Video: Simplifying Expressions Involving Powers of Complex Numbers

Simplify (1 βˆ’ 𝑖)Β³ βˆ’ (1 + 𝑖)(1 βˆ’ 𝑖) + (1 + 𝑖)Β³.

03:14

Video Transcript

Simplify one minus 𝑖 cubed minus one plus 𝑖 times one minus 𝑖 plus one plus 𝑖 cubed.

To answer this question, we’re simply going to begin by distributing each set of parentheses. Let’s start with one minus 𝑖 cubed. Now, we could use the binomial theorem to work out one minus 𝑖 cubed. But actually, it’s straightforward enough to write this as one minus 𝑖 times one minus 𝑖 times one minus 𝑖 and then multiply each pair of parentheses in turn. We’ll begin by multiplying the first term in these first two parentheses.

One times one is one. We’ll now multiply the outer terms. One multiplied by negative 𝑖 is negative 𝑖. We multiply the inner terms and we get another negative 𝑖. And then, we multiply the last terms and we get 𝑖 squared. Now, in fact, 𝑖 squared is equal to negative one. So, we change 𝑖 squared to negative one. And then we remember, of course, that we’re going to be multiplying all of this by one minus 𝑖. Before we do though, let’s see if we can simplify.

Well, one minus one is zero. So, we have negative two 𝑖 times one minus 𝑖. To distribute these parentheses, we simply multiply negative two 𝑖 by one and negative two 𝑖 by negative 𝑖. That gives us negative two 𝑖 plus two 𝑖 squared. But of course, we said that 𝑖 squared was equal to negative one. So, we get negative two 𝑖 plus two times negative one which is negative two minus two 𝑖. Let’s repeat this process with one plus 𝑖 times one minus 𝑖.

Again, we multiply the first term in each expression. To get one, we multiply the outer terms. That’s negative 𝑖. We then multiply the inner terms, which is 𝑖. And finally, we multiply the last term in each expression. 𝑖 multiplied by negative 𝑖 is negative 𝑖 squared. But Of course, we now know we can replace that with negative one. And then, we see that negative 𝑖 plus 𝑖 is zero. So, we obtain one minus negative one, which is simply two. So, one plus 𝑖 times one minus 𝑖 is simply two.

Now, you might have noticed that one plus 𝑖 and one minus 𝑖 are complex conjugates of one another. That is, if a complex number 𝑧 is of the form π‘Ž plus 𝑏𝑖, where π‘Ž and 𝑏 are real constants, then the complex conjugate of 𝑧 β€” sometimes denoted 𝑧 star or 𝑧 bar β€” is π‘Ž minus 𝑏𝑖. You just change the sign of the imaginary part. And then, you can find the product of a complex number and its conjugate by finding the sum of the squares of the real and imaginary parts of the original complex number. So, 𝑧 times 𝑧 star is simply π‘Ž squared plus 𝑏 squared.

We’re now going to distribute this final set of parentheses. It’s one plus 𝑖 times one plus 𝑖 times one plus 𝑖. When we distribute the first set of parentheses, we get one plus 𝑖 plus 𝑖 plus 𝑖 squared. We replace 𝑖 squared with negative one. And then, we see that one minus one is zero. So, we have two 𝑖 times one plus 𝑖. And once again, we distribute by multiplying two 𝑖 by one and then by 𝑖. And that gives us two 𝑖 plus two 𝑖 squared, which we can then write as two 𝑖 plus two times negative one, which is negative two plus two 𝑖.

Now that we distributed each of these sets of parentheses, we’re going to replace each term in our original expression. We get negative two minus two 𝑖 minus two plus negative two plus two 𝑖. Now, in fact, we don’t need the brackets. We can write this as negative two minus two 𝑖 minus two plus negative two plus two 𝑖. Well, negative two 𝑖 plus two 𝑖 is zero. And so, we have negative two minus two plus negative two, which is simply negative six.

And so, we fully simplified our expression. It’s negative six.

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