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 + i)^2 - (3 - i)^2

Accepted Answer

Hello everyone in this video we're going to be looking at this practice problem where we need to simplify the expression four plus I squared minus one minus I squared and we want to make sure that our answer is expressed in standard form. So you can see that we're dealing with complex numbers here because we have numbers that have both imaginary and real components. And there's also two sort of factors. We have the four plus I squared and then we have the one minus I squared. So in order to simplify this problem, I'm going to solve for these factors separately and then we'll combine them at the end. So I have four plus I squared. And I'm going to expand the exponent on that. So it becomes four plus I squared is equal to four plus I times four plus I. And whenever we multiply complex numbers, it's very similar to multiplying polynomial. So I just want to make sure that each factor in the first set of parentheses gets multiplied to the each factor in the second set of parentheses. So at four times four is 16. 4 times I is for I I times four is 4 I and then I times I is I squared. And I can combine these middle terms. So I'm left with 16 plus eight. I plus I squared. And if I move on to my second factor, I have one minus I squared. I'm going to expand that as I did before. So it becomes one minus I times one minus I. And if I multiply I have one, times one is 11 times negative I is negative, I negative I times one is negative I negative I times negative I is positive I squared. And I can combine those middle terms again. So I'm left with one minus two. I plus I squared. And whenever you're dealing with complex numbers, it's important to remember that a complex numbers represented by I which is equal to the square root of negative one. So if I do I squared, I have a square root of negative one, times square root of negative one. And whenever you have two square roots that are being multiplied by each other and they have the same inside you get the inside back. So I'm left with I squared is equal to negative one. So you can see that by squaring I were left with a real number negative one. So I'm going to go back to my expressions and wherever I see I squared, I'm going to replace with negative one. And if I do that, my first expression becomes 16 plus eight. I plus negative one. And you can see that the 16 and the negative one can be combined. Certainly use me with 15 plus eight. I and I can do the same for my second expression. So I have one -2. I plus -1. I can combine my one and my negative one and I'm left with zero. So it actually just leaves me with this middle term of negative two. I So now I've sold for both components of my equation and I can rewrite it. So my original equation was four plus I squared minus one minus I squared. But I determined that four plus I squared is 15 plus eight. I and I determined that one minus I squared is negative two. I. So now I can distribute this negative here and it becomes 15 plus eight. I plus two. I. And then I can combine these last two terms to leave me with 15 plus 10. I. And we can verify that this is in standard form because recall that standard form is the real component followed by the imaginary component. And in this case we have the real followed by the imaginary. So this becomes my final answer. And you can see that this aligns with answer choice D. 15 plus 10 I. So hopefully this video helped and see you next time.

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