De Moivre's Theorem and Powers of Complex Numbers

Transcript

0:00Introduction

Hello everyone! First, one quick challenge. Root three plus iota, to the power ten. How will you find it? Multiply it by itself ten times? Oh, that is a lot of work, and one sign mistake ruins everything, right? Today we will do it in one line, with De Moivre's theorem. And look, in JEE, powers of complex numbers come up again and again. So, let's begin.

0:28De Moivre's theorem

So, take cos theta plus iota sin theta; theta, this Greek letter, is the angle, and iota is the imaginary unit. Last time we saw that multiplying by this number adds theta to the angle. Now raise it to the power n, the small n up there. What happens to the angle? Yes, exactly! The angle becomes n theta. For every integer n, cos theta plus iota sin theta, to the power n, equals cos n theta plus iota sin n theta. This is De Moivre's theorem, and it holds for negative powers too. Okay? In Euler form, it is even shorter. Remember, e to the iota theta is just short for cos theta plus iota sin theta; this e is the base of natural logarithms. So e to the iota theta, whole to the power n, equals e to the iota n theta. Very clean, isn't it? Now put a modulus r in front: r e to the iota theta, whole to the power n. The angle gets multiplied by n. Does r also get multiplied by n? No! The modulus gets raised to the power n: r to the n, times e to the iota n theta. So, modulus to the power n, argument times n. That's it, that is the whole theorem.

1:51Useful consequences

Alright, a few results follow straight away. With a minus sign: cos theta minus iota sin theta, to the power n, equals cos n theta minus iota sin n theta. The minus sign just stays, okay? And one over cos theta plus iota sin theta is cos theta minus iota sin theta. Very handy in fractions. Now let z equal e to the iota theta. Tell me, what is z plus one over z? Two cos theta! Because z is cos theta plus iota sin theta, and one over z is cos theta minus iota sin theta. Add them, and the iota sin theta parts cancel. Simple as that. And for any power n: z to the n, plus one over z to the n, equals two cos n theta. This one is very useful in JEE problems, okay?

2:47The classic JEE trap

Now, the biggest trap of this topic. Sin theta plus iota cos theta, to the power n. Just apply De Moivre: sin n theta plus iota cos n theta, right? No, no, careful! The slashed equal sign means not equal. De Moivre's theorem is only for cos theta plus iota sin theta: the same angle in both places, and a plus sign. Check with n equals two and theta equals zero. Sin theta plus iota cos theta is then just iota, so the left side is iota square, minus one. The right side is iota. Not equal! So how do we write sin theta plus iota cos theta in the correct form? Take iota common! It equals iota, times cos theta minus iota sin theta; multiply back and check. And cos theta minus iota sin theta is e to the minus iota theta. So we get iota e to the minus iota theta. Always convert first, then apply De Moivre. Got it?

3:49Topper's trick: triple-angle formulas in two lines

Now a trick that toppers use. Put n equals three: cos theta plus iota sin theta, whole cube, equals cos three theta plus iota sin three theta. Expand the left side by the binomial theorem. Then how do you get cos three theta out of it? Yes, compare real and imaginary parts! Call cos theta c, and sin theta s. c plus iota s, whole cube, is c cube minus three c s square, plus iota times three c square s minus s cube. Real parts: cos three theta equals c cube minus three c s square. Put s square as one minus c square: four c cube minus three c, that is, four cos cube theta minus three cos theta. Imaginary parts: sin three theta equals three c square s minus s cube. Put c square as one minus s square, and you get three s minus four s cube, that is, three sin theta minus four sin cube theta. Two lines, both formulas done! The same method works for higher multiple angles too. Really powerful.

5:02Example 1: (root 3 + i)^10, (1 - i)^8 and a cis power

Now, remember our challenge? Let's finish it. Part one: root three plus iota, to the power ten. Part two: one minus iota, to the power eight. Part three: cos pi by twelve plus iota sin pi by twelve, to the power eight; pi is this Greek letter. Pause the video, and try it. Done? Okay, let's check. Step one, always: polar form. Root three plus iota has mod two, the modulus, and arg pi by six, the argument; the root sign means square root. So it is two e to the iota pi by six. Now De Moivre. Two to the ten is one thousand and twenty four, and the angle is ten pi by six. Its cos is half, and its sin is minus root three by two. So we get five hundred and twelve, minus five hundred and twelve root three iota. See? Our challenge, solved in one line! Part two. One minus iota is root two, e to the minus iota pi by four. Root two to the eight is sixteen, and the angle becomes minus eight pi by four, that is minus two pi. One full turn, back where we started. So the answer is just sixteen. Part three is already in cos plus iota sin form. Power eight: the angle becomes eight pi by twelve, which is two pi by three. So the answer is minus half, plus root three by two iota. Clear?

6:35Example 2: simplify a quotient with sin theta + i cos theta

Now a slightly harder one, JEE level. Simplify: cos theta plus iota sin theta, to the power four, divided by sin theta plus iota cos theta, to the power five. Pause and try. Hint: check the denominator first. Alright, the denominator first. By De Moivre, it is sin five theta plus iota cos five theta, right? No, no! That is exactly the trap we just saw. Convert first: sin theta plus iota cos theta is iota e to the minus iota theta. So the denominator is iota to the five, times e to the minus five iota theta. And iota to the four is one, so iota to the five is just iota. The numerator, by De Moivre, is e to the four iota theta. Divide, and the angles subtract: four theta minus minus five theta is nine theta. And one over iota is minus iota. So the quotient is minus iota, e to the nine iota theta. Last step, back to cos and sin. Minus iota times cos nine theta is minus iota cos nine theta. And minus iota times iota sin nine theta gives plus sin nine theta, because iota square is minus one. So the answer is sin nine theta minus iota cos nine theta. See? Convert first, and the trap disappears.

8:07Quick revision: power rule, sin + i cos trap, cos 3 theta

So, quick revision. Power n: raise the modulus to the power n, multiply the argument by n. De Moivre is only for cos theta plus iota sin theta, so write sin theta plus iota cos theta as iota e to the minus iota theta first. And for multiple angles, expand by the binomial theorem and compare real and imaginary parts. Now go and try the practice problems on this page. All the best!