In Exercises 95–99, perform the indicated operations and write th...

Question

In Exercises 95–99, perform the indicated operations and write the result in standard form.(1 + i)/(1 + 2i) + (1 - i)/(1 - 2i)

Accepted Answer

Welcome back, I'm so glad you're here. We've got this complex expression of complex numbers. This first fraction here, three plus I divided by three plus five I. And that's added to another fraction 3 - divided by 3 -5 i. And we are to get this whole thing in simplified and into standard form. Standard form of a complex number is A plus B. I. And so we know that when we are simplifying complex numbers with fractions, we want to get the eye values out of our denominators and how do we do that? We recall from previous lessons that to get an I. Value out of the denominator, we're going to multiply it by its complex conjugate. The complex conjugate is one in which we take our A values and they remain RB values and our values remain. And we're just going to flip the sign, you might be starting with a negative and going to a positive, you might be starting with a positive and going to a negative. So in this case we actually have one of each. We've got a positive complex number in the denominator and we have a negative complex number in the denominator. So we will multiply them by the complex conjugate is in this case for this three plus five I we keep the A value which is three. We keep the B value which is five, we keep the eye and we flip the sign to negative and what we multiply to the denominator. We have to multiply by the numerator two because in effect anything over itself is just one. So we're just multiplying it by one. Now we take a look at this one on the right hand side three minus five I. And we're going to keep the A. Value which is three. Keep the B. Value which is five and I. And flip the science now that's positive. And what we multiply to the denominator. We also multiply to the numerator so now we could just go ahead and start all that fancy foiling. But if we recall from previous lessons when we multiply a number by its complex conjugate a complex number by its complex conjugate. We are going to get a result of a squared plus B squared. Well this is really exciting because here we've got three plus five I times three minus five I occurring in both of our fractions. And so for both fractions we're going to get okay that a value is three so three squared plus the B. Value. That's five squared. So we've got nine plus 25. We've got 34 for both fractions. Denominators wasn't that kind of them? 34 are both denominators and we do have to go ahead and foil for the enumerators. So the first we got three times three for this first one that's nine. Outsides three times I plus three I insides negative five times three is minus 15 I and last negative five I times I is minus five I squared. Now we'll do this second fraction first three times three again is nine outside three times five I is plus 15. I insides negative three times or negative I times three is minus three I and last negative I times positive five I is minus five I squared. Again. We've got a couple of like terms going on for each numerator and then we can recall from previous lessons that by definition I squared equals negative one. So for both of these we've got a negative five times a negative one. So let's rewrite some of that with simplification. We've got a nine plus three I minus 15 I is going to be a negative 12 I so we can write nine minus I And we've got a negative five times in negative one. All over 34 plus again nine positive 15 I minus three I is plus 12 I minus five times negative one. All over 34. Well this is great. As we said before that 34 is the denominator for both fractions which means we can combine these two And we have 34 in our denominator and we're just going to add straight across so we can look and notice that we've got a negative 12 I in a positive 12. I those two are just going to cancel out who and in both cases are negative five times negative one. Well that's just going to be plus five. So bringing that over we've got nine plus five plus nine plus five and nine plus five is 14, so 14 plus 14, which is 28 over And those are both divisible by two. So we've got 14 back down to 14 again over 17. And that's as simplified as it's going to get we look at our answer choices and that matches with answer choice b. Well done, we'll catch you on the next one.

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In Exercises 95–99, perform the indicated operations and write the result in standard form.(1 + i)/(1 + 2i) + (1 - i)/(1 - 2i)

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