in-26-37-find-each-product-32-i33-i

Question

In $26-37,$ find each product.
$$
(3+2 i)(3+3 i)
$$

Heather Zimmers · Accepted Answer

to multiply complex numbers. We use the same method we would use to multiply by no meals, which is the foil method. So we multiply the first and we get nine. Multiply the outsides and we get nine. I multiply the insides and we get six I and multiply the last and we get six. I squared. Now remember that I squared is negative one. So we&#x27;re going to change that term. And we&#x27;re also going to add the like terms. Nine I and six I. So we have nine plus 15 I plus six times negative one. Let&#x27;s simplify six times. Negative one. That&#x27;s negative. Six. So we have nine plus 15 I minus six. Now we can combine the nine in the minus six. So we have three plus 15 I

Jeremy Strong · Answer

for this problem Where asked the multiple of the complex number seven plus times the complex number two plus three. Now using the foil method, our first terms are seven and two. So seven times too will give us 14. Our outer terms are seven and three on multiplying. These thing ever is going to give us 21. Uh, our inner terms are the ah, and the two multiplying them together will give us plus Thio. And then our last terms are I and three I. And when we multiple these together, it is going to close three square. All right now, three squared may be rewritten as three times negative one because I square it is the spirit of negative one squared. So we&#x27;re just left with negative one. Now we have 14 and we have monness three, because three times the negative one, it&#x27;s going to give us minus three. And then we have 21 I plus two I as our imaginary component, so 14 minus three is going to give us 11 and then 21 I plus two is going to give us 23 i for a final answer of 11 plus 23

Jeremy Strong · Answer

in this problem, we arrest most by the complex number five Modest three out by the complex number five plus three to begin. You may notice that these two complex numbers are complex contra hits. But for now, we will ignore this and we will continue the multiplication as we normally would. So to begin, we have to foil this expression. So we have five times, which is gonna give us 20 five. Our outer components are plus three and plus five, which is gonna give us plus 15. I, our inner components are modest area and positive five to give us monos 15 I and our outer components are modest. Three out and positive three out to give us modest non, uh, squared. Now Ah, squared is a very negative one squared. So this actually one to cancel to give us a negative one. Now, since we have a mon isn on multiple about modest one, this term will cancel out to go. It&#x27;s a plus. None. Now we have 25 plus nine to give us 34 and then we have plus 15 out modest 59 which is going to give us plus zero I and that is going to give us a final answer of simply 34. Now it is worth mentioning that when you multiply complex number by its complex congregate, there actually is a simple formula for doing so. The formula is a squared plus B squared where a it&#x27;s the real component and B is the imaginary component. In this case, a equaled five equals five and B equals three. So by the formula, our answer should be of squared plus three squared, which is equal to 25 plus nine, which is going to give us 34 which is the answer we got.

Heather Zimmers · Answer

to multiply complex numbers. We can use the same method we would use to multiply by no meals, which would be the foil method. And we could just jump right in and start using the foil method now. But I&#x27;m going to be a little bit more strategic than that. I&#x27;m going to take a look at the second factor and noticed that we could factor 1/10 out of both terms. So I&#x27;m going to think of this as three minus I times three plus high times 1/10 and I&#x27;m going to start by multiplying the three plus I in the three minus side together. So using the foil method, we have three times 39 for the 1st 3 times I for the outside three i For the inside, we have negative three I and for the last we have minus I squared and that is still all multiplied by 1/10. So let&#x27;s take a look at that. We have a positive three I and a negative three I those at 20 And don&#x27;t forget that I squared equals negative one. So we have nine minus negative one times 1/10. Well, that&#x27;s the same as 10 times 1/10. So the answer is one

Heather Zimmers · Answer

to multiply complex numbers. We use the same technique we would use to multiply by no meals, and that would be the foil method. So we multiply the first and we get 12. We multiplied the outside&#x27;s when we get negative. Eight. I we multiply the insides and we get negative. 15 I and we multiply the last and we get plus 10. I squared. Now let&#x27;s combine like terms. Negative eight I and negative 15 I and we have 12 minus 23 I and remember that I squared is negative one. So we have plus 10 times negative one. So if we simplify that, we have 12 minus 23 I minus 10. And let&#x27;s go ahead and add those like terms together, and our final answer will be to minus 23 I

Heather Zimmers · Answer

to multiply complex numbers. We use the same method we would use if we were multiplying by no meals. And that would be the foil method. So we multiply the first and we get negative, too. Welcome by the outsides, and we get negative. Four. I multiply the insides and we get negative one I and multiply the last time we get negative to I squared. Let&#x27;s combine the inside and outside terms so we have negative two minus five I and then remember that I squared is negative one. So we have minus two times negative one. So we&#x27;ll simplify that. We have negative two minus five I plus two and we can add the negative to in the plus two and they just give us zero. So in standard form we have zero minus five I

Problem

In $26-37,$ find each product. $$ (2-3 i)^{2} …

Problem 27 Medium Difficulty

In $26-37,$ find each product.
$$
(3+2 i)(3+3 i)
$$

Answer

$3+15 i$

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Absolute Value - Example 1

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$3+15 i$

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Ann Xavier GantertAlgebra 2 and Trigonometry

Grace H.

Maria G.

Kristen K.

Michael J.

Ann Xavier Gantert

Algebra 2 and Trigonometry