Reta Final EFOMM - Números complexos | Prof. Baltar

[00:00:03]Hey everyone, how are you all doing? Good evening everyone.

[00:00:07]Welcome everyone to the Military Theorem channel!

[00:00:09]Today, Professor [Name] will be tackling complex number problems geared towards the Efon exam.

[00:00:17]Efon's little test is coming soon, isn't it? Oh, wanting to make money. You guys want to go to Efon to make money, don't you?

[00:00:23]I think I don't know. So let's change the gas today. This is a very important topic, okay, everyone? I've gathered a lot of cool complex number problems here for us to solve. But first confirm me, please. Does everyone listen to me? Does everyone see me as doing well?

[00:00:41]Is the audio and video quality good, guys? All very well? Everything standard? Whoever arrives, leave a message in the chat so I know I 'm not alone. Leave a like on the live stream, share the link with everyone, okay, guys? So that's it. Look, for today I've selected questions about complex numbers, as I've already told you, all of them from previous ITA exams.

[00:01:04]Yes, ITA, the Flight Attendant Training Institute, as we already know. Is it closed?

[00:01:11]Older questions, but, man, in my opinion, they're the very essence of the Beauty pageant, guys. Awesome! Oh, there aren't many people watching the live stream today, huh?

[00:01:22]Not many people watching the live stream today, huh? So, let's share this live stream with everyone. It cost. Okay, everyone, so, as usual, here's a list for today's class, alright? So, after class, I'll be making this list available to our Military Theorem students at the TM Squadron.

[00:01:44]a list with 11 questions. Obviously, we won't have time to do everything here, but I'll be making this list with the solution available to everyone in the TM squad. Is it closed? Okay, screen sharing is all there is to it. The first question is on the screen.

[00:02:08]Just let me adjust my pen here so we can get started. Is it closed? Let's go. Look, folks, consider the equation. And then we have this equation, right, which involves complex numbers that, when you look at them at first glance, might scare you. And then the question went something like this: Given that x is a real number, what is the sum of the squares of the solutions to this equation? Okay, let's go, everyone.

[00:02:42]First, look, I'm going to repeat the left side of my equation here, okay? Here on the left side I'll have 16 multiplying 1 - xi divided by 1 + x.

[00:03:05]All of this cubed will equal the amount shown here on the right. Observe the following: I'm going to operate on these two fractions inside the parentheses, right? Since I have the 'i', right, since I have imaginary data involved in the parentheses, I'm going to consider the product of these denominators to be the least common multiple (LCM). So it's going to be here, like this: 1 - i x 1 + i in my denominator, right? This will be 1² minus i². However, we know that i² is -1, so I'll have 1 - -1.

[00:03:44]Who will be my deciding factor, guys? It's going to be number two. Great, Balto. I understood.

[00:03:50]Continuing now, then, look, up there it will be 1 + i, 1 + i, 1 - i divided by 1 - i, look, there will be something left over on top. So I'll have 1 + i x 1 + i, that is, 1 + i² - 1 + i. 1 - i divided by 1 + i will leave 1 - i, which will then be multiplied by 1 - i, resulting in 1 minus i².

[00:04:23]This here, obviously all raised to the fourth power.

[00:04:30]Show. So far we haven't done anything extraordinary.

[00:04:33]Nothing out of the ordinary for now. OK, Baltar.

[00:04:36]Let's continue then. Oh, on the left I 'll simply repeat it. So, I'll have it like this, look. 16 multiplied by 1 minus xi over 1 plus xi. All of this here, cubed, is going to equal who, folks? Look with me, then. 1 + i. Let's do it in draft form so as not to clutter our development. 1 + i² - 1 - i². This is a difference in squares, right? This will be the product of the sum and the difference. So it will be 1 + i + 1 - i, the sum 1 + i repeats and -1 - i, that is, + i.

[00:05:29]See, in the first parenthesis, the "do" will remain, right? Because the "i" will cancel out with the "i".

[00:05:37]Times here in the second parenthesis, look, there will be 2i left over. Hmm. Consequently, in my numerator, what will I get as a result, folks? I'm going to have the 4 I.

[00:05:53]This here, look, divided by 2 raised to the fourth power. Wow, Btar, what luck, huh? It will be? Look what's happening now, folks. 4i divided by 2 will become 2i.

[00:06:10]Is it or isn't it? This here, look, is the 2i.

[00:06:15]So, here's what's going to happen in practice. 16 multiplied by 1 - xi divided by 1 + xi cubed will be 2 to the power of 4. 2 to the power of 4 is 16. Hmm. i to the power of 4. Well, folks, i to the power of 4 is 1, right? 1 raised to the power of 4 equals 1. Perfect. So it will simply be 16 over there on the right.

[00:06:48]Look, I have 16 on the left and I have 16 on the right.

[00:06:53]So, what do we end up with in the end?

[00:06:56]We have to divide 1 - xi by 1 + xi. All of this here, look, the cube equals 1.

[00:07:11]Now comes the first very strong idea from today's lesson, man. The way things are now, depending on how you're dealing with the problem, you're going to have big problems, you're going to have a high chance of messing up, because you're going to keep developing, developing, developing, you're going to put in a lot of work and in the end you're not going to get anywhere.

[00:07:36]But look, does everyone agree that 1 - xi divided by 1 + xi is a complex number? Yes definitely. So let's make a change of variable. I'll do it like this, look. Let 1 - xi divided by 1 + xi = a z. Why?

[00:08:06]Because it's going to be a complex number.

[00:08:08]Counter. Beauty? So, what does our equation look like? In the end? It's going to stay like this, see?

[00:08:15]Z cubed equals 1. My friends, cube roots of the imaginary unit.

[00:08:25]Yes or no? You know that 'o' is the same as 'o' for zero, right? One. So I'm going to do it like this, see? 6 out of zero. Okay, so?

[00:08:41]The second law of the tree, I think that's how you say it, right? Tell me this, let's review it here, if you don't remember, look, I'm going to do a complete review here because we're going to use the first one in a little while too.

[00:08:57]So, look, if I have a complex number Z of the form Ris of theta, Moave's first law tells me about exponentiation.

[00:09:14]He says that z raised to the power of n will be absolutely equal to his r raised to the power of n multiplied by whose sis? De n teta.

[00:09:26]This is Moave's first law.

[00:09:28]Moave's second law deals with root extraction. So you're going to talk about the nth root of Z.

[00:09:37]Then you're going to do it like this, see? Raí nésima do R, que é o modifica seu, vezes o sis. Pay close attention now, everyone, to theta + 2kπ divided by m.

[00:10:01]So, look, this K goes from zero to n - 1.

[00:10:15]Therefore, observe the following: if I have a cube root, my k will be zero, one, and two. Each of these falls will give me a different root. Awesome! Thus, applying the second law of a tree to this equality here, look, of z³ ig. What can we say for sure, guys? We can state the following, master. The cube root of z³ will obviously be z.

[00:10:46]Notice that his 'r' is a 'i', right? The cube root of 1 is 1 itself. So it will be z multiplied by 6. I'll expand it now, see, the cosine of 0 + 2kπ, I don't need to include the zero, so it will look like this. 2k divided by 3 plus the i that multiplies the sine of 0 + 2kπ/3.

[00:11:17]Notice that each k will give me a root, right?

[00:11:21]Hmm, I understand, master. So it's going to stay like this, see? If I set k = 0, then k = 0 will give me a real root, right?

[00:11:31]Note that for k = 0, then my z will be as follows. The cosine of 0 is equal to the sine of 0. Hmm, I understand. It's not good. Awesome!

[00:11:45]Let me bring our mobile law down here to give us space to finish resolving the issue.

[00:11:53]Perfect.

[00:11:55]And if I mention the K as equal now, remember?

[00:11:56]If N is 3, then K will be 0, 1, and 2. So let's go. What if my K equals 1? How does it look, master? If k equals 1, then my Z will be exactly equal to the cosine of 2 x 1 x pi / 3.

[00:12:23]2 pi so 3, right? Oh, 2 pi is 360 di times 3 is 120. So it will be the cosine of 120º.

[00:12:37]plus the 'i' that multiplies the sine of 120º.

[00:12:43]So, my Z2 will be, guys, the cosine of 120 minus the cosine of 60, right? So it will be -1/2 plus the sine of 120 equals the sine of 60, so it will be √2/2 x i.

[00:13:02]And then I have K = 3, and K = 2. So, this is what differentiates the men from the boys, right?

[00:13:11]If you want, you can replace it.

[00:13:13]However, according to the complex root theorem, we know that if a root is complex, its conjugate will also be complex. So, folks, if I have three roots, the first is Z2, which is a complex number, everyone agrees that Z3 can only be Z3.

[00:13:32]So, instead of substituting K = 2, I'm just going to take the conjugate of Z2. So it's going to stay like this, then. I'll have Z3 absolutely equal to -1/2 less, right? Because it's the conjugate √2/2 that multiplies the 'e'. Okay, so our three have been decided. But seriously, for God's sake, be very careful, everyone. Look at that question, huh? Given that x is a real number, the sum of the squares of the solutions to this equation is... a lot of people come here, man, and add z², z²², z³², mark the answer, chicken out.

[00:14:14]Remember that Z is actually the result of a substitution, a change of variable, right? So now I need to undo this trade. Could it be? Okay, let's go. I'm going to delete that Moaba law, then you guys take a screenshot during the live stream.

[00:14:39]But what did we see? We saw the following, folks. We saw that Z is actually who? It is 1 minus XI divided by 1 + XI. Okay, I understand, Balta. So, each z will give me an X.

[00:14:56]Will it? Let's go! The first case will be 1 minus xi over 1 + xi being equal to z1. But who was Z1? One from here, so I have that 1 - xi will be equal to 1 + xi.

[00:15:22]Consequently, 2xi will be equal to zero; the only way for this to happen is if x = 0. There you go, I found my first root.

[00:15:34]So, this is my first root.

[00:15:37]We're in this together. Beauty? X1 = 0. Guys, is your internet working well or is my image freezing? Last week I tried to do a live stream, and it was freezing up a lot. If it's bad, please let me know, okay?

[00:15:51]It's easy to keep up, right? Anything you need, just call me. Okay, master, we've found our first root. Let's move on to our second root now.

[00:16:02]Perfect. Thanks for the feedback, guys.

[00:16:05]Second root. Oh, now I'm going to have to use the Z2.

[00:16:10]So I'm going to have the following, look.

[00:16:15]Let's go. 1 minus xi divided by 1 + xi will equal what, master? Z2, that is, -1/2 + √2/2 which multiplies I. In this case, note that it will be interesting for us to work with top, datão. That it?

[00:16:50]We're going to have to work with the division on the left side, right, so we can compare the real part with the real part, and the imaginary part with the imaginary part on both sides of the equation.

[00:17:01]So, who exactly is this Z? How can I get that Z, guys?

[00:17:07]Can I multiply by the conjugate of the denominator on both the top and bottom, yes or no?

[00:17:12]Here I'm just trying to rewrite it in order to separate part A from part A, part more from part imaginary.

[00:17:20]When I do that, what will I get on top of it?

[00:17:24]Okay, then? Up there it's going to be a draft here, look, 1 man the hissing sound, isn't that right?

[00:17:31]It will be the square of the first minus twice the first times the second plus the square of the second x² e². But since it 's -1, it will be minus x².

[00:17:46]So, 1 minus x² is actually the same as... what, master? It's the same as 1 minus 2xi minus x² over the bottom. I have the product of the sum and difference, which will be 1 qu minus xi.

[00:18:14]1.

[00:18:16]x² is x² and x² is -1. So it will become minus -x, which will give +x.

[00:18:26]In practice, how can we think about this Z of ours, folks? So, what do you mean? Look at this little scratch here with me again so we can save space up there. It will give you, look, 1 - x² over 1 + x - 2xi over 1 + x, right? I'm separating part a from part a from nothing with imaginary part.

[00:18:53]But up there, look at this, everyone. 1 - x² is the same as 1 + x x 1 - x. Oh, notice that I'm going to cancel out the 1 + x above and below. Yes or no?

[00:19:07]There you go.

[00:19:09]So, who gets our Z, folks?

[00:19:13]1 - X - 2X divided by 1 + x.

[00:19:28]This is our Z. Consequently, it's over now. Want to see, guys? That's how it's going to stay, guys. I'm going to rewrite my Z here. It's going to look like this, look. 1 - x - 2x / 1 + x will have to be equal to whom, guys?

[00:19:49]Vezi, i, in this case, right? It will have to be equal to -1 over 2 + √2 of 2 that multiply there. Hmm, I understand. B. But part al has to be the same as part al. So let's just set it up right here.

[00:20:05]1 - x must equal -1/2. Oops.

[00:20:13]So, what does x2 equal, guys? 1 + 2, that is, 3 / 2.

[00:20:24]3 divided by 2. We're in this together. Beauty. So this will be our x2. Balta. So how am I supposed to get there now?

[00:20:38]Hey guys, wait a minute, you know what I think I messed up? Tell me here!

[00:20:51]I messed up here. Man, doing it fast is tough, isn't it?

[00:20:56]Hold on a minute. Let me redo these, guys.

[00:21:04]Up there it's going to be, help me not to point anything out there, like, 1 - xi², right? Oh, this plus this, this will give the square of the first minus twice the first times the second plus the square of the second, that is, plus x² i², but i² is -1, so it will be, look, - x² up there. Beauty?

[00:21:31]This will then be our numerator. And the denominator is 1 + x, or 1 + xi in this case, right? E 1 + xi will become 1² which is 1 minus x² e². It will be 1 + x² and not x as I had written.

[00:21:53]So what will happen to our z? That's how it's going to stay, folks. 1 - x² divided by 1 + x² - 2x/ 1 + x² x i. Now that's clear, right? It's OK. Just that correction.

[00:22:17]So, how will this work in practice? Let's get this over with now, so we can finish this matter. Oh, here I'm going to have 1 - x² over 1 + x² - 2x over 1 + x² equal to what, guys? Being equal to -2 + a√2/ 2 x i. There's an "i" here too, right? I think I forgot. Is it closed?

[00:22:49]Nice. Beauty. So, let's equate the part with the partial. When I do this, I'll have, look, left partial 1 - x² over 1 + x² right partial -1/2. Hmm. When I cross-multiply, I'll have 2 - 2x² = a -1 - x².

[00:23:20]Who will this give me, guys? You're going to give me this, just listen. It's going to stay like this, you see. It will get smaller. That's right. It will be -x² = -3.

[00:23:56]x² will be equal to 3. Therefore, x will be equal to plus or minus √3.

[00:24:04]Notice that here we've solved the riddle of the next two roots, right?

[00:24:09]Therefore, master, x1 [clearing throat] equals 0, and x2 equals √3.

[00:24:18]However, if an irrational number is a root, then its conjugate will also be a root. So, who's next?

[00:24:28]It will be at least a√3. You didn't even need to calculate the second one with just that little trick. Ready. So, there you have it, all the roots of this monstrous equation have been determined.

[00:24:40]If I want to know the sum of the squares, look, 0² is 0, √3² is 3, √3² is 3, this here will give a sum of 6. (Answer key: " Badass.") Look at that! Okay. How are you all doing? Wow, guys, six people on the live stream, is that too fast? Are you making sense? Feel free to ask any questions, everyone. Make yourselves at home, okay?

[00:25:10]Any questions?

[00:25:14]Great, awesome!

[00:25:16]Possession for next time.

[00:25:23]Perfect. Okay, cool.

[00:25:34]Question two, second phase of the ITA exam in 2004.

[00:25:41]Look at the mess now, huh?

[00:25:47]So, let's go.

[00:25:51]This statement reads: "Given z = 1 + i divided by √quad 2." Calculate this, look at the mess, folks.

[00:26:04]Guys, here, actually, he gave a little help, right? He could have simply asked you to calculate this here, on the left, but he was being nice, wasn't he? Look how nice he is. He told you to calculate the one on the left; just think about this. Of course, right, guys? So, how do you interpret a summation? The sum, in fact, is a sum of several parts, right? So, if I have it like this, let's suppose he didn't, let's suppose he hadn't given that part on the right. Let's suppose he only orders this much here. The absolute value of the summation of z raised to the power of n, going from 1 to 60. How does this look, folks? Let's suppose he didn't turn right, oh. I'm going to take all the numbers raised to the power of n and add them together. But with which ones? n ranges from 1 to 60.

[00:26:57]So I'll have z raised to the power of 1, z raised to the power of 2, z raised to the power of 3. This continues until I reach whom? no Z60. So, once you interpret the summation, you notice that Vor didn't need to give you that part on the right; he could have simply asked you to calculate the absolute value of the summation, and that would have been it.

[00:27:18]Beauty? Nice. So, let's go, everyone. This is a small question that, if you look closely, what you have inside the module is a geometric progression, right?

[00:27:27]It's a geometric progression whose first term A1 is Z, and the common ratio... damn, it's unbelievable just to look at it!

[00:27:46]I can't kill someone just by looking, man.

[00:27:49]Whose ratio is worth Z, right? Why? The first term is Z.

[00:27:56]Multiplying by z gives z squared.

[00:27:59]I multiply by z, I arrive at z³. And so on.

[00:28:04]And so it goes, folks. Remember, how do we sum the first n terms of a geometric progression?

[00:28:12]A1 x Q raised to the power of n - 1.

[00:28:18]All of that divided by q - 1.

[00:28:24]However, notice that in this case, what do I want? The module, right? So, look, this is what I want to do, guys, this is what I want to do.

[00:28:35]So, what will I get here in practice? So, look, the absolute value, A1 is Z, which multiplies the common difference, and Z is N. N is 60, right? Because there are 60 terms.

[00:28:55]-1 divided by q - 1, that is, z - 1.

[00:29:07]Look at this, everyone. I can open this like this, see? The absolute value of Z times the absolute value of Z to 60 - 1 over the absolute value of Z - 1.

[00:29:27]Is it or isn't it? And that.

[00:29:31]We're in this together. Beauty. Nice.

[00:29:34]Awesome, master. Okay, now we need to see how we can manipulate this Z to make it more user-friendly, so we can substitute it into the expression we need to determine. Well then, first, everyone, take note of the following. Look, everyone, if I separate part A from part A, part plus from part plus, it's going to look like this.

[00:29:56]1 over a √2 + 1 so √2 x i.

[00:30:09]Or you can think of it this way: Z will be equal to √2 over 2 + √2 over 2 times I. Hmm. Does that remind you of anyone? Remember, right?

[00:30:34]Let's put it into the arcana plan to help identify who is having the most difficulty.

[00:30:40]Okay, folks, this is the real axis, this is the imaginary axis.

[00:30:49]Hmm.

[00:30:51]When √2/2 √2/2, look, partial, imaginary part, both are equal to √2/ 2.

[00:31:00]So, look, part a here, imaginary part here, look, both equal to √2/2. Notice, right, that our argument, look, will be the 45th, look, because I have part a and imaginary part equal.

[00:31:23]And that's it. So I thought to myself, look, here, √2/2, here √2/2. That's going to be a diagonal of a square, right? It's going to be the side times √2, it's going to be √2 / 2 times √2, that's going to give 1. Hmm. Okay, I understand.

[00:31:45]So, folks, when I look at the letter Z, how can we write the letter Z, master?

[00:31:55]R is 1 times the x of 45, which is the same as pi/4.

[00:32:08]So you see, the absolute value of Z is 1, right?

[00:32:14]I understand, Baltar.

[00:32:18]I understand, master. So, what happens now with this game? Let's see.

[00:32:28]Look at our expression, folks, the absolute value of Z is 1. So notice that this guy is about to die. Why, Balta? Because he is one. Therefore, a multiplicative identity element is irrelevant.

[00:32:42]OK? Nice. Show.

[00:32:46]Let's think about z raised to the power of 60 - 1.

[00:32:50]What would z raised to the power of 60 - 1 be, everyone?

[00:32:56]Let's think about it.

[00:32:57]Look, if I'm thinking about z raised to the power of 60 - 1 and then substituting it into the absolute value, first let's think about z raised to the power of 60.

[00:33:07]Now we're going to use the first law of a tree that I told you about. If I want to raise zero to the power of 60, who do I do? R is 1 raised to the power of 60 times the x of pi / 4 x 60. Is n't that what I taught you today? Yes, Balta. Hmm, I understand. So I'll have it like this, see? Z raised to the power of 60 = 1 raised to the power of 60, which is 1 itself. What does this give us? Hey, take a look at the draft with me, guys.

[00:33:39]Pi x 60 / 4.

[00:33:42]This will give you 15, right? 15 pi.

[00:33:46]Yes or no? 15 pi. Balta perfect. Who is 15 pi?

[00:33:52]15 pi will be 180 x 15, which will equal zero. No, let me use the calculator here so we can save screen time. Just a minute.

[00:34:06][Clearing throat] X 15. That's going to be 2700.

[00:34:23]So, look, this is going to be 2700.

[00:34:29]But who is 2700 on the first lap?

[00:34:31]Let's divide here by 360.

[00:34:35]2700 divided by 360 equals 7.

[00:34:42]So, 0.42 carries over 4. 21 + 4.25 here equals 180. In other words, 2700 is congruent to pi, folks.

[00:35:00]So, who's this going to be? It will be the cosine of pi plus the i that multiplies the sine of pi.

[00:35:13]However, the sine of pi is zero. Hmm.

[00:35:17]So, what exactly is z raised to the power of 60, folks? The cosine of pi. What is the cosine of pi? -1.

[00:35:26]Is it or isn't it?

[00:35:30]Is it or isn't it?

[00:35:32]Z raised to the power of 60 is basically -1.

[00:35:36]So, what will my numerator look like now?

[00:35:43]-1 -1 will give -2 inside the module baltar is 2.

[00:35:48]Now we need to work with this guy here. We still need to work on the Z-1 module. It's almost finished. It's almost over.

[00:36:00]Come on, everyone. First, huh?

[00:36:03]First, let's remember that Z is this guy right here.

[00:36:12]Guys, what's Z-Men all about? Z-1 is going to be this, look.

[00:36:20]It will be √2/2 - 1. I combined partial partial + √2/2 x i. If I want to know the magnitude of this, what do I do? The word "baixo" gives you the square root, right, of the squared part plus the squared imaginary part. Simple as that. All good?

[00:36:46]So, what will our absolute value of Z + 1 look like?

[00:36:51]Part A squared will give the square of the first. It will be, look, √2/ 2² which is 2/ 4, it will give 12 minus twice the first times the second plus the square of the second. This is the squared part. Plus the imaginary part squared √2/ 2² will give 1/ 2.

[00:37:19]That is, teacher, this one here will be absolutely equal to the square root of 2 + equals 1. 1 + 1 equals 2.

[00:37:32]Together, [clearing throat] of course we are.

[00:37:35]Great, awesome! So, this is our absolute value sum, obviously, right, everyone? Oh, what will the value of n be, from 1 to 60 of z raised to the nth term inside the absolute value, guys? The top digit was 2, the bottom digit was √² 2 - √2. Then you can rationalize, right?

[00:38:05]But since this is discursive, it's okay to leave it like this, okay?

[00:38:08]resolving.

[00:38:12]Phew, we're in this together. Any questions, any doubts, or is it clear to everyone? Guys, my point of view is difficult, okay?

[00:38:31]Difficult.

[00:38:35]Beauty?

[00:38:38]Two tough questions, right? Did you understand? Do you have any questions, or can I take the next one?

[00:39:03]Please tell me, guys.

[00:39:13]You guys aren't saying anything. I'll get the next one.

[00:39:25]Perfect. Let's move on to the next one. Okay, so here 's the thing, huh?

[00:39:44]The sum of the roots of the equation z³ + z² - absolute value of z² + 2z =, where z belongs to the set of complex numbers, is equal to what? Hey there!

[00:40:04]Here, folks, we have to remember the following property: One of the most important properties of complex numbers is that the product of a complex number and its conjugate is always equal to the square of its absolute value.

[00:40:19]Beauty? So, look, whenever you make the Z sign, wow, something happens in your eye.

[00:40:24]Sorry, everyone. Whenever you multiply Z by Z bar, this will be exactly equal to the absolute value of Z squared.

[00:40:37]Rain or shine. To begin, I will rewrite this equation as follows. Look closely, then. Z³ + Z².

[00:40:50]Instead of the absolute value of Z², I'll put it like this: Z x Z bar + 2Z = 0. Perfect, master. I understood.

[00:41:04]Here I can highlight the Z.

[00:41:09]So, I'll just say that factoring out z³ divided by Z gives Z, which gives Z, Z² divided by Z gives Z, Z bar divided by Z gives Z, bar 2 divided by Z gives 2.

[00:41:30]Folks, the product of two things equals zero, meaning one or the other will be zero.

[00:41:38]So, here I already have a first root.

[00:41:42]I already have a Z1, just like Zero, right? That when this guy wants to be personal, or now I have to think about the other partner.

[00:41:55]So let's go to him, then. Z² + Z + Z bar + 2 equals 0. And here, folks, let's make a restricted equation for Z.

[00:42:15]Let Z equal + B. So how does this work then, folks? A + B² will give a² + 2a bi +². So it will be, like this, -b².

[00:42:41]Hmm, I understand. Baltar plus Z A + B + Z A - BI + 2 SN equals 0. Hmm, I understand. Now cancel out bi with -bi. Perfect. So here I have, let's put together part by part, imaginary part by imaginary part. I'll have the following, then.

[00:43:21]Look at this, everyone.

[00:43:23]a² - b² + 2a + 2.

[00:43:35]Okay? I put together all the real parts plus 2 AB. What should this be equal to?

[00:43:50]equal to zero.

[00:43:53]We're in this together. Beauty? Awesome! Or perfect.

[00:43:59]So, folks, notice what needs to happen.

[00:44:04]C = 0. What does this mean?

[00:44:06]That 0 + 0i.

[00:44:11]For that to happen, I need you to... Let me see if there are any messages here in the chat.

[00:44:16]All very well. Ah, great. Now that I've seen your message. How much do I need both part 'a' and part 'imagined' to be worth, guys? Zero.

[00:44:26]Is it or isn't it? So this has to be zero.

[00:44:32]This has to be zero.

[00:44:37]We're in this together. Beauty.

[00:44:40]He wants the sum of the roots, right? Okay.

[00:44:46]Let's go find our next ones, and let 's go, everyone! I'll start by analyzing this 2ab here. Just look.

[00:44:56]First, 2ab has to be zero.

[00:45:02]For this to be true, A must equal 0 or B must equal 0, or both must equal zero. I don't know. Let's see. Let's go. In the case where A equals zero, in this case, folks, ITA 2004, okay?

[00:45:29]In the case where A equals zero, note that Acer already guarantees that I will zero out my imaginary part. Now I need to reset my real money. If 'a' is zero, oh, how will my part of the wing be, guys? A quad 0 2a 0. So I'll have to have - b a quad + 2 being zero. In other words, I'll need b² to equal 2. So, what value of b will I need?

[00:46:08]More or less the square root of 2. Hmm.

[00:46:12]So I've found two more roots now.

[00:46:15]Look here. Z1 is zero. I'll put it here in the corner for you all. Z1 equals 0. Z2, remember, everyone, that Z is A + BI, right?

[00:46:26]I'm assuming A = 0. So there won't be an A.

[00:46:31]bi √2, Z3 will be -√2i.

[00:46:43]Ready.

[00:46:45]This is done with A being zero. What if I do, master with a B?

[00:46:55]So, how does this look? If B = 0, it will look like this. a² eh - b² vai 0 a + 2a + 2 = 0. You'll have to solve it here.

[00:47:16]Delta will be b² 4 - 4ac.

[00:47:23]Hmm. It'll be -4. In other words, folks, for Big, it's a real equal that satisfies this, right? Therefore, it doesn't make sense to be the Big. So, we're actually only going to have these three roots, see? Zero, √2i, and -√2i.

[00:47:40]If he wants the sum of the roots, well, just add them all up, right? It's going to be a zero, guys.

[00:47:46]Key template.

[00:47:50]We're in this together.

[00:47:55][snoring] Beautiful or not?

[00:48:13]Wow, but if I remember correctly, that alpha template.

[00:48:21]Let me see if I made any mistakes in the calculations here, guys. Just a minute, okay? Because I remember doing it before, giving her the alpha signal, but we're on the right track. We're on the right track. Just a moment. Hold on a second. Oh, here I've swapped the Z in the Z² modulus. I made the switch to ZZ Z slash. So it became Z³ + Z². Here's the damn thing, look.

[00:48:49]Damn, that's a mistake right at the beginning of the question, man. That's messed up, you know? Here, folks, it's minus the absolute value squared. I overdid the z-slash; it should be less z- slash, okay? Oh, beautiful?

[00:49:05]Let's correct it, but this time more briefly, okay, everyone? This was supposed to be one less, look. That 's messed up, huh?

[00:49:15]You can be quicker, folks. We've got the idea. Oh, that's a powerful idea, you know? She appears. Many people reach this final equation and don't know what to do.

[00:49:30]So, let's take this opportunity to review things here. Look, this was supposed to be less, not more.

[00:49:37]Less.

[00:49:39]OK. Beauty. Less. So, let's go. I'm going to highlight the Z.

[00:49:45]Then I'll have, look, Z³ times Z will give Z².

[00:49:49]Z² divided by Z will give Z minus, not plus. Zz/ divided by Z will give Z bar 2 divided by Z will give 2 = 0. So, look, we've already arrived at a first root, right? I arrive at Z1 = 0.

[00:50:08]Now let's go in search of the next ones.

[00:50:09]Therefore, I'm going to think of z² + z - z bar + 2 = 0. Let's go.

[00:50:25]Hey, awesome, Guilherme! It's a real gem, you know?

[00:50:29]Good papyrus, bro. We're in this together.

[00:50:32]So, it's going to stay like this, see? Let's talk about it the same way that Z is the best A.

[00:50:38]Be careful not to get the calculations wrong, everyone.

[00:50:40]So, the z²ado will look like this, see?

[00:50:42]The square of the first plus twice the first times the second plus the square of the second. It's going to be -b², right? Because it will become b² i², it will give - b² plus z + bi - z bar. This is the z-bar, folks. It's the men bi, I want him less, right?

[00:51:05]So it will be - a - bi will give + bi.

[00:51:11]Everything changes, oh. + 2 equals 0. So, the same principle applies.

[00:51:20]Notice that now it's going to cancel out whoever is A with A, right?

[00:51:23]Let's combine part a with part a, it will give a + 2.

[00:51:36]Part plus nothing, which is part plus nothing. It's going to stay like this, see?

[00:51:40]Well, sometimes "i" here is a plus, right, guys?

[00:51:57]Another one. This here being equal to zero.

[00:52:02]For that to happen, it's 0 + 0i. And here's that final little tip I shared with you all. So here's what I'm going to have, look. These two expensive ones have to be zero.

[00:52:12]This one here has to be worth zero.

[00:52:17]And this one here, likewise, also has to be worth zero.

[00:52:21]Beauty? So, let's go. Let's look at the second guy, he's cuter, he's more elegant. So, let's go. If I think like that, 2ab + b equals 0, then please don't divide the equation by b, for God's sake. You don't know if b is zero, don't divide, factor it out. You'll have, like this, 2a + 1 = 0. So, for this to happen, folks, I can have b = 0 or a = -, right? That's the idea?

[00:53:02]Oh my God, people.

[00:53:05]Here, look, uh, bi, it's going to be 2B, okay?

[00:53:09]Please. I'm glad I realized the mistake. Now, 2ab + 2b equals 0. Look, we can divide everything by 2, so we'll have ab + b = 0.

[00:53:31]b multiplied by a + 1 = 0. Now that's correct.

[00:53:37]So I'll have B equal to A = -1.

[00:53:45]Let's go. If B equals 0.

[00:53:48]First case, folks, look at this. If B equals zero, what will our A be? Then you look at this expression here, see? I'll have a quad minus b², which will give 0 + 2, equaling 0.

[00:54:12]Therefore, I'll have a² = -2.

[00:54:18]Consequently, my 'a' will be more or less 2i.

[00:54:27]Okay, that's the beauty.

[00:54:42]Given that b = 0, what will my z values ​​be?

[00:54:48]Here it is. Eh, √2 x i - √2.

[00:55:00]Now let's consider a = -1.

[00:55:04]If I consider a = -1, what will our b be? Look at that equation again. It's going to stay like this, see?

[00:55:13]- 1² will give 1 - b² + 2 = 0.

[00:55:21]Perfect. So I'll have b² here, which will be equal to 3.

[00:55:33]Consequently, what about the good side, guys? Approximately √3.

[00:55:43]So then, who will get our Z?

[00:55:45]Remember that from a to a.

[00:56:23]Now, folks, just add them up and be happy. Oh, when I add everyone up, this guy cuts out with this one, this guy cuts out with this one. Who am I, basically?

[00:56:34]-2. Now that's the correct answer, alpha.

[00:56:39]We're in this together. Beauty.

[00:56:45]OK.

[00:56:48]Awesome or not, guys? Any questions, any doubts?

[00:56:52]Can I take the next one?

[00:56:57]The next one's punk, huh?

[00:57:02]The next one is punk. I'll just grab a glass of water real quick.

[00:57:15]The next one is rot.

[00:57:20]The classic problem of complex numbers.

[00:57:23]Guys, think for a second while I go get some clean water. Just a moment. I will be back in a moment.

[00:57:56]Cha.

[00:58:17]M.

[00:58:40][snoring] Alright, let's go for another super classic idea so we can stay sharp on this stuff.

[00:59:03]Question five goes like this: "Look, everyone, let alpha and beta belong to the set of complex numbers, such that modulus alpha modulus beta = 1 and modulus of α - βa = √2.

[00:59:21]Then, α² + beta² is equal to what?

[00:59:28]Well, everyone, this is a classic problem, okay? When you have problems like this, where you equate moduli and work with the relationship between complex numbers, it's interesting to work with the trigonometric form.

[00:59:42]Oh, B, how am I going to work with the trigonometric form if I don't know the argument, I only know the modulus? Exactly. You'll have to assume.

[00:59:50]Okay? Great.

[00:59:53]So, I'll have the following. I'll say that my alpha, observe here how the modulus is one, when I do this, R 6 theta, its modulus is one, only the s will remain, right? So, I'll say here, my alpha is basically what, everyone?

[01:00:13]x of theta 1, that is, I'll say that my alpha is basically the cosine of t1 plus the i that multiplies the sine of theta 1.

[01:00:33]And beta, master, beta, the same thing, our rho will be one, right? So I wo n't worry about it. So it will now be the cosine of theta 2 plus the i that multiplies the sine of theta 2. Okay. I understand. Great. Adam.

[01:00:59]Good.

[01:01:00]With that, when I do alpha minus beta, when I do alpha - beta, what will I have in practice?

[01:01:16]Look, partial with part.

[01:01:19]cosine theta 1 minus cosine theta 2 + sine theta 1 minus sine theta 2 x i. Isn't that right, folks? α beta.

[01:01:44]Notice that I obtain a new complex number.

[01:01:48]And then see that, very important. Look, folks, see how we actually do it, huh, folks? Uh, modulus of a complex number, square root, part a squared plus part imaginary squared.

[01:02:07]So, I'm going to do the following. Look. If the modulus of alpha - beta is equal to ra squared [clearing throat] 2, let's think about this modulus squared, that is, taking the square root, it will look like this: cosine of theta 1 minus the cosine of theta 2 squared plus sine of theta 1 minus the sine of theta 2 also squared. What will this be equal to? Let's put √2 squared.

[01:02:50]Okay? Let me see if there's anything here in the chat.

[01:02:54]My YouTube closed. Done. Standard.

[01:02:57]Solving, I have the square of the first minus twice the first times the second plus the square of the second.

[01:03:16]Okay?

[01:03:18]Solving the second square of the first minus twice the first times the second plus the square of the second.

[01:03:38]This equals 2. I only developed it on the left.

[01:03:41]However, my friends, see that cosine² times sine² times theta 2 equals what?

[01:04:11]One.

[01:04:14]Hmm. Okay.

[01:04:17]Minus. Then I'll factor out the two, look.

[01:04:23]Cosine theta 1, cosine theta 2 + sine theta 1, sine theta 2. That's what's left, right? That will be equal to 2. Wow, master, what luck, huh?

[01:04:40]1 + 1 2. Look, that will leave zero, isn't that right?

[01:04:50]Hi, little master. When I look at these parentheses, my eyes pop.

[01:04:55]The itching cosine theta 1, cosine theta 2 + sine theta 1, sine theta 2. Does it remind you of anything? It reminds you of B. It reminds you of the cosine of theta 1 - theta 2, what will that be? Zero. Why zero going back? Because, folks, 2 times that is 0 plus 2 dividing. Look at this important conclusion. The cosine of theta 1 minus the sine of theta 1 is absolutely equal to zero.

[01:05:32]Will this help me in any way to solve the problem? Obviously yes.

[01:05:38]How, master? Let's see.

[01:05:41]Okay everyone, look at this.

[01:05:46]He wants the value of alpha² + beta², right? Yes or no? But look at this, everyone. We saw that alpha is what? The 6 of theta 1.

[01:06:05]We saw that beta is what, everyone? The 6 of theta 2.

[01:06:12]So, when I do α² + beta²a², I'll have t² of 2 ta1 plus when I do beta², I'll have t²² of 2 ta.

[01:06:42]Isn't that right?

[01:06:56]Okay? So that's what I want to determine.

[01:07:00]But look what happens. What luck.

[01:07:04]What is the sig 2 t1, my friends? This is like this. Look, everyone. It will be the cosine of 2 theta 1 + the i sine of 2 theta 1 [clearing throat] plus which is the 6 of 2 ta everyone cosine of 2 ta plus the i of the sine of 2θ Let's do the following, let's join part a with part a, part Imaginary part with imaginary part.

[01:07:45]Guys, and remember this with me, for God's sake, okay? Project vacation, people.

[01:07:52]Sine of P plus sine of Q. What is this?

[01:07:58]This is 2 sine of P + Q / 2 cosine of P - Q / 2.

[01:08:14]And if I think of the cosine, putting cosine of P + cosine of Q as 2 cosine of P + Q / 2 cosine of P - Q / 2. Is n't that right?

[01:08:38]Look how the question ends now, okay?

[01:08:40]I'm going to add this guy with this guy, see, partial with partial, sum of cosines 2, which multiplies the cosine 2 t1 + 2θ2 div by 2 will give theta 1 + t2 times the cosine of the difference. 2 1 - 2θ2 div by 2 will give theta1 - t2. I don't know if you've noticed what's going to happen.

[01:09:12]I added the sines.

[01:09:15]Now let's add the cosines, Excuse me. Now let's add the sines. It'll stay there, right? That multiplying 2 times sine 2 t1 + 2 t2 divided by 2 gives theta 1 + theta 2.

[01:09:35]Hold on, man. I'm finishing up the lesson here quickly. Sorry, everyone.

[01:09:39]Theta 1 + theta 2 times the cosine of the semi-difference 2 t1 - 2 t2 divided by 2 gives theta1 - theta 2. A little break.

[01:09:54]But we concluded that the cosine of theta 1 minus theta 2 will be 0.

[01:10:01]Zero, master.

[01:10:03]Zero.

[01:10:05]What's the result? Zero.

[01:10:09]The question is over.

[01:10:11]Bravo is our answer key. We're all in this together, everyone. Any questions, doubts?

[01:10:22]Classic.

[01:10:24]Let me take a look at the chat.

[01:10:34]Okay, everyone.

[01:10:42]Alright or not, guys? Speak up, please.

[01:10:56]Oh, man, if you don't say anything, I'll end the live stream then. So, thanks, everyone. Stay with God. Have a good rest everyone. See you next week. Thanks, guys. Bye, bye. Have a good weekend. I'm off.