Argument, Polar Form and Euler Form

Transcript

0:00Introduction

Hello everyone! Let's start with one small puzzle. Find the angle of minus one minus iota. y by x is one, and tan inverse of one is pi by four. But pi by four points up and to the right, and minus one minus iota sits down and to the left! Wait, something is wrong here, isn't it? Today we find the mistake, and write any complex number with a length and an angle. JEE uses this idea a lot. So, let's begin.

0:33Argument: the angle of the arrow

Take z equals x plus iota y, not zero; the crossed equals sign means not equal. In the first picture, the blue arrow is the vector O P, the little arrow on top. The green curve turns anticlockwise from the positive real axis to it. That angle, theta, is an argument of z. The alpha in the caption, we meet in a minute. Its length is mod z, so cos theta is x by mod z, and sin theta is y by mod z. Now turn the arrow one more full round, two pi. Does the point move? Not at all! Same point, same z. So arguments of z differ by multiples of two pi. We pick one: the principal argument, arg z, between minus pi and pi; the curved E means belongs to. Round bracket, minus pi not included; square bracket, pi included. And zero has no arrow, so arg of zero is not defined. Okay?

1:33Principal argument: quadrant first

So how do we find it? First, alpha equals tan inverse of mod y by x; the bars make it positive. It is the acute angle with the real axis, from zero to pi by two; square brackets, both ends included. Now look at the four pictures. First quadrant: just alpha. Second: pi minus alpha. Below the real axis we turn clockwise, so the angle is negative. Third: minus pi plus alpha; the figure writes minus of, pi minus alpha. Fourth: minus alpha. Now, on the negative real axis: pi, or minus pi? Pi, yes! Minus pi is left out, remember the round bracket. So, positive real axis zero, positive imaginary pi by two, negative real pi, and negative imaginary minus pi by two.

2:26The trap: tan inverse y/x works only in quadrants 1 and 4

Now, back to our puzzle. For minus one minus iota, y by x is minus one by minus one, that is one. Tan inverse of one is pi by four. So the argument is pi by four, right? No, no, careful! x and y are both negative: third quadrant, not first! Alpha is pi by four, so the argument is minus pi plus pi by four: minus three pi by four. Puzzle solved! Tan inverse of y by x works only in quadrants one and four. Got it?

2:56Polar form and Euler form

Now the payoff. Let r be mod z, and theta be arg z. Then x is r cos theta, y is r sin theta, and the double arrow means so: z equals r into, cos theta plus iota sin theta. The polar form! Short form: r cis theta, from cos, iota, sin. And Euler's formula: e to the iota theta equals cos theta plus iota sin theta. So z equals r e to the iota theta: the Euler form. Now put theta equals pi. What is e to the iota pi? Cos pi plus iota sin pi: minus one, exactly! And e to the iota pi by two is iota. Mod of e to the iota theta is always one, and its conjugate, the long bar on top, is e to the minus iota theta.

3:47Moduli multiply, arguments add

Why so useful? Multiply two of them; the small one and two are labels, the dot means times. The r's multiply, the powers of e add: moduli multiply, arguments add! So, arg of z one z two is arg z one plus arg z two. For a quotient, subtract. And for z to the power n, n times arg z. z bar, the mirror image, has argument minus arg z; so does one by z. And minus z is z turned through pi: arg z, plus or minus pi. Alright, now take z one equals z two equals minus one. Each argument is pi, so the sum is two pi. So arg of z one z two is two pi, right? Careful! z one z two is one, and arg one is zero. So with principal values, add or subtract two pi to come back into range. Clear?

4:44Topper's trick: the half-angle factorisation

Now a topper's trick. In one plus e to the iota theta, take e to the iota theta by two common. The bracket: e to the minus iota theta by two, plus e to the iota theta by two. By Euler's formula, what is it? Two cos theta by two, yes, exactly! The sin parts cancel. So one plus e to the iota theta is two cos theta by two, times e to the iota theta by two. So one plus cos theta plus iota sin theta becomes two cos theta by two, into cos plus iota sin of theta by two. Looks like polar form! Looks like. Remember that for Example two.

5:22Example 1: modulus and principal argument of six numbers

Now, on to Example one. Find the modulus and principal argument of: one plus iota root three, minus root three plus iota, minus two minus two iota, three minus three root three iota, minus five, and minus four iota. Quadrant first! Pause the video, and try. Done? Okay, let's check. Part one: r is root of one plus three, two. First quadrant, alpha is tan inverse root three, so arg is pi by three. Part two: r is root of three plus one, two. Second quadrant: x negative, y positive. Alpha is pi by six, so arg is pi minus pi by six, five pi by six. Part three: r is root of four plus four, two root two. Third quadrant, like our puzzle! Alpha is pi by four, so arg is minus pi plus pi by four, minus three pi by four. Part four: r is root of nine plus twenty seven, six. Fourth quadrant, alpha is pi by three, so arg is minus pi by three. Last two, on the axes. Minus five: r is five, arg is pi. Minus four iota: r is four, arg is minus pi by two. That's it! Quadrant first, every time, and the trap can never catch you.

6:44Example 2: argument of 1 + cos 6pi/5 + i sin 6pi/5

Now a harder one, JEE level. Find the modulus and argument of one plus cos six pi by five, plus iota sin six pi by five. Pause and try. Hint: the topper's trick. Done? Okay, let's check. Theta is six pi by five, so theta by two is three pi by five. By the trick, z is two cos three pi by five, into cos plus iota sin of three pi by five. So the modulus is two cos three pi by five, and the argument is three pi by five. Right? No, no, careful! Remember, looks like. Three pi by five is more than pi by two, so its cos is less than zero. But a modulus is never negative. So, not polar form yet! So, push a minus sign into the bracket: minus two cos three pi by five, times, minus cos three pi by five, minus iota sin three pi by five. Now the first factor is positive. Minus cos three pi by five is cos two pi by five. And the minus on the bracket? Just like minus z: turn through pi. So subtract pi from the angle. So mod z is two cos two pi by five, and arg z is three pi by five minus pi: minus two pi by five, in range. That's the method: the trick, plus one careful check!

8:05Quick revision: quadrant rule, polar form, half-angle trick

So, quick revision. Quadrant first, then alpha; principal argument between minus pi and pi. Polar form r e to the iota theta: moduli multiply, arguments add. Half angle trick, but keep the modulus positive. Now try the practice problems on this page. All the best!