# Question Video: Complex Conjugates of Sums and Products Mathematics • 12th Grade

Consider π§ = 5 β πβ(3) and π€ = β(2) + πβ(5). Calculate π§^* and π€^*. Find π§^(*) + π€^(*) and (π§ + π€)^*. Find π§^(*)π€^(*) and (π§π€)^*.

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### Video Transcript

Consider π§ equals five minus π root three and π€ equals root two plus π root five. Calculate π§ star and π€ star. Find π§ star plus π€ star and π§ plus π€ star. Find π§ star π€ star and π§π€ star.

In this question, weβve been given two complex numbers, π§ and π€. Now, a complex number is of the form π plus ππ, where π and π are real constants. We say that π is the real part of the complex number, whereas π is its imaginary part. Then, we denote the complex conjugate of π§ as π§ star. And we find this by changing the sign of the imaginary part. So, for a complex number π plus ππ, its conjugate is π minus ππ.

So, letβs look at our first complex number. Its real part is five, whereas its imaginary part is negative root three. Remember, to find its conjugate, we change the sign of the imaginary part. So, the real part of π§ star is still five, whereas its imaginary part is now positive root three. And so, the complex conjugate of π§, π§ star, is five plus π root three.

Weβll repeat this process for π€. This time, the real part of π€ is the square root of two, whereas the imaginary part is root five. We find the conjugate of π€, π€ star, by changing the sign of the imaginary part. So, the imaginary part of the conjugate is negative root five. And we can say that π€ star is the square root of two minus π root five. Now that we found the conjugates of π§ and π€, letβs answer the second part of this question.

The first part of the second part of this question asks us to find the sum of the conjugates of our complex numbers. So, thatβs the sum of five plus π root three and root two minus π root five. Now, to add complex numbers, we simply add the real parts and then separately add the imaginary parts. The real parts are five and the square root of two. And the imaginary parts are root three and negative root five. So, we can say that the sum of our conjugates is five plus root two plus the square root of three minus the square root of five π.

Now, the second part of this asks us to find the conjugate of the sum of our original complex numbers. So, letβs simply begin by finding the sum of our original complex numbers. Once again, we find the sum of these complex numbers by adding their real parts and then separately adding their imaginary parts. The real parts are five and the square root of two, whereas the imaginary parts are negative root three and root five. So, π§ plus π€ is five plus the square root of two plus the square root of five minus the square root of three π.

Now, weβre looking to find the complex conjugate of this. And remember, we find this by changing the sign of the imaginary part. So, the real part is still five plus the square root of two. And we see that π§ plus π€ star, the conjugate of the sum of our original complex numbers, is five plus root two minus root five minus root three π. We can distribute the negative one across our second set of parentheses, and we can write this as five plus root two plus the square root of three minus the square root of five π. And so, we found that sum of the conjugates and the conjugate of the sums.

Finally, letβs look at the product of our complex numbers. We begin by finding the product of the conjugates. So, thatβs five plus π root three times root two minus π root five. Weβll distribute these parentheses using the FOIL method. We multiply the first terms and we get five root two. We multiply the outer terms and we get negative five π root five. Multiplying the inner terms gives us π root six. Remember, the square root of three times the square root of two is the same as the square root of three times two. And then, we multiply the last terms and we get negative π squared root 15. But we also know that π squared is equal to negative one. So, this last term becomes plus root 15. Collecting together the real and imaginary parts and we find that π§ star times π€ star is five root two plus root 15 minus five root five minus root six π.

Next, weβre going to find the conjugate of the products of our original numbers. So, letβs just simply find π§π€. Thatβs five minus π root three times root two plus π root five. Distributing as we did before, and we get five root two plus five π root five minus π root six minus π squared root 15. And once again, that last term becomes plus root 15. Once again, we collect the real and imaginary parts. And we find that π§π€ is five root two plus root 15 plus five root five minus root six π. The conjugate of this is found by changing the sign of the imaginary part. Well, the imaginary part is five root five minus root six. So, π§π€ all star, the conjugate of the product of our complex numbers, is five root two plus root 15 minus five root five minus root six π.

And so, we found the product of our conjugates and the conjugate of the products, and theyβre the same. In fact, we can generalize our results. And we can say that the sum of the complex conjugates of two complex numbers is equal to the conjugate of the sum of those numbers. Similarly, the product of the conjugates is equal to the conjugate of the products.