WEBVTT
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In this video, weβre going to learn how to calculate the modulus of a complex number.
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We will learn what we mean by the term modulus before deriving a standard formula we can use for all cases.
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We will then consider the properties of the modulus in relation to operations on complex numbers β such as addition, multiplication, and division β before solving simple equations involving the modulus of a complex number.
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We have seen that we can represent complex numbers on a two-dimensional plane.
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We call this plane the Argand diagram or Argand plane, after the amateur mathematician who discovered it in the early 1800s.
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We can use it to graph a complex number of the form π§ equals π plus ππ, whose real part is π and whose imaginary part is π.
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To do this, we locate the real part π on the real axis.
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Thatβs the horizontal axis.
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And then we move up or down to locate the imaginary part, thatβs π, on the imaginary axis, the vertical axis.
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Our complex number can therefore be represented by the point π, π as shown.
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We add a straight line connecting this point to the origin.
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And we can see that we can now calculate extra pieces of information.
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We can work out the length of this line segment.
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We call this thing modulus of the complex number.
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And itβs denoted as shown.
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So how do we calculate the length of this line segment?
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Well, we can create a right-angled triangle with this side as the hypotenuse.
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The base of this triangle is π units in length.
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And the height of the triangle is π units in length.
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Itβs a right-angled triangle.
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So we can use the Pythagorean theorem to calculate the length of the hypotenuse.
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This states that for a right-angled triangle with side lengths π, π, and π, where π is the hypotenuse, π squared plus π squared equals π squared.
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We said that the length of the hypotenuse in our triangle is the modulus of π§.
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So the modulus of π§ squared must be equal to π squared plus π squared.
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Weβre going to solve this for the modulus of π§ by finding the square root of both sides.
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And we see that weβve created a formula for the modulus of π§ given by π plus ππ.
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Itβs the square root of π squared plus π squared.
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This is sometimes called the absolute value of π§.
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And, of course, since it represents a length, we know that the modulus of π§ will always be greater than zero.
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Now, in fact, itβs often also called the magnitude of the complex number due to the geometric interpretation of a complex number as a vector.
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Weβre now going to consider an example of the application of this formula.
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What is the modulus of the complex number three plus seven π?
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The definition of the modulus of a complex number of the form π plus ππ is the square root of π squared plus π squared.
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π is the real part of the complex number whilst π is the imaginary part.
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Letβs compare this form to our complex number three plus seven π.
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This complex number has a real part of three and an imaginary part of seven.
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Donβt confuse this with seven π.
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The imaginary part is essentially the coefficient of π.
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We can substitute these values into the formula for the modulus.
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And we get the square root of three squared plus seven squared.
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Three squared is nine and seven squared is 49.
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So the modulus is the square root of 58.
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We would usually try to simplify this surd.
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However, there are no factors of 58 which are also square numbers.
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So weβre done; itβs in its simplest form.
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And we can say that the modulus of the complex number three plus seven π is root 58.
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In our next example, weβll look at the relationship between the modulus of a complex number and the modulus of its conjugate.
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As we move forward, see if you can recall the relationship between the representation of a complex number and its conjugate on the Argand plane.
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Consider the complex number π§ is equal to negative four plus π root five.
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Part one: Calculate the modulus of π§.
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Part two: Calculate the modulus of the conjugate of π§.
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Part three: Determine the product of π§ with its conjugate.
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For part one, weβve been given a complex number.
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And weβve been asked to find its modulus.
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Remember, for a complex number of the form π plus ππ, the modulus is found by the square root of π squared plus π squared.
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π is the real part of the complex number whilst π is the imaginary part.
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Letβs compare this form to our complex number, negative four plus π root five.
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Its real part is negative four and its imaginary part is the square root of five.
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Substituting these values into the formula for the modulus, and we see that the modulus of π§ is found by the square root of negative four squared plus the square root of five squared.
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Negative four squared is 16.
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And the square root of five squared is five.
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So we see that the modulus of π§ is the square root of 21.
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For part two, weβre being asked to find the modulus of the complex conjugate of π§.
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Now, to find the conjugate, we changed the sign of the imaginary part.
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In this example, the conjugate of π§ is negative four minus π root five.
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This time the real part of our number is negative four.
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And the imaginary part is negative root five.
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Substituting these values into our formula for the modulus, and we see that the modulus of the conjugate is the square root of negative four squared plus negative root five squared.
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Now, in fact, negative four squared is, once again, 16.
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And negative root five squared is five.
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So we see that the modulus of the conjugate of π§ is also root 21.
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For part three, weβre being asked to find the product of π§ with its conjugate.
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We saw that the conjugate of π§ was negative four minus π root five.
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So the product of π§ with its conjugate is negative four plus π root five multiplied by negative four minus π root five.
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And we can use algebraic manipulation of binomials to find the product.
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We might use the grid or the FOIL method.
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Letβs have a look at the grid method.
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We start by multiplying negative four by negative four, and thatβs 16.
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Then, we multiply negative four by π root five.
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And we get negative four π root five.
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And you may have written this in a different order.
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But Iβve chosen to put the square root of five at the end of this term so itβs clear that weβre not finding the square root of π as well.
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Next, we multiply negative π root five by negative four.
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And we get four π root five.
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And when we multiply the remaining two terms, we get negative π squared multiplied by the square root of five squared.
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Well, in fact, the square root of five squared is simply five.
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And π squared is equal to negative one.
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So we have negative negative one multiplied by five.
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And that simplifies to five.
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Next, we collect like terms.
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And we notice that when we do, we add four π root five and negative four π root five.
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And, actually, we get zero.
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So our expansion becomes six [16] plus five, which is equal to 21.
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And the product of our complex number with its conjugate is 21.
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In this example, weβve seen that not only is the modulus of a complex number equal to the modulus of its conjugate, but also that the square of the modulus of a complex number is equal to the product of a complex number with its conjugate.
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Now, actually, earlier, I said have a think about the relationship between the representation of a complex number and its conjugate on the Argand plane.
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And when we do, we see that this first rule makes a lot of sense.
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It represents a reflection in the π₯-axis or the real axis.
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And we can therefore see that the length of the line segment joining π§ with the origin is exactly the same as the length of the line segment joining the conjugate of π§ with the origin.
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And this confirms to us that the modulus of π§ must be equal to the modulus of the conjugate of π§.
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Weβll now look at the relationship between addition and the modulus of a complex number.
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Consider two complex numbers, π€ equals negative one plus seven π and π§ equals five minus three π.
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Part one: Calculate the modulus of π€ plus the modulus of π§ to two decimal places.
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And there are two other parts to this question which weβll consider in a moment.
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Part one is asking us to find the modulus of π€ and the modulus of π§.
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And then find the sum of these two values.
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Remember, the definition of the modulus of a complex number of the form π plus ππ is the square root of π squared plus π squared.
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If we compare this to our first complex number π€, we can see the π, the real part, is negative one.
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And we see that π, the imaginary part, is seven.
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So the modulus of π€ is found by the square root of negative one squared plus seven squared.
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Negative one squared is one and seven squared is 49.
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So the modulus of π€ is the square root of 50.
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And, actually, weβre going to be evaluating our answer to two decimal places.
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So it doesnβt really matter if we write this in its simplest form or not.
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But if we did, we would see that the modulus of π€ is five root two.
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Weβll repeat this process for the modulus of π§.
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The real part of π§, thatβs π, is five.
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And the imaginary part is negative three.
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So the modulus of π§ is the square root of five squared plus negative three squared.
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Thatβs the square root of 25 plus nine which is root 34.
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So the sum of these two numbers is five root two plus root 34 which is 12.9020 and so on.
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Correct to two decimal places, thatβs 12.90.
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Part two: Calculate the modulus of π§ plus π€ to two decimal places.
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This time, we need to add the complex numbers first and then find the modulus of their sum.
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To add two complex numbers, we add their real parts and then separately add their complex parts.
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This is a little bit like collecting like terms.
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Five plus negative one is four and negative three plus seven is four.
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So we have four plus four π.
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The real part of this number is four.
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And the imaginary part is also four.
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So the modulus is the square root of four squared plus four squared, which is the square root of 32.
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And if we evaluate this on the calculator, we see that the modulus of the sum of these two complex numbers is 5.66, correct to two decimal places.
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Part three: Which of the following relations do π€ and π§ satisfy?
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a) The modulus of π€ plus the modulus of π§ is equal to the modulus of π§ plus π€.
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b) The modulus of π€ plus the modulus of π§ is greater than or equal to the modulus of π§ plus π€.
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c) Itβs less than or equal to the modulus of π§ plus π€.
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d) Itβs equal to two times the modulus of π§ plus π€.
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And e, the square root of the modulus of π€ plus the modulus of π§ is equal to the modulus of π§ plus π€.
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Itβs clear to see that these two numbers are not equal.
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So we can straightaway rule out a.
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In fact, we see that the modulus of π€ plus the modulus of π§ is indeed larger than the modulus of π§ plus π€.
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So it looks like option b is correct.
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Weβll check the other three.
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Clearly, weβve seen that c cannot be correct.
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And in fact, if we double the modulus of π§ plus π€, we get 11.32.
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So d has to be incorrect.
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And if we find the square root of the sum of their moduli, itβs 3.59, which once again shows us that e is also incorrect.
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So the correct answer is b.
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In fact, this statement holds for all complex numbers.
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We can say that for two complex numbers π§ one and π§ two, the sum of their moduli will always be greater than or equal to the modulus of their sum.
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And what about multiplication and division?
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Letβs have a look at the relationship between multiplication and division and the modulus of a complex number.
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Consider the complex numbers π§ equals three minus four π and π€ equals negative 15 plus eight π.
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Part one: Find the modulus of π§ and the modulus of π€.
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And there are two other parts to this question which weβll consider in a moment.
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Remember, for a complex number π§ equals π plus ππ, the modulus of π§ is the square root of π squared plus π squared.
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If we compare this to our first complex number, we see that it has a real part of three and an imaginary part b of negative four.
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So the modulus of π§ is the square root of three squared plus negative four squared.
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The square root of three squared plus negative four squared is the square root of 25, which is simply five.
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π€ has a real part of negative 15 and an imaginary part of eight.
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So the modulus of π€ is the square root of negative 15 squared plus eight squared.
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This is the square root of 289, which is 17.
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Part two: Calculate the modulus of π§π€.
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How does this compare to the modulus of π§ multiplied by the modulus of π€?
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This time, we need to calculate the product of the complex numbers π§ and π€.
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We can use algebraic manipulation of binomials to work out three minus four π multiplied by negative 15 plus eight π.
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We could use the grid method or the FOIL method.
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Letβs look at the FOIL method.
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We begin by multiplying the first term in each bracket.
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Thatβs negative 45.
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When we multiply the outer terms, three multiplied by eight π is 24π.
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Negative four π multiplied by negative 15 is 60π.
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And when we multiply the last two terms, we get negative 32π squared.
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But, of course, π squared is equal to negative one.
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So negative 32π squared is the same as positive 32.
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And we see then that the product of π§π€ is negative 13 plus 84π.
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And this means that the modulus of their product is the square root of negative 13 squared plus 84 squared, which is 85.
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Now, if we compare this to the moduli that we calculated earlier, if we multiply those, we get five multiplied by 17 which is also 85.
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And we can say that the modulus of π§π€ is the same as the modulus of π§ multiplied by the modulus of π€.
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Letβs clear some space and have a look at part three.
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Calculate the modulus of π§ divided by π€.
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How does this compare with the modulus of π§ divided by the modulus π€?
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This time, we need to work out π§ divided by π€.
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Thatβs three minus four π divided by negative 15 plus eight π.
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Just like rationalizing the denominator of a fraction which has a radical as part of the denominator, we can multiply both the numerator and denominator of this fraction by the conjugate of the denominator.
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To find the conjugate, we change the sign for the imaginary part.
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So the conjugate here is negative 15 minus eight π.
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And once again, we multiply as normal.
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On the numerator, we get negative 45 minus 24π plus 60π plus 32π squared, which becomes negative 32 because π squared is equal to negative one.
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And we get negative 77 plus 36π as our simplified numerator.
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On the denominator, we get 225 plus 120π minus 120π minus 64π squared, which is 289.
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The modulus is the square root of negative 77 over 289 squared plus 36 over 289 squared.
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And we can simplify this by taking out a factor of 289 squared.
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And when we do, weβre left with one over 289, cause we need to square root that, multiplied by the square root of 5929 plus 1296.
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This becomes 85 over 289.
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This simplifies to five seventeenths.
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And, actually, if we look carefully, if we were to divide the modulus of π§ by the modulus of π€, we would get five seventeenths.
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And we see that the modulus of π§ divided by the modulus of π€ is the same as the modulus of π§ divided by π€.
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Weβll finish by looking at a brief example at how the properties weβve looked at can help us solve equations involving the modulus.
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If π§ is equal to one over the conjugate of π§, where π§ is a complex number, what is the modulus of π§?
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To solve this equation, weβll begin by finding the modulus of both sides.
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And we know that the modulus of the quotient of two complex numbers is the same as the quotient of their respective moduli.
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So the right-hand side of this equation becomes the modulus of one divided by the modulus of the conjugate of π§.
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Now, in fact, we know that the modulus of one is one.
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And we also know that the modulus of a complex number π§ is the same as the modulus of the conjugate of π§.
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So we have the modulus of π§ is equal to one divided by the modulus of π§.
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Weβre going to multiply both sides of this equation by the modulus of π§.
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And then we find the square root of both sides.
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We would normally take both the positive and negative square root.
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But modulus represents a length and therefore must always be positive.
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So we can say that the modulus of π§ is equal to the square root of one which is, of course, simply one.
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In this video, weβve seen that we can find the modulus of a complex number by taking the square root of the sum of the squares of the real and imaginary parts of that number.
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And it represents the distance of π§ from the origin.
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And weβve learned a number of properties about the modulus of π§.
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And, in fact, there is one more.
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We can extend the property for multiplication to see that the modulus of π§ raised to the power of π is the same as the modulus of π§ raised to the power of π.