Part A: Complex Numbers

Transcript

0:00Introduction

Hello everyone! Okay, one puzzle before we start. Take the equation, z bar equals z to the power five. How many complex numbers satisfy it? Most students say five, because of the power five. Some say six. Well, the real answer surprises almost everyone, and we will crack it by the end of this class. Today is our mixed problem class on complex numbers: the tools, the tricks, and the traps that JEE loves to set. So, let's begin.

0:32Tool 1: big powers of omega collapse

So, tool number one. This curly letter is omega, a non-real cube root of one, from chapter two. Two facts to keep ready: omega cube equals one, and one plus omega plus omega square equals zero. Okay? Now try this. The sixth power of, one minus omega plus omega square, plus the sixth power of, one plus omega minus omega square. And please, don't expand anything! Look inside the first bracket. What is hiding there? One plus omega plus omega square, yes! Write the bracket as, one plus omega plus omega square, minus two omega. The first part is zero, so the bracket is just minus two omega. Same way, the second bracket is one plus omega plus omega square, minus two omega square. So it is minus two omega square. Really simple. Now the sixth powers. Minus two, to the power six, is sixty four; the minus sign dies in an even power. So we get sixty four omega to the six, plus sixty four omega to the twelve. Six and twelve are multiples of three, so both these powers of omega are one. Sixty four plus sixty four: one hundred and twenty eight. Remember the idea: powers of omega come back to one every three steps.

1:54Tool 2: the modulus is a distance

Tool number two: the modulus is a distance. Mod of z minus a, written with these two vertical bars, is the distance of the point z from the point a. Clear? Now, mod of z minus three plus four iota equals five; the i on screen is iota. Find the greatest and least values of mod z, that is, the distance of z from the origin. So, a circle of radius five, with centre minus three plus four iota, right? No, no, careful! This is the classic sign trap. Take the minus sign outside a bracket: mod of z, minus the bracket, three minus four iota. So the centre is three minus four iota, and the radius is five, okay? Now, how far is this centre from the origin, the point O on screen? Mod of three minus four iota: the square root of nine plus sixteen, that is, five. Look at that, exactly equal to the radius! So the circle passes through the origin itself. The origin is one of our points z, and there mod z is zero. So the least value is zero. And for the farthest point, go from the origin, through the centre, and five more units to the other side. Five plus five: ten. So greatest ten, least zero. Draw the picture, and the algebra disappears.

3:16Tool 3: on the unit circle, z bar = 1/z

Tool number three: small, but very powerful. The bar on top gives z bar, the conjugate of z: same real part, and the sign of the imaginary part flipped. From chapter one: z times z bar equals mod z square. Got it? Now suppose mod z is one, so z lies on the circle of radius one around the origin, the unit circle. Then what is z bar? One over z, exactly! z times z bar becomes one; the double arrow means, implies. So z bar equals one over z. On the unit circle, the conjugate and the reciprocal are the same number. Keep this ready; example two needs it.

3:59Example 1: z^2025 for z = (root 3 + i)/2

Alright, example one. z equals root three plus iota, over two. Find z to the power two thousand and twenty five. Pause the video, and try it first. Done? Okay, let's check. The real part is root three by two, and the imaginary part is half. So mod z is the square root of three fourths plus one fourth, that is, one. And arg z, the angle from the positive real axis, is pi by six. So in Euler form, z equals e to the power iota pi by six. Right? Now, just like omega in tool one, look for a power that comes back to one. What is the smallest power of z that equals one? Twelve, that's right! Twelve times pi by six is two pi, one full turn; the dot on screen means, times. So z to the twelve is e to the two pi iota, which is one. So divide two thousand and twenty five by twelve. Twelve times one hundred and sixty eight is two thousand and sixteen, and nine is left over. So z to the two thousand and twenty five is, z to the twelve, all to the power one hundred and sixty eight, times z to the nine. The first part is just one, so only z to the nine is left. That is e to the power nine iota pi by six, or three pi by two: a three quarter turn, straight down to minus iota. No need for two thousand multiplications; just the remainder, nine. Got it?

5:30Example 2: how many z satisfy z bar = z^5

Now the harder one, and it is our puzzle from the start. How many complex numbers z satisfy, z bar equals z to the power five? Pause the video and try it. Hint: take the modulus of both sides first. Done? Okay, let's check. Take the modulus of both sides. Mod of z bar is just mod z, and mod of z to the five is mod z, to the power five. So mod z equals mod z to the power five. Take mod z common: mod z, times, mod z to the four minus one, equals zero. And mod z is a real number, never negative. So mod z is zero, or mod z is one. Two cases, okay? Case one: mod z equals one. Now remember tool three. What can we write for z bar? One over z, yes! So one over z equals z to the five. Multiply by z: z to the six equals one. These are the sixth roots of unity from chapter two: six different numbers, so six solutions. So the answer is six, right? Write it and move on? No, no, wait! We forgot case two: mod z equals zero. That means z equals zero itself. Zero bar is zero, and zero to the five is zero. So it works too! See, this is the JEE trap: students lose the zero solution. So the total is six plus one: seven. Our puzzle is solved: not five, not six, but seven complex numbers. Remember: modulus first, then the cases, and never forget z equals zero. Clear?

7:14Quick revision: big powers, modulus as distance, z bar

So, quick revision. One: for big powers, find the power that gives one, then use the remainder. Two: mod of z minus a is a distance, so draw the circle; and on the unit circle, z bar is one over z. Three: for equations with z bar, take the modulus first, and never forget z equals zero. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!