Question Video: Simplifying Expressions Involving Powers of Complex Numbers Mathematics

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

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Video Transcript

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

To simply this expression, we’ll begin by distributing each set of parentheses. Let’s begin by distributing one minus 𝑖 cubed. Now, we could write this as one minus 𝑖 times one minus 𝑖 times one minus 𝑖 and then distribute the first pair of parentheses and then multiply all of that by one minus 𝑖.

The binomial theorem tells us that for positive integer values of 𝑛, π‘Ž plus 𝑏 to the 𝑛th power is the sum from π‘˜ equals zero to 𝑛 of 𝑛 choose π‘˜ times π‘Ž to the 𝑛th power times 𝑏 to the power of 𝑛 minus π‘˜. So, in the distribution of one minus π‘Ž cubed, the first term is one cubed. Then, the second term is three choose one times one squared times negative 𝑖 to the power of one.

The third term is three choose two times one times negative 𝑖 squared. And then, the third term is negative 𝑖 cubed. Let’s simplify this a little. One cubed is one, and three choose one is three. So, our second term becomes three times one times negative 𝑖, which is negative three 𝑖. Our third term is three choose two, which is three times 𝑖 squared. But of course, we know that 𝑖 squared is negative one. So, this simply becomes negative three.

And our third term is negative 𝑖 cubed. But since we can write 𝑖 cubed as 𝑖 squared times 𝑖, and we know 𝑖 squared is negative one, this becomes negative negative one times 𝑖, which is simply 𝑖. And so, we simplify fully. And we find that one minus 𝑖 cubed is negative two minus two 𝑖.

We’ll now repeat this process for our second term. That’s one plus 𝑖 times one minus 𝑖. Now, we could distribute the parentheses using something like the FOIL method. Or we can recall that the product of a complex number and its conjugate is the sum of the squares of the real and imaginary parts of that complex number. And remember, the complex conjugate of one plus 𝑖 is found by changing the sign of the imaginary part, so it’s one minus 𝑖. Its real part is one, and its imaginary part is the coefficient of 𝑖, so it’s also one. So, one plus 𝑖 times one minus 𝑖 is one squared plus one squared, which is simply two.

Let’s move on to the third pair of parentheses. We’ve got one plus 𝑖 cubed. Once again, we use the binomial theorem. And we get one cubed plus three choose one times one squared times 𝑖 plus three choose two times one times 𝑖 squared plus 𝑖 cubed. We simplify a little, and we get one plus three 𝑖 plus three 𝑖 squared plus 𝑖 cubed. And then, we use again the fact that 𝑖 squared is negative one. So, our third term becomes negative three. And our fourth term becomes negative 𝑖.

Simplifying this time, and we find that one plus 𝑖 cubed is negative two plus two 𝑖. We can now replace each parentheses in our original expression with these. We get negative two minus two 𝑖 minus two plus negative two plus two 𝑖. Then, we see that negative two 𝑖 plus two 𝑖 is zero. So, this becomes negative two minus two minus two, which is negative six. And so, when we simplify one minus 𝑖 cubed minus one plus 𝑖 times one minus 𝑖 plus one plus 𝑖 cubed, we get negative six.

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