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In this video, weβre going to learn how to work with purely imaginary numbers.
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Knowing how to work with these numbers is an important foundation towards working effectively and confidently with complex numbers.
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Weβll begin by learning how to evaluate and simplify imaginary numbers including when finding the product of these numbers.
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Then, weβll discover how to solve equations which have imaginary solutions.
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Rafael Bombelli was the mathematician regarded as the inventor of complex numbers.
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Whilst other mathematicians were solving equations by recognizing purely real solutions, Bombelli saw the usefulness of working with the square root of negative numbers and he introduced the rules of arithmetic for imaginary numbers that we still use today.
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Interestingly, Bombelli shied away from giving a special name to the square root of negative numbers, instead choosing to deal with them as he would any other radical.
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He called what we now know as π βplus of minus.β
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And he used the term βminus of minusβ to describe negative π.
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But what is the definition of this imaginary number that we now call π?
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In its most basic form, π is defined as the solution to the equation π₯ squared equals negative one.
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This means that π squared is equal to negative one.
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And we can in turn say that π is equal to the square root of negative one.
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π is called an imaginary number, essentially because itβs not part of the set of real numbers.
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What this means is any real multiple of π β in other words, ππ, where π is a real number β is also a purely imaginary number.
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So why do we use them?
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Why donβt we just stick with a set of real numbers that we already know so well?
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Well, as weβve already seen in the definition for π, there are some equations which have no real solutions.
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Imaginary numbers allow us to actually solve these equations.
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Letβs look at an example of this.
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Weβll begin by considering an equation with very little in the way of rearranging required.
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Solve the equation π₯ squared equals negative 16.
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To solve an equation like this, we do begin by solving as we would any equation with real solutions by performing a series of inverse operations.
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In this case, weβre going to find the square root of both sides of the equation.
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Before we do though, we choose to rewrite negative 16 slightly.
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Weβre going to write it as 16π squared.
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And weβll see why we do this in a moment.
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But for now, it works because π squared is equal to negative one.
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And this means that 16π squared is 16 times negative one which is negative 16.
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And now that we have the equation π₯ squared equals 16π squared, we can now find the square root of both sides of the equation, remembering that we can take both the positive and negative root of 16π squared.
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The square root of π₯ squared is π₯.
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So π₯ is equal to the positive and negative square root of 16π squared.
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And during this next step, itβs going to become evident why we chose to write negative 16 as 16π squared.
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We can split the square root of 16π squared into the square root of 16 times the square root of π squared.
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The square root of 16 is four and the square root of π squared is simply π so in turn we can see that π₯ is equal to plus or minus four π.
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The solutions to the equation π₯ squared equals negative 16 are four π and negative four π.
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And we should now be able to see why we did write negative 16 as 16π squared.
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It made these final steps a little easier to deal with.
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And of course, we can check these solutions by substituting them back into the original equation.
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Letβs try this for π₯ equals four π first.
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π₯ squared is four π squared.
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And of course, thatβs four π times four π.
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Multiplication is commutative.
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It can be performed in any order.
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So we can rewrite this as four times four times π times π.
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Four multiplied by four is 16 and π times π is π squared.
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And since π squared is negative one, π₯ squared is negative 16 as required.
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We can repeat this process for π₯ equals negative four π.
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π₯ squared is negative four π times negative four π, which can in turn be written as negative four times negative four times π times π.
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Once again, since negative four times negative four is positive 16, we get 16π squared which is negative 16 as required.
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Next, weβll have a look at an equation which requires just a little more work to solve it.
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Solve the equation two π₯ squared equals negative 50.
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To begin solving this equation, weβre going to divide both sides by two.
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Negative 50 divided by two is negative 25.
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So π₯ squared is equal to negative 25.
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We now rewrite negative 25 as 25π squared.
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And remember we can do that because π squared is negative one.
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And then, we find the square root of both sides of this equation.
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Of course, we can find both the positive and negative square root of 25π squared.
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So π₯ is equal to plus or minus root 25π squared.
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We can then write the square root of 25π squared as the square root of 25 multiplied by the square root of π squared, which is simply five π.
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So π₯ is equal to plus or minus five π.
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The solutions to the equation two π₯ squared equals negative 50 are five π and negative five π.
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In these next examples, weβll look at how to extend the rules of arithmetic and algebra for real numbers to help us solve problems involving purely imaginary numbers.
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Simplify two π squared multiplied by negative two π cubed.
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When we square a number, we multiply it by itself.
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So two π squared is the same as two π multiplied by two π.
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And since multiplication is commutative, we can write this as two times two times π times π.
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And in fact, this is a little bit like evaluating an algebraic expression.
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We multiply two by two to get four and we multiply π by π to get π squared.
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But remember π is not a variable.
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Itβs the solution to the equation π₯ squared equals negative one such that π squared is negative one.
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So four π squared is four multiplied by negative one which is negative four.
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Next, weβre going to evaluate negative two π cubed.
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But weβre not going to write it out as negative two π times negative two π times negative two π.
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Instead, weβre going to use the rules for exponents that weβre used to.
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And weβre going to write it as negative two cubed times π cubed.
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Negative two cubed is negative eight.
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But what about π cubed?
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Now it might look a little scary.
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But itβs just the same as writing π squared times π.
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And π squared is negative one.
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So our expression becomes negative eight multiplied by negative one times π which is simply eight π.
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Our final step is to replace two π squared and negative two π cubed with negative four and eight π, respectively.
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And then, weβll evaluate that just as we would any algebraic expression.
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It becomes negative four times eight π which is negative 32π.
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Weβve just seen that we can apply some of the rules for manipulating algebraic expressions to help us evaluate those involving imaginary numbers.
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And we just saw the result that π cubed is equal to negative π.
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At this stage, it might be useful to consider what happens with other powers of π, π to the power of four or five for example.
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We can evaluate π to the power of four by thinking of it as π squared times π squared.
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And then since π squared is negative one, we say that π to the power of four is negative one multiplied by negative one and that simply one.
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And at this point, we can begin to generalise.
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Weβre going to raise this entire equation to the πth power.
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And this works for integer values of π.
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When we do, we see that π to the power of four π is equal to one to the power of π.
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But actually, one to the power of anything is just one.
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So we can see that π to the power of four π equals one.
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We might then choose to multiply both sides of this equation by π or π to the power of one.
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Remember when we multiply two numbers with the same base, here thatβs π, we add the powers.
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So π times π to the power of four π is π to the power of four π plus one and π to the power four π plus one is equal to π.
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Letβs do this again.
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When we do, we see that π to the power of four π plus two is equal to π squared.
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But π squared is simply negative one.
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So π to the power of four π plus two is equal to negative one.
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Weβll repeat this process one more time.
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And we see that π to the power of four π plus three is negative π.
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And now we stop.
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Why?
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Well, if we were to multiply by π again, weβd have π to the power of four π plus four.
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But four is a multiple of four.
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So this will have the same result as π to the power of four π.
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And this cycle repeats infinitely.
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Thereβs a nice graphic we can use to help us to find any power of π.
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For integer values of π, we can use this cycle to define any power of π.
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Letβs look at the potential of these results by simplifying an expression in terms of powers of π.
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Simplify π to the power of 30.
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To simplify this expression, we really donβt want to write π out 30 times and evaluate each pair.
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Instead, weβll recall the cycle that helps us remember the identities for various powers of π.
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Letβs compare our number π to the power of 30 to this cycle.
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We need to represent the power 30 in the form four π plus π.
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And to correspond to the powers of π in our cycle, π would be zero, one, two, or three.
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Now in fact, 30 can be written as four multiplied by seven plus two.
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So π to the power of 30 corresponds to the part of the cycle where π is to the power of four π plus two.
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According to this, π to the power of four π plus two is equal to negative one.
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And this means that π to the power of 30 is negative one.
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Now another method we could have chosen would still have been to write π to the power of 30 as π to the power of four times seven plus two.
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And we know by the rules of exponents that this is the same as π to the power of four to the power of seven times π to the power of two.
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π to the power of four is one and π squared is negative one.
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So our expression becomes one to the power of seven multiplied by negative one which is once again negative one.
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So weβve seen how this cycle can save us time when working with positive powers of π.
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And in fact, itβs important to remember that these sets of rules for simplifying powers of π do actually work for negative powers too.
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Letβs look at a more in-depth example of this.
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Given that π is an integer, simplify π to the power of 16π minus 35.
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Remember the cycle that helps us remember the identities for various powers of π is as shown.
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So we can do one of two things.
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Our first method is to use the laws of exponents to essentially unsimplify our expression a little.
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We know that π₯ to the power of π times π₯ to the power of π is the same as π₯ to the power of π plus π.
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So we can reverse this and say that π to the power of 16π minus 35 is equal to π to the power of 16π times π to the power of negative 35.
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π to the power of 16π can actually be further written as π to the power of four to the power of four π.
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This corresponds to the part of our cycle π to the power of four π.
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So we can see that π to the power of 16π can be written as one.
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And what about π to the power of negative 35?
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This one is a little more complicated.
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Weβre going to write negative 35 in the form four π plus π, where π can take the values zero, one, two, or three to correspond with the values in our cycle.
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Itβs the same as four times negative nine plus one.
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Remember four times negative nine is negative 36 and adding one gets us negative 35.
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And we chose negative nine instead of negative eight as we needed π to be zero, one, two, or three.
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And we certainly donβt want it to be a negative value.
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So π to the power of negative 35 will have the same result as π to the power of four π plus one in our cycle; thatβs π.
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So π to the power of 16π minus 35 is one multiplied by π which is π.
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Letβs have a look at the alternative method.
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Here we would have jumped straight into writing the power β thatβs 16 π minus 35 β in the form four π plus π, where π again is zero, one, two, or three.
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We can write 16π as four times four π and negative 35 as four times negative nine plus one.
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We can factor this expression and we see that 16 π minus 35 is the same as four multiplied by four π minus nine plus one.
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So we can see that once again π to the power of 16π minus 35 will have the same result as π to the power of four π plus one in our cycle; thatβs π.
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Our very final example involves one of the laws of radicals weβve briefly looked at in this lesson.
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Thatβs the square root of π times π is equal to the square root of π times the square root of π.
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We need to be extremely careful with this rule.
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As whilst it works for all positive real numbers, the same cannot be said for negatives.
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Simplify the square root of negative 10 times the square root of negative six.
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Weβll begin by expressing each radical in terms of π.
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Remember π squared is equal to negative one.
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So we can say the square root of negative 10 is the same as the square root of 10π squared.
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And similarly, the square root of negative six is the same as the square root of six π squared.
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And at this point, we can split this up.
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We get the square root of 10 times the square root of π squared.
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And since the square root of π squared is π, we can see that the square root of negative 10 is the same as root 10 π.
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And similarly, the square root of negative six is root six π.
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Next, we multiply these together.
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Multiplication is commutative.
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So we can rearrange this a little and say itβs equal to the square root of 10 times the square root of six which is root 60 times π squared.
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And since π squared is negative one, we see that the square root of negative 10 times the square root of negative six is negative root 60.
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And in fact, we need to simplify this as far as possible.
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There are a number of ways to do this.
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We could consider 60 as a product of its prime factors.
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Alternatively, we find the largest factor of 60 which is also a square number.
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In fact, that factor is four.
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So this means that the square of 60 is the same as the square root of four times the square root of 15 which is equal to two root 15.
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And we fully simplified our expression.
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We get negative two root 15.
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Letβs look at what would have happened had we applied the laws of radicals.
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We would have said that the square root of negative 10 times the square root of negative six is equal to the square of negative 10 times negative six which is equal to the square root of positive 60 or two root 15 and thatβs patently different to our other solution.
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In this video, weβve learned that many of the rules of arithmetic and algebra that weβre so confident with can be extended into the world of imaginary and complex numbers.
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Weβve also seen that some of the rules requires to be a little bit more careful, such as generalising the law for multiplying radicals when these radicals include negative numbers.
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We also saw how the integer powers of π form a cycle and that allows us to simplify any real power of π fairly quickly.