Question Video: Multiplying Complex Numbers Involving Powers of the Imaginary Units Mathematics

Simplify (5 βˆ’ 𝑖^(83))(5 βˆ’ 𝑖^(69))(5 βˆ’ 𝑖^(61)).

04:39

Video Transcript

Simplify five minus 𝑖 to the 83rd power multiplied by five minus 𝑖 to the 69th power multiplied by five minus 𝑖 to the 61st power.

In this question, we’re asked to simplify the product of three complex numbers. However, we can see that all three of our products share something interesting. They contain 𝑖 raised to a very high exponent. We might’ve been tempted to just start distributing our parentheses straight away. However, in this case, we’re going to simplify what’s inside our parentheses first.

To do this, let’s start by recalling that we define 𝑖 to be the square root of negative one. And this gives us a very useful property. We can ask the question, what happens if we square both sides of this equation? We get that 𝑖 squared is equal to negative one. And we can immediately use this to simplify all three of our factors.

Let’s start with negative 𝑖 to the 83rd power. By writing this product out in full and simplifying, all by using our laws of exponents, we can see this is equal to negative one multiplied by 𝑖 to the 82nd power multiplied by 𝑖. But 𝑖 to the 82nd power can be written in terms of 𝑖 squared. Either by writing the product out in full and simplifying or by using our laws of exponents, we can see 𝑖 to the 82nd power is 𝑖 squared all raised to the 41st power. But we know 𝑖 squared is equal to negative one. So this simplifies to give us negative one times negative one to the 41st power multiplied by 𝑖.

And now we can evaluate this expression by using the fact that negative one raised to an even power is equal to one and negative one raised to an odd power is equal to negative one. This means negative one raised to the power 41 is negative one. And then we multiply this by negative one to just get one. So this entire expression just simplified to give us 𝑖. So let’s keep note of what we’ve shown. So far, we’ve shown that negative one times 𝑖 to the 83rd power is just equal to 𝑖.

We can now do the same for our second term. This time, we’re going to want to simplify negative one times 𝑖 to the 69th power. We’ll start by rewriting this as negative one times 𝑖 to the 68th power multiplied by 𝑖. Then, once again, we use the fact that 𝑖 to the 68th power is equal to 𝑖 squared all raised to the 34th power to rewrite our expression. Then all we use is the fact that 𝑖 squared is equal to negative one. This gives us negative one times negative one to the 34th power multiplied by 𝑖. And because negative one raised to the 34th power is negative one raised to an even power, this is just equal to one. So this all simplifies to just give us negative 𝑖.

Once again, let’s keep note of this. Negative one times 𝑖 to the 69th power is equal to negative 𝑖. So let’s clear some space and do the same for our last factor. This time, we want to simplify negative one times 𝑖 to the 61st power. We’ll start by writing this as negative one times 𝑖 to the 60th power multiplied by 𝑖. Then we rewrite 𝑖 to the 60th power as 𝑖 squared all raised to the 30th power. Then we just use the fact that 𝑖 squared is equal to negative one to write this as negative one times negative one to the 30th power multiplied by 𝑖. And of course negative one raised to the 30th power is just equal to one. So this all simplifies to give us negative 𝑖.

And let’s keep note of this last simplification because we’ll need it for later. So let’s clear some space, so we’re now ready to start simplifying our expression. First, we’ll start with our expression in full. That’s five minus 𝑖 to the 83rd power multiplied by five minus 𝑖 to the 69th power times five minus 𝑖 to the 61st power.

Next, we’re going to use our three simplifications. This lets us write our first factor as five plus 𝑖, our second factor as five minus 𝑖, and our third factor also as five minus 𝑖. But remember, the question wants us to simplify this expression. And we could simplify this by distributing all of our parentheses.

Now, we could do this factor by factor by using the FOIL method. However, there’s an easier method. We can see that the first two factors in this expression is actually a factoring of a difference between squares. It’s of the form π‘₯ plus 𝑦 multiplied by π‘₯ minus 𝑦. And we know this is equal to π‘₯ squared minus 𝑦 squared. So instead of multiplying this out in full, we already know their products will be five squared minus 𝑖 squared. This gives us five squared minus 𝑖 squared multiplied by five minus 𝑖.

Next, we can simplify this by remembering that 𝑖 squared is equal to negative one. This gives us 25 minus negative one all multiplied by five minus 𝑖. And of course subtracting negative one is the same as adding one. And 25 plus one is equal to 26. So we get 26 multiplied by five minus 𝑖. And finally, we’ll distribute 26 over our parentheses. And since 26 multiplied by five is 130, this gives us our final answer of 130 minus 26𝑖.

Therefore, we were able to simplify the expression five minus 𝑖 to the 83rd power times five minus 𝑖 to the 69th power multiplied by five minus 𝑖 to the 61st power. We were able to show it’s equal to 130 minus 26𝑖.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.