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.

(8 + 9i)(2 - i) - (1 - i)(1 + i)

Accepted Answer

Hello everyone in this video we're going to be looking at this practice problem where we're going to be simplifying the expression six plus five I times three minus I minus three minus I. Times three plus I. And we want to make sure to express our answer in standard form. So you can see here that we're dealing with complex numbers because we have these imaginary components or the I. Components in all of the terms or in all of the parentheses present in our expression. So I'm going to try to break it down to make it easier to simplify. I'm going to deal with the left hand side of the subtraction and the right hand side of the subtraction separately. So if I look at the left hand side of the subtraction, I have six plus five I Times 3 -1. And when we multiply complex numbers, it's very similar to multiplying polynomial. So we can follow the foil method Where we do the first terms six times 3, 18. The outer terms six times negative I is negative six. I The inner terms five i times 3, 15 I and then the last terms five I times negative I negative five I squared. And we can combine these middle terms and that leaves us with 18 plus nine I minus five I squared. So now I can do the same to my right hand side. My right hand side is 3 -1 times three plus I. And if I do the foil method again I have three times times I is three I negative I times three is negative three I and negative I times positive I is negative I squared. And you can see that these middle terms cancel out because we have a positive three I and a negative three I. So I'm left with just nine minus I squared. And I squared is a bit of an odd term here because you see that we have an iron and I squared but I is equal to the square root of a negative one. So if we have an I squared, that is the square root of negative one times the square root of negative one. And that gives us -1. So notice how by squaring I goes back to being a real number negative one. So I'm going to go back to my expressions and wherever I see I squirt I'm going to substitute in the negative one. So my first or the left hand side becomes plus nine, I minus five times negative one. And if I do that I have negative five times negative one which is positive five and 18 plus five is 23. So my left hand side becomes 23 plus nine nine. And I can do the same to my right hand side where I have nine minus I squared and it becomes nine minus negative one, which leaves me with nine plus one. So 10. So now I've simplified both the left hand side and the right hand side of that subtraction. So I'm going to recombine them into the expression where I have my first term 23 plus nine nine minus My second term, which turned out to be just 10. So if I solve for this, I'm left with 13 plus nine nine and we can check that it's in standard form because recall that standard form is the real component, followed by the imaginary component. And in this case that's how we have written. So this becomes our final answer. And you can see that this aligns with answer choice A 13 plus nine I So hopefully this video helped and see you next time.

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