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In this video, weβll learn how to find the πth roots of unity and explore their properties.
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Weβll begin by learning what we mean by the πth roots of unity and find their general form.
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Weβll then calculate their sum and find the properties of their reciprocal before discovering the application of the πth roots of unity and their geometric properties.
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If π§ is an πth root of unity, then it satisfies the relation π§ to the power of π equals one.
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We can use de Moivreβs theorem to help us solve this equation and thus find the general form for the πth roots of unity.
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de Moivreβs theorem for roots states that, for a complex number written in polar form, π cos π plus π sin π, its πth roots are given by π to the power of one over π times cos of π plus two ππ over π plus π sin of π plus two ππ over π.
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And π takes integer values from zero through to π minus one.
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To solve the equation π§ to the power of π equals one, weβll begin by writing the number one in polar form.
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One whose real part is one and whose imaginary part is zero is a fairly easy number to write in polar form.
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If we represent one on an Argand diagram, we see it can be represented by the points whose Cartesian coordinates are one, zero.
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The modulus of this number, thatβs π
in the general form for our complex number, is the length of the line segment that joins this point to the origin.
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So its modulus must be one unit.
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The argument is the measure of the angle that this line segment makes with the positive real axis.
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And thatβs measured in a counterclockwise direction.
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We can see that its argument must be zero.
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And we can therefore say that one is equal to one times cos of zero plus π sin of zero.
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The πth roots of unity, in other words, the πth roots of one, are therefore given by one times cos of zero plus two ππ over π plus π sin of zero plus two ππ over π for values of π between zero, two, π minus one.
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We can simplify this expression somewhat.
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And we see that the πth roots of unity are given by cos of two ππ over π plus π sin of two ππ over π, which in exponential form is π to the two ππ over π π.
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And thatβs for values of π between zero and π minus one.
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Itβs important to realise that this is the definition of the πth roots of unity.
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It should be learned and recalled when finding the πth roots of unity.
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Now that weβve derived this is absolutely not necessary to use de Moivreβs theorem every time.
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In our next two examples, weβre going to look at how to apply this formula to find the πth root of unity.
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Find the cubic roots of unity and plot them on an Argand diagram.
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Finding the cubic roots of unity is like saying what are the solutions to the equation π§ cubed equals one.
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To find them, we can use the general formula for the πth roots of unity.
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Thatβs cos of two ππ over π plus π sin of two ππ over π for integer values of π between zero and π minus one.
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Since weβre finding the cubic roots of unity, in this example, our value of π is three, which means π will take the values zero, one, and two.
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Letβs begin with the case when π is equal to zero.
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This root is cos of zero plus π sin of zero.
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Well, cos of zero is one.
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And sin of zero is zero.
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So the first root is one.
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And it makes complete sense if we think about it that a solution to the equation π§ cubed equals one would be one.
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Next, we let π be equal to one.
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This root is cos of two π by three plus π sin of two π by three.
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And in exponential form, thatβs π to the two π by three π.
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Finally, we let π be equal to two.
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The root here is cos four π by three plus π sin of four π by three.
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Notice though that the argument for this root is outside of the range for the principal argument.
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We therefore subtract two π from four π by three to get negative two π by three.
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So our third and final root is cos of negative two π by three plus π sin of negative two π by three.
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Or in exponential form, thatβs π to the negative two π by three π.
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Now that we have the cubic roots of unity, we need to plot them on an Argand diagram.
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There are two ways we could approach this problem.
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We could convert each number to algebraic form.
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Thatβs one, negative a half plus root three over two π, and negative a half minus root three over two π.
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These are plotted on the Argand diagram as shown.
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Alternatively, we couldβve used the modulus and argument of each root.
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Either way, letβs notice that the points that represent these complex numbers are the vertices of an equilateral triangle.
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This triangle is inscribed in a unit circle whose centre is the origin.
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Actually, an interesting geometrical property of the πth roots of unity is that, on an Argand diagram, they are all evenly spaced about the unit circle whose centre is the origin.
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They form a regular π-gon.
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It has a vertex at the point whose Cartesian coordinates are one, zero.
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Weβll investigate this property a little bit further along in the video.
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Find the sixth roots of unity.
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Finding the sixth roots of unity is the same as solving the equation π§ to the power of six equals one.
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Weβll once again use the general formula for the πth roots of unity.
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They are cos of two ππ over π plus π sin of two ππ over π, when π takes integer values from zero to π minus one.
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In this example, weβre looking to find the sixth roots of unity.
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So π is six.
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And π takes integer values from zero through to five.
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The first root is found when π is equal to zero.
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This is cos of zero plus π sin of zero, which is one.
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Now actually, we just saw that, on an Argand diagram, the points which represent the πth roots of unity form a regular π-gon with a vertex at the point one, zero.
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Thatβs this root.
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For the second root, we let π be equal to one.
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This is cos of two ππ over six plus π sin of two ππ over six.
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The argument here simplifies to π by three.
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And we could also write this in exponential form as π to the π by three π.
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When π is equal to two, our root is cos of four π by six plus π sin of four π by six.
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And this argument simplifies to two π by three.
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When π is equal to three, we have cos of six π over six plus π sin of six π over six, which is negative one.
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And then, when π is equal to four, we have cos of eight π over six plus π sin of eight π over six.
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Now here, the argument simplifies to four π by three.
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And this is outside of the range for the principal argument.
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We therefore subtract two π from four π by three to get negative two π by three.
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And in exponential form, our fifth root is π to the negative two π by three π.
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Finally, when π is equal to five, we get cos of 10π over six plus π sin of 10π over six.
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This time, the argument simplifies to five π by three, which is once again outside of the range for the principal argument.
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Five π by three minus two π is negative π by three.
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And we therefore see that, in exponential form, our final root is π to the negative π by three π.
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And we have the sixth roots of unity.
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In exponential form, they are one, π to the π by three π, π to the two π by three π, negative one, π to the negative two π by three π, and π to the negative π by three π.
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We could plot the sixth roots of unity on an Argand diagram.
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And weβd see that the points representing these roots form the vertices of a regular hexagon inscribed in a unit circle as shown.
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At this stage, itβs also worth noting an extra definition.
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We say that a primitive root of unity is a root which is not also a πth root of unity, where π is less than π.
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So in this example, the primitive roots are when π is equal to one and π is equal to five since the roots when π is equal to two and π is equal to four are also cubic roots of unity.
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And when π is equal to zero and π is equal to three, these are the square roots of unity.
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In other words, theyβre the square root of one.
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We will look in more detail at the relationship between different roots of unity later in this video.
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Itβs useful to know that we often use the symbol π to denote the primitive root of unity with the smallest strictly positive argument.
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In the case of the sixth roots of unity, this would be π to the π by three π.
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Interestingly, its other roots are all powers of π.
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Now, itβs outside of the scope of this video to explore this property more.
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But it is an interesting one that you might wish to investigate.
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Now we have a couple of definitions and the process that we need to take to find the πth roots of unity, letβs look at the properties of the sum of the πth roots of unity.
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Find the sum of the sixth roots of unity.
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Now, we already calculated the sixth roots of unity.
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In polar form, they are as shown.
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Weβre going to change these to algebraic form.
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The second root is a half plus root three over two π.
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The third root is negative a half plus root three over two π.
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The fifth root is negative a half minus root three over two π.
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And the final root is a half minus root three over two π.
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And so their sum is as shown.
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And we can find the sum of complex numbers written in algebraic form by adding together their real parts and separately adding their imaginary parts.
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Weβll begin with the real parts.
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One plus negative one is zero.
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A half plus negative half is zero.
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And negative a half plus a half is also zero.
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And what about the imaginary parts?
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Well, root three over two minus root three over two is zero.
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And again, root three over two minus root three over two is zero.
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And so the sum of the sixth roots of unity is zero.
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Now, this isnβt actually a process that you would need to take each time.
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Itβs very much a means to an end since we can actually generalize this.
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In general, the sum of the πth roots of unity, when π is greater than one, is always zero.
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And this is another result that needs to be learned and applied where necessary.
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For our next example, weβre going to look at how to find the reciprocal for the πth roots of unity.
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Let π§ be an πth root of unity and π be a positive integer.
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Which of the following is the correct relationship between the reciprocal of π§ and π§?
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Is it a) the reciprocal of π§ is equal to π§.
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Is it b) the reciprocal of π§ is equal to negative π§.
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Is it c) the reciprocal of π§ is equal to the negative conjugate of π§.
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Or is it d) the reciprocal of π§ is equal to just the conjugate of π§.
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To work out which of these is the correct relationship, weβre first going to evaluate π§ to the power of negative one or the reciprocal of π§.
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Since π§ is an πth root of unity, we can say that π§ can be written as π to the ππ, where π is two ππ over π and π takes integer values from zero to π minus one.
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This means π§ to the power of negative one is the same as π to the ππ to the power of negative one, which is the same as π to the negative ππ.
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Next, we recall the property of the conjugate of a complex number written in an exponential form.
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We know that the conjugate of ππ to the ππ is ππ to the negative ππ.
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And this means that π§ to the power of negative one is equal to the conjugate of π§ since we defined π§ to be π to the ππ.
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And we can therefore see that the reciprocal of π§ or π§ to the power of negative one is equal to the conjugate of π§.
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The correct answer is d).
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And this definition can be somewhat extended a little bit.
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We can say that the reciprocal of the πth root of unity is its complex conjugate.
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But that is also an πth root of unity.
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Weβre going to look at one more definition.
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And then, we have an example of that definition to the geometric properties of the πth roots of unity.
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Weβre going to begin by considering how the πth roots of unity are related for different values of π.
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What is the relationship of the cubic roots of unity to the sixth roots of unity?
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We have briefly considered this idea.
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Weβve already seen that the cubic roots are one, π to the two π by three π, and π to the negative two π by three π.
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And we also saw that the sixth roots of unity are one, π to the π by three π, π to the two π by three π, negative one, π to the negative two π by three π, and π to the negative π by three π.
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We can see that all the cubic roots of unity are also sixth roots of unity.
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And we looked at this one when we were discussing the concept of the primitive roots of unity.
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Letβs extend this idea to a definition.
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We can say that if π is equal to the product of some other π and π, then the πth roots of unity are also the πth roots of unity.
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And similarly, the πth roots of unity must also be πth roots of unity.
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We can say that the common roots of π§ to the power of π minus one equals zero and π§ to the power of π minus one equals zero must be the roots of π§ to the power of π minus one equals zero, where π is the greatest common divisor of π and π.
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Letβs consider an example using the applications of the properties of the πth roots of unity.
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Two regular polygons are inscribed in the same circle.
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The first has 1731 sides.
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And the second has 4039.
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If the two polygons have at least one vertex in common, how many vertices in total will coincide?
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Remember, the geometrical interpretation of the πth roots of unity on an Argand diagram is as the vertices of a regular π-gon inscribed within a unit circle whose centre is the origin.
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This means then that we can say that, to solve this problem, we need to find the number of common roots of π§ to the power of 1731 minus one equals zero and π§ to the power of 4039 minus one equals zero.
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Remember, the common roots of π§ to the power of π minus one equals zero and π§ to the power of π minus one equals zero are the roots of π§ to the power of π minus one equals zero, where π is the greatest common divisor of π and π.
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So we know that the common roots of our two equations are the roots of π§ to the power of π minus one equals zero, where π is the greatest common divisor of 1731 and 4039.
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And this means if we can find the value of π, the greatest common divisor of 1731 and 4039, that will tell us how many common roots there actually are.
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As a product of their prime factors, they can be written as three times 577 and seven times 577, respectively.
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So their greatest common divisor and the value of π is 577.
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And this means that as long as the polygons have one vertex in common, they will actually have a total of 577 vertices that coincide.
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In this video, weβve seen that we can find the πth roots of unity and express these in either polar or exponential form.
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In exponential form, they are π to the two ππ over π π, where π takes integer values from zero through to π minus one.
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We saw that if we represent these roots on an Argand diagram, the points that represent them from the vertices of a regular π-gon inscribed within a unit circle whose centre lies at the origin.
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We also saw that the sum of the πth roots of unity is zero for values of π greater than one.
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And we saw that the reciprocal of an πth root of unity is equal to the complex conjugate of that root.
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And that is also in itself an πth root of unity.
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Finally, we saw that we can find the common roots of π§ to the power of π minus one equals zero and π§ to the power of π minus one equals zero by finding the roots of π§ to the power of π minus one equals zero, where π is the greatest common divisor of π and π.
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And we considered briefly the geometrical application of this fact.