The Rotation Theorem

Transcript

0:00Introduction

Hello everyone! First, one puzzle for you. Picture a square drawn around the circle mod z minus one equals root two, with one vertex at two plus root three iota. Find the other three vertices. Distance formula, slopes, a full page? No way! Toppers do it with three multiplications. Today: the rotation theorem. And look, JEE has asked this exact question. So, let's begin.

0:31Multiplying by e^(iθ) is a turn

So, take any complex number z, and multiply it by e to the iota theta; theta, this Greek letter, is an angle. Moduli multiply and arguments add, right? So what happens to mod z, with the two vertical bars, and to arg z, the argument? Yes, exactly! The modulus stays: mod of z e to the iota theta equals mod z. And arg of z e to the iota theta equals arg z plus theta. Clear? Same distance from the origin, but the angle grows by theta. That is a turn! So z e to the iota theta is z rotated about the origin O, through theta, anticlockwise. As simple as that. So multiply by iota, and the point turns ninety degrees anticlockwise; the little circle means degrees, and the cross means multiply. By minus one: one hundred eighty degrees. By minus iota: ninety degrees, clockwise. Now, to turn about another point z nought, with a small zero below: just shift it to the origin, turn, and shift back. The new point, z prime, with the little dash, is z nought, plus, z minus z nought, times e to the iota theta.

1:49The rotation theorem

Alright, the main theorem. In the figure, point A is z one, B is z two, and C is z three. The blue arrow is z two minus z one, the orange one is z three minus z one: two arms from A, okay? Theta is the angle from AB to AC, anticlockwise. So: how do you get arm AC from arm AB? Yes! Turn it through theta, and stretch it by AC over AB. Exactly what the box in the figure says. z three minus z one, over z two minus z one, equals mod z three minus z one, over mod z two minus z one, times e to the iota theta. The mods are the lengths AC and AB. Got it? So the argument of this ratio is theta: the angle from the AB vector to the AC vector; the arrow on top means vector. Note the order, from the bottom arm to the top arm. Now, suppose AB is perpendicular to AC; the upside down T means perpendicular. The turn is a right angle. So what kind of number is this ratio? Purely imaginary, exactly! Real stretch, times a quarter turn. The double arrow means if and only if: perpendicular exactly when the ratio is purely imaginary. With me?

3:10JEE trap: the order decides the sign of θ

Now the JEE trap. Here is our ratio, with angle theta from AB to AC. Suppose you flip it: z two minus z one, over z three minus z one. Same two arms, so the argument is still theta, right? No, no, careful! The angle goes from the vector in the denominator to the vector in the numerator, anticlockwise. Flip the fraction, and you get minus theta. See, this is how students land on the other answer in a two-answer question. Same rule for two lines: the angle between lines z one z two and z three z four is arg of z four minus z three, over z two minus z one. Bottom to top, always.

3:55Equilateral triangles

Now, the most famous use: the equilateral triangle z one, z two, z three. All sides equal. So at z one, what takes side z one z two onto side z one z three? That's right! Only a turn, no stretch. The angle is sixty degrees, that is pi by three, with pi the Greek letter, one way or the other. So z three minus z one, over z two minus z one, is e to the plus or minus iota pi by three. At z two, z one minus z two over z three minus z two gives the same number, same sign. Cross multiply, expand, and we get one symmetric line: z one square plus z two square plus z three square equals z one z two plus z two z three plus z three z one. Learn this by heart. And with a vertex at the origin: zero, z one and z two are equilateral exactly when z one square plus z two square equals z one z two. Just put z three equals zero, right?

5:03Topper's trick: squares and regular polygons

Now a topper's trick. You know the centre c of a square, and one vertex v. How do you get the next vertex? Distance formula? Never! Just turn about the centre! For a regular polygon with n sides, the other vertices are c plus, v minus c, times e to the two pi iota k by n, for k equals one, two, and so on. For a square, n is four, and this is just iota to the power k. That's it! Multiply v minus c by iota again and again, and add c back. Remember our puzzle from the start? This is exactly what it needs.

5:42Example 1: rotate 3+4i through 90 and 60 degrees

Let's do Example one. The point three plus four iota is rotated about the origin through ninety degrees anticlockwise, and separately through sixty degrees. Find the new points. Pause the video, and try it first. Done? Okay, let's check. Ninety degrees anticlockwise: multiply by iota. Three iota plus four iota square, and iota square is minus one. So, minus four plus three iota. Simple! Now the sixty degree turn. What do we multiply by? Yes, it is e to the iota pi by three! That is cos sixty degrees plus iota sin sixty degrees: half, plus root three by two, iota; this sign is the square root. Multiply term by term. Three by two, plus three root three by two iota, plus two iota, and four iota times root three by two iota gives two root three iota square: minus two root three. Collect the parts. Real part: three minus four root three, over two. Imaginary part: four plus three root three, over two. Ugly, right? But completely correct.

6:56Example 2: other vertices of a square around a circle

Now a harder one, JEE level: our puzzle from the start. One vertex of the square circumscribing the circle mod z minus one equals root two is two plus root three iota. Circumscribing means drawn around. Find the other vertices. Pause and try. Hint: the topper's trick. Done? Okay, let's check. The square's centre is the circle's centre: c equals one. And v is two plus root three iota, so v minus c, the arm from centre to vertex, is one plus root three iota. Now the next vertex. Easy, isn't it? Just multiply two plus root three iota by iota, right? No, no, careful! That turns it about the origin, but our square turns about its centre, c. Turn the arm, then add c back. For k equals one: one plus, iota times one plus root three iota, that is one plus iota minus root three: one minus root three, plus iota. For k equals two: iota square, that is minus one. One minus one minus root three iota: just minus root three iota. The opposite vertex. And for k equals three: iota cube, that is minus iota. One minus iota plus root three: one plus root three, minus iota. Look, three multiplications, no distance formula. Puzzle solved!

8:28Quick revision: rotation, angle order, equilateral, squares

So, quick revision. Multiplying by e to the iota theta turns a point by theta; about z nought, add z nought back. The ratio's argument goes from the bottom arm to the top arm. Equilateral: the sum of the squares equals the sum of the pair products. Squares: turn about the centre with iota. You did really well today. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!