In Exercises 65–70, perform the indicated operation(s) and write ...

Question

In Exercises 65–70, perform the indicated operation(s) and write the result in standard form.(2 - 3i)(1 - i) - (3 - i)(3 + i)

Accepted Answer

Hello everyone in this video we're going to be looking at this practice problem where you need to simplify the expression three minus I times two minus four I minus two minus I times two plus I. And in this case you can see that we're going to be dealing with complex numbers because we have an imaginary component the eyes. And in order to simplify it I'm going to break it down into a left hand side and the right hand side. So I'm gonna have to terms I'm gonna say this is A. And then I have a minus here -7. So I just broke it down between the left hand side and the right hand side of that subtraction. So I have my a term is 3 -1 Times 2 - I and then I have my right hand term two minus I times two plus I. So just looking at the left hand term, you can see that we're gonna be doing a multiplication across two complex numbers. And for this it's very similar to polynomial. So we can follow the foil method where we multiply the first terms three times to six. The outer terms three times negative four. I negative 12. I the inner terms negative I times two negative two. I and the last terms negative I times negative four. I positive for I squirt. And you can see that these middle terms can be combined and we're left with 6 -14. I plus four. I squirt. I'm gonna go ahead and do the foil method to my right hand side. So I have two times two is 42 times I is to I negative I times two is negative two I and negative I times positive I is negative I squared. And in this case you can see that the middle terms cancel out because we have positive to I and a negative two I so that becomes zero and I'm left with four minus I squared. And in this case whenever you're working with complex numbers it's important to know that I is equal to the square root of -1. So if we have I squared we have the square root of negative one times the square root of negative one. And when we have those together we have negative one squared. So we can pull the negative one out and we're left with ice squared Is equal to -1. So I can go ahead and substitute a negative one wherever I see I squared. And if I do that to my first term I have six minus 14 I plus four times negative one substituting I squared. And if I do that multiplication, I have four times negative one, negative four and I can combine the six and the negative four, leaving me with two minus 14. I and if I substitute negative one into my second component as well, I'm left with four minus negative one. So that becomes four plus one. So five. So now I've simplified my left hand side and my right hand side to their most basic and now what I'm going to recombine them so I have A minus B. And my A. Term is two minus 14 I minus my B. Term, which is just five. And if I combine them I have a negative 3 -14 I. And this is already in standard form because I have my real number first followed by my imaginary component or imaginary number. So this becomes my final answer And you can see that negative 3 -14. I aligns with answer choice b. So hopefully this video up and see you next time.

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In Exercises 65–70, perform the indicated operation(s) and write the result in standard form.(2 - 3i)(1 - i) - (3 - i)(3 + i)

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