Question Video: Finding the Square of Complex Numbers in Algebraic Form | Nagwa Question Video: Finding the Square of Complex Numbers in Algebraic Form | Nagwa

# Question Video: Finding the Square of Complex Numbers in Algebraic Form Mathematics • First Year of Secondary School

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Given that π = 5β(3) β 5π, determine the algebraic form of πΒ².

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### Video Transcript

Given that π is equal to five root three minus five π, determine the algebraic form of π squared.

In this question, weβre given a complex number π. And weβre asked to determine the algebraic form of π squared. To do this, weβre first going to need to recall what we mean by the algebraic form of a complex number. We recall when we say a complex number is in algebraic form if itβs in the form π plus ππ, where both π and π are real numbers. So, in fact, we can see that π is already given in algebraic form since five root three and negative five are both real numbers.

So, the question wants us to find an expression for π squared in algebraic form. Remember, this is five root three minus five π all squared. And to do this, remember, when weβre squaring a number, weβre multiplying it by itself. In other words, π squared is equal to five root three minus five π multiplied by five root three minus five π. Now, we can see weβre multiplying two complex numbers given in algebraic form together. However, another way of looking at this is weβre multiplying two binomials together. So, one way of doing this will be to use the FOIL method.

First, the FOIL method tells us we need to multiply the first two terms of our binomials together. This gives us five root three multiplied by five root three. Next, the FOIL method tells us we need to add on the products of the outer two terms together. So, we add on five root three multiplied by negative five π. The next step in the FOIL method is to add on the products of the inner two terms together. And in this case, thatβs negative five π multiplied by five root three. Finally, the last step in the FOIL method is to add the product of the last two terms together. And in this case, thatβs negative five π multiplied by negative five π.

So, now, weβve distributed over our parentheses, and we have five terms. And we can simplify each of these terms separately. First, to multiply five root three by five root three, we have five times five is equal to 25 and root three multiplied by root three is equal to three. So this gives us 25 multiplied by three, which is equal to 75. Now, to simplify our second term, we have five root three multiplied by negative five π.

We want to find the coefficient of π. So, we need to multiply five root three by negative five. And of course negative five times five is negative 25. So, this simplifies to give us negative 25 root three π. And in fact, we get exactly the same story with our third term. This also simplifies to give us negative 25 root three π. We want to do the same with our fourth term. So, we start by multiplying the negative fives together, and that gives us positive 25. However, we still have two factors of π. So, we need to multiply this by π squared. So, now, weβve shown this is equal to 75 minus 25 root three π minus 25 root three π plus 25π squared.

However, thereβs more simplification we can do. We can see that our second and third term are equal. So, we can just add these together to get negative 50 root three π. And thereβs still more simplification we can do. We can notice in our third term we have a factor of π squared. And remember, π is the square root of negative one. So, π squared will be the square root of negative one squared, which is, of course, just equal to negative one. So, we can replace π squared with negative one, which means weβre now subtracting 25 in our expression. And now, we see we have 75 minus 25, which is, of course, equal to 50, giving us our final answer of 50 minus 50 root three π.

And we could leave our answer here; however, there is one thing worth pointing out. We can see weβre squaring a binomial expression. And because multiplication in this fashion is commutative, we couldβve also done this by using the binomial formula. By applying this, we have π plus ππ all squared is equal to π squared plus two times πππ plus ππ all squared. However, this isnβt all we can do. We can also distribute the square over our final parentheses. This will give us a factor of π squared and a factor of π squared.

And of course, we know π squared is equal to negative one. So instead of adding ππ all squared, we can subtract π squared. And this would then give us a second way of answering this question. We could just substitute the values of π and π into this formula. Either method would work and would give us the correct algebraic expression for π squared.

Therefore, we were able to show if π is equal to five root three minus five π, then we have two methods for determining the algebraic form of π squared. Using both methods, weβll get 50 minus 50 root three π.

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