogous to the real case, using both logarithm
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and exponentiation. For instance, to calculate i^i, we have to substitute z and w to be i in the definition, and recall the logarithm of complex numbers explained just now. We can calculate the logarithm of i, and after slight manipulation, we get e^(-pi/2), a real number! And transcendental as well! But before we wrap this chapter up, we need to revisit conjugation. Recall that if z = a + bi, then its conjugate is a - bi. Let’s say we now only know z
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and z bar, and how do we find a and b, the real and imaginary parts of z? This is a relatively simple system of equations to solve, and you could and should verify these formulas. A more geometric way is to consider the complex plane. The real part is the average position of z and z bar, while the difference of z and z bar has length twice the imaginary part of z, but since the vector points vertically upwards, the difference is purely imaginary, so we have to divide by i.
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Now let’s get back to Euler’s formula. cos theta is the real part, sin theta is the imaginary part, and since its conjugate is simply e^(-i theta), just like what we did a moment ago, we can write cos and sin theta in terms of the exponentials. However, from the way we established it, this is only true when theta is real. So these formulas only make sense when theta is real, but we *define* cosine and sine of complex numbers using these
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formulas so that they remain true even when theta isn’t real, because now we know how to take e to any complex power. A final thing about conjugation, which is also a sort of summary of this chapter. Conjugation has all these properties, which looks a lot, but basically they say that the operations we have covered in this chapter work the same way in the reflected world, because conjugation is just reflecting in the real axis, and I do encourage you to verify all these properties
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of conjugation, not just algebraically, but also geometrically whenever you can. But in any case, these four types of operations are the major ones talked about in this long chapter. [Chapter 4: de Moivre’s theorem] Previously, we mentioned that if we multiply the complex numbers, we simply have to multiply their moduli and add up the arguments. So if we multiply n of them, we have a similar procedure, still multiplying their moduli, and adding up the arguments, and so in particular,
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if all the moduli are 1, then we only get the adding arguments bit, and if we further assume that all the arguments are the same, being theta, then we get this: multiplying e^(i theta) by itself n times is e^(i n theta). Using Euler’s formula, we can rewrite the exponentials in the Cartesian form, and this is de Moivre’s theorem. Note that n has to be an integer. But why dedicate an entire chapter to this theorem? Remember that in
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the very beginning, the reason to start the study of complex numbers was the appearance of these square roots, but we also need to deal with the cube roots. And de Moivre’s theorem is going to be involved in finding cube roots because it is generally useful in finding nth roots of 1. Let’s make de Moivre’s theorem slightly more useful. We consider the nth power of a more general complex number, so we also have the modulus to consider, but the added
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modulus only changes the modulus of the result to r^n. Technically, the above is *the* de Moivre’s theorem, but since the bottom one is just one step away, let’s call the bottom one de Moivre’s theorem from now on. The nth roots of 1 satisfy this equation by definition, and 1 could be represented in this polar form. Now compare it with the right hand side. Here, the modulus is 1, so r^n = 1, and n theta = 0. For the first equation, r is the modulus,
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so it has to be real and positive. As long as n is not 0, there is only one solution: r = 1. What about the other equation? Well again, as long as n is not 0, theta can only be 0. Since ultimately x is r(cos theta + i sin theta), by substituting the values of r and theta, x equals 1, which definitely isn’t surprising at all. However, that is not the only way to write 1. The angle could also have been 2 pi, or
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4 pi, or 6 pi, or in general, 2k pi, where k is an integer. With these new representations, the first equation about the modulus wouldn’t change, but the second equation would change, because now, n theta is not just 0 anymore, but any multiple of 2pi, and so theta is 2k pi / n, which means that the nth root can actually have a lot of different possible arguments. Well, actually exactly n of them, namely, from when k = 0 all the way to n-1,
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and these n different answers are called nth roots of unity. How do these roots of unity look on the complex plane? Well, when n=1, there is only one root, namely 1, and when n gets large, it’s easier to see the pattern of these roots of unity: they all lie on the unit circle and are evenly spaced across the circumference. Usually, the first one above 1 is important, denoted by zeta with subscript n, because now, all of the other roots are powers of zeta_n. So instead of using this procedure
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to find all nth roots, say, if we want to find cube roots of 2 + 11i, we only need to find one of them, and then the others are obtained by multiplying powers of zeta_3. Why? When we actually consider the solutions cubed, we then have (2+i)^3, with an additional factor, but by definition, zeta_3 cubed is 1, so all these roots cube to the same place. This means there are three cube roots, but sometimes we have a preferred or a standard
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one. How do we choose? Well, we simply consider their arguments, compared with the complex number we want to take the cube root of, remembering that argument lives in the range (-pi, pi]. Then we choose the one with the argument exactly a third of the original. In this case, we choose 2+i with an argument exactly a third of the original, and this choice using the arguments is called the principal cube root, and this is just one of the shenanigans of
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de Moivre’s theorem. There is another, which is to prove Euler’s formula. Wait, what? In this video, we have literally derived de Moivre’s theorem from Euler’s formula, and now you want to derive Euler’s formula from de Moivre’s theorem? This is a circular argument! BUT there is a way out *if* we didn’t use the Euler’s formula in the first place! In fact, it is possible to only use compound angle formulas and induction,
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which does not rely on Euler’s formula at all, so this is not really a circular argument. The bottom right to left implication is what the remainder of this video is about. The proof would start from the limit definition of the exponentials. Let’s look at how this limit definition works visually. First of all, we identify where 1 and 1 + i theta / n are. We have chosen n to be 2 for now. We then construct a triangle
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with 0, 1 and 1 + i theta /2 as vertices, and as we have illustrated before, rotate and stretch the triangle to get its square. But that is only the case when n = 2, we should increase the value of n, and these triangles together gradually become a sector. But how do we explain this phenomenon? Well let’s focus on a very tiny bit of the picture before we draw all the triangles. Let’s add a part of the unit circle. While this point is not
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exactly on the circle, it will be closer and closer when n increases. Since the distance here is theta / n, the angle subtended, by definition of radians, would also be theta / n. So we can say that it is approximately cos theta / n + i sin theta / n, equal in the limit. Now e^(i theta) being the limit of nth power of (1 + i theta / n) can be rewritten like this. By de Moivre’s theorem, the angle is multiplied by n, so the limit is cos theta
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+ i sin theta: Euler’s formula! This proof was again in “Visual complex analysis” mentioned before, and also mentioned by Mathologer in this video. And that’s it! We have covered, quite comprehensively, the basics of complex numbers. And this video is mostly made so that it is a reference for viewers for any of my future videos. In case you’re new here, this is part of a video series on complex analysis, and we have gone
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through two of them. The next video will focus on all sorts of complex functions, where you might see stuff like this! If you think you know what this is, comment below! Thanks to these patrons, and as always, like and subscribe with notifications on, comment and share the video as well! See you next time!
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