SAT Math Part 15 - Simplifying Complex Imaginary Numbers - By The Organic Chemistry Tutor
Transcript
00:00 | 52 . Which is the following is equivalent to the | |
00:04 | expression shown below . So now we're dealing with imaginary | |
00:08 | numbers . Let's begin by distributing the negative sign . | |
00:13 | So this is eight plus five I minus four I | |
00:18 | squared . And then the two negative signs will make | |
00:22 | the next term positive . So this is gonna be | |
00:24 | plus seven I . And then plus five . Now | |
00:30 | let's combine like terms . So we can add these | |
00:33 | 28 plus five is 13 . Next we can combine | |
00:39 | size I and 79 That's going to be 12 . | |
00:44 | I . Now what is I squared I . Is | |
00:51 | the square root of negative one I squared . Is | |
00:54 | going to be the square root of negative one squared | |
00:57 | which is negative one . So negative four I squared | |
01:02 | is negative four times negative one which becomes positive for | |
01:08 | Yeah . So now let's add these two numbers 13 | |
01:11 | and four 13 plus 4 17 . So that's we | |
01:17 | have the answer 17 plus 12 . I answer choice | |
01:22 | C . Number 53 . What is the product of | |
01:28 | the complex numbers three plus four I . & 5 | |
01:32 | -2 . i . The product tells us that we | |
01:38 | need to multiply these two complex numbers . Yeah so | |
01:43 | let's foil . We have 3-plus 5 which is 15 | |
01:50 | . And then three times negative two . I That's | |
01:54 | -6 i . And then four times 5 That's gonna | |
02:00 | be 20 i . And then four i . times | |
02:03 | negative two i . Which is negative eight I squared | |
02:09 | . Now our next step is to combine like terms | |
02:14 | -6 . I plus 20 I . That's going to | |
02:17 | be 14 I and I squared I . Is the | |
02:23 | square root of negative run . So I squared is | |
02:26 | going to be the square root of negative one squared | |
02:30 | which becomes negative one . So negative eight I squared | |
02:36 | is negative eight times negative one . So this becomes | |
02:40 | 15 plus 14 I plus eight . So let's add | |
02:47 | those two numbers . 15 plus eight is 23 . | |
02:54 | So it's gonna be 23 plus 14 I . Which | |
02:58 | corresponds to answer choice D . 54 . Which of | |
03:04 | the following is equivalent to the expression shown below . | |
03:09 | So in this problem we are dividing two complex numbers | |
03:14 | . We're dividing two plus three I By five plus | |
03:17 | 2 . i . So how can we do this | |
03:22 | ? In order to simplify this expression ? We need | |
03:25 | to multiply the top and the bottom by the conjugate | |
03:29 | of the denominator . So what is the conjugate of | |
03:35 | the economy ? It's going to have a five . | |
03:40 | It's going to have a to I . But instead | |
03:44 | of the positive sign , we're going to use a | |
03:47 | negative sign . Now at this point , what we | |
03:52 | wanna do is for you . So let's multiply two | |
03:55 | times five which is 10 and then two times negative | |
03:59 | two . I that's going to be negative four I | |
04:01 | . And then three items 5 Which is 15 i | |
04:06 | . And then three times negative two , Yvette's negative | |
04:08 | six I squared on the bottom . Let's do the | |
04:12 | same thing . So we have five times five which | |
04:15 | is 25 , five times negative two i . And | |
04:19 | then two items 5 . And then to I times | |
04:23 | negative two I squared , which is negative for I | |
04:26 | squared . Next let's combine like terms , We can | |
04:31 | add those two and negative 10 . I plus 10 | |
04:33 | . I will cancel negative four plus 15 . That's | |
04:38 | going to be positive 11 . And we have negative | |
04:41 | six I squared I squared is -1 . So negative | |
04:45 | six times negative one is positive six . Now negative | |
04:51 | four I square . That's negative four times negative one | |
04:55 | . So this will be positive for Yeah . Now | |
05:01 | 10 plus six is 16 And 25 plus four is | |
05:06 | 29 . Now we need to put the answer in | |
05:10 | a plus B format . That's in standard form , | |
05:13 | A . Is a real number , B . I | |
05:15 | . Is the imaginary part of the complex number . | |
05:19 | So we're gonna divide each number by 29 . So | |
05:23 | we're gonna have 16 divided by 29 And then 11 | |
05:27 | divided by 29 times i . So we can see | |
05:31 | that answer choice . D . Is the right answer | |
05:35 | number . 55 simplify What is I raised to the | |
05:40 | to 83 equal to ? Well , there's some things | |
05:46 | that we need to know before we could find the | |
05:48 | answer . I is equal to the square root of | |
05:53 | -1 . And we know that I squared this is | |
05:57 | equal to OK . That to just doesn't look right | |
06:01 | , let's do that again . I squared is negative | |
06:03 | one . I . Q . Is going to be | |
06:06 | I squared times I . So this is negative one | |
06:11 | times I are simply negative I . I . to | |
06:15 | the 4th power is I squared squared . So that's | |
06:21 | negative one squared which is one . So to summarize | |
06:26 | it , here's what we have . I . is | |
06:29 | the square root of -1 . I squared is equal | |
06:33 | to -1 . I to the third is negative I | |
06:38 | and I to the fourth is one . But we're | |
06:43 | going to show the work for every step . Just | |
06:47 | keep that in mind . Mhm . So what we | |
06:51 | can do is break down to 83 into 280 plus | |
06:57 | tree . Next Let's divide to 80 x four , | |
07:03 | 28 , divided by 4 7 . So to 80 | |
07:06 | , divided by four is 70 . So we can | |
07:09 | write this as I to the 4th raised to the | |
07:13 | to 80 . I mean to the 70 power Because | |
07:18 | four times 70s to 80 I . Q . I'm | |
07:21 | gonna write that as I square times I now I | |
07:26 | to the fourth I'm going to write it as I | |
07:29 | squared squared . Yeah , I squared is -1 . | |
07:43 | So this is what we now have . This is | |
07:48 | for those of you who want to show the work | |
07:50 | for every step . Now negative one squared is positive | |
07:55 | one and negative one times I is negative . I | |
08:03 | 1 to the 7th power is still one . So | |
08:07 | this is one times negative I , which gives us | |
08:10 | a final answer of negative I so B is the | |
08:14 | correct answer . |
Summarizer
DESCRIPTION:
OVERVIEW:
SAT Math Part 15 - Simplifying Complex Imaginary Numbers is a free educational video by The Organic Chemistry Tutor.
This page not only allows students and teachers view SAT Math Part 15 - Simplifying Complex Imaginary Numbers videos but also find engaging Sample Questions, Apps, Pins, Worksheets, Books related to the following topics.