The n-th Roots of Unity and of Any Complex Number

Transcript

0:00Introduction

Hello everyone! Let's take one small challenge before we start. Cos two pi by seven, plus cos four pi by seven, plus cos six pi by seven. Two pi by seven is not a standard angle, right? So how will you add these? And yet, the answer is a clean, simple fraction! The secret is the n-th roots of unity. And in JEE, they come up again and again. So, let's begin.

0:26The n-th roots of unity

So, start with the simplest equation: z to the power n equals one; n is the small power up there. Tell me, how many roots does it have? Yes, exactly n roots! We call them the n-th roots of unity, because unity means one. Clear? And here they are. Alpha k, alpha with a small k below it, equals e to the power two pi iota k by n; e is Euler's number, alpha and pi are Greek letters, and iota is the imaginary unit. In cos and sin form: cos two k pi by n, plus iota sin two k pi by n. And k runs from zero up to n minus one, so n roots, okay?

1:10A regular n-gon on the unit circle

Now a shortcut. Call the root with k equals one simply alpha: e to the two pi iota by n. Then alpha k is just alpha to the power k, so the roots are one, alpha, alpha square, up to alpha to the power n minus one. Really neat, right? And look at the picture. On the left, the cube roots from our last lesson, one, omega and omega square: an equilateral triangle. On the right, n equals eight: one, alpha, alpha square, alpha cube and the rest, all on the unit circle, with the same angle, two pi by n, between neighbours. So the n-th roots of unity form a regular n-gon in the unit circle. Beautiful, isn't it?

1:55Roots of any complex number

Alright, now any non-zero complex number w. Write it as capital R, e to the iota phi: R is the modulus, and phi, this Greek letter, is the argument. First guess for its n-th root: R to the power one by n, times e to the iota phi by n. Is that the only one? No, no! That is just one of them. Phi plus two k pi points in the same direction as phi, right? So the n-th roots are R to the power one by n, times e to the iota, phi plus two k pi, by n. With k from zero to n minus one, again exactly n roots, all with modulus R to the power one by n. Remember this; Example two needs it.

2:41Sum and product of all n roots

Now the properties, and JEE loves these. Add all n roots: one, plus alpha, plus alpha square, up to alpha to the power n minus one. What do you get? Zero! Yes, exactly. For n greater than or equal to two, okay? Why? It is a geometric series: alpha to the n, minus one, over alpha minus one. Alpha to the n is one, so the top is zero; and alpha is not equal to one, that is the slashed equal sign. That's it, zero! Next, the product of all n roots; the dots mean multiplication. It equals minus one, to the power n minus one. See, the power is n minus one, not n. Students mix this up, so be careful. Clear? And since the sum is zero, its real and imaginary parts are both zero. So the sum of cos two k pi by n, for k from zero to n minus one, is zero; the big Greek letter sigma means sum. Same for the sins. Our challenge needs this one.

3:48The JEE trap: is the root 1 in the sum?

Now, the classic JEE trap. Our sum uses all n roots, from k equals zero, including the root one. But now a question starts from alpha: alpha plus alpha square, up to alpha to the power n minus one. Roots of unity, so zero, right? No, no, careful! The root one is missing. One plus this sum is zero, so this sum is minus one. Always check where the sum starts. Got it?

4:16Topper's trick: factorise zⁿ-1

Now a trick that toppers use. The roots of z to the n, minus one, are exactly the n roots of unity. So it factorises: z minus one, times z minus alpha, and so on, up to z minus alpha to the power n minus one. Divide both sides by z minus one. Left side: one plus z, up to z to the power n minus one. Right side: the remaining factors. Now put z equals one. What does the left side become? One plus one plus one, n times: just n! So one minus alpha, times one minus alpha square, and so on, equals n. JEE has asked exactly this. It is a real classic.

5:01Example 1: cos 2pi/7 + cos 4pi/7 + cos 6pi/7

Now, remember our challenge? Let's finish it. Find cos two pi by seven, plus cos four pi by seven, plus cos six pi by seven. Pause the video and try it. Hint: take n equals seven. Done? Okay, let's check. These angles belong to which roots of unity? Yes, the seventh roots! Alpha to the k equals cos two k pi by seven, plus iota sin two k pi by seven, for k from zero to six. Their sum is zero, so their real parts add to zero: cos zero, plus cos two pi by seven, and so on, up to cos twelve pi by seven. But wait. Seven cosines here, and only three in our sum. What about the other four? Look, cos zero is just one. And alpha to the k and alpha to the seven minus k have equal real parts; for example, cos twelve pi by seven equals cos two pi by seven. So the last three cosines repeat our three. So, one plus two times our sum equals zero; the double arrow means implies. Our sum is minus half! Ugly angles, clean answer, as promised. And if you wrote zero, that is our trap again: the root one, cos zero, is not in our sum. With me?

6:26Example 2: number of z with |z| = 1 and |z^26 - i| = 1

Now a harder one, JEE level. How many complex numbers z satisfy mod z equals one, and mod of z to the twenty six, minus iota, equals one? The vertical bars mean modulus. Pause and try. Hint: put w equals z to the twenty six. Done? Okay, let's check. Put w equals z to the twenty six. Mod z is one, so mod w is also one, and mod of w minus iota is one. Now think geometrically. These are two circles of radius one: one with centre zero, and one with centre iota. So w lies on both circles. Where do they meet? At plus or minus root three by two, plus half iota; the root sign means square root. In Euler form, e to the iota pi by six, and e to the five iota pi by six. Just two values of w. So the answer is two, right? Oh no, careful! We want z, not w. Remember the roots of any complex number: each w has exactly twenty six distinct twenty sixth roots. And mod w is one, so all of them have mod z equal to one. Two values of w, twenty six roots each: two times twenty six, fifty two. See how the roots of any complex number saved us? Simple, isn't it?

7:54Quick revision: the n-gon, sum 0 or -1, roots of any number

So, quick revision. z to the n equals one has n roots, e to the two pi iota k by n: a regular n-gon on the unit circle. The sum of all n roots is zero; without the root one, the sum is minus one. And the n-th roots of R e to the iota phi are R to the one by n, e to the iota, phi plus two k pi, by n, for k from zero to n minus one. Now go and try the practice problems on this page. All the best!