The Argand Plane and the Modulus

Transcript

0:00Introduction

Hello everyone! Let's start with one small puzzle. Iota square is minus one, we all know that. So if mod z square were z square, mod iota, whole square, would also be minus one. But mod is a length. Can a length, squared, be negative? Hmm, something is wrong, right? Today we catch the mistake. And look, modulus questions come again and again in JEE, okay?

0:28Every complex number is a point

So, take z equals x plus iota y. It is fixed by two real numbers, x and y: an ordered pair. So we draw z as the point P, x comma y; the double arrow means each gives the other. We can also see z as the vector O P, the little arrow on top. Clear? This is the Argand plane. The x axis is the real axis, the y axis the imaginary axis; the curly R and curly I in the figure mean real and imaginary. Theta, the angle, comes later; z bar is the mirror image of z in the real axis. Today, let's measure the length r.

1:07Modulus: the distance from the origin

Alright, now the length. The blue arrow goes x across and y up, and the dashed lines make a right angled triangle, right? So how far is P from the origin? Yes, Pythagoras! Root of x square plus y square. This is the modulus of z, written with two vertical bars, and read mod z. In words: root of, real part of z whole square, plus imaginary part of z whole square. Being a distance, mod z is greater than or equal to zero, never negative. And it is zero only at the origin; the double headed arrow means if and only if. Keep our puzzle in mind!

1:46Properties of the modulus

Now the most important property. Take z equals x plus iota y, and its conjugate, z bar, the z with a bar on top: x minus iota y. Multiply them. What do you get? x square plus y square, yes! Because iota square is minus one. And that is mod z square. So, z z bar equals mod z square. Write it down; it is the key to our puzzle. For two complex numbers z one and z two, the small one and two are just labels. Mod of a product is the product of the mods, the same for a quotient, and mod of z to the power n is mod z to the power n. Got it? And this one is pure geometry: mod of z one minus z two is the distance between the points z one and z two. See, it turns algebra into pictures.

2:37The trap: |z|² is not z²

So, back to our puzzle. Iota is the point zero comma one. So, what is mod iota? One, exactly! Root of zero square plus one square. So mod iota, whole square, is one. Positive, like every length squared. But iota square is minus one. So mod z square is not z square in general; the crossed equals sign means not equal. That was our puzzle's mistake! Always true: mod z square equals z z bar. With me?

3:09Triangle inequality

Now this figure. Blue arrow: z one. Orange arrow: z two, moved to the tip of z one. Green arrow: their sum. The dashed lines complete the parallelogram, so the three lengths form a triangle. Can one side of a triangle be longer than the other two together? Never, of course! So mod of z one plus z two is less than or equal to mod z one plus mod z two. And one side is at least the difference of the other two: mod of, mod z one minus mod z two. Plus or minus: it works for the difference too. This is the triangle inequality. Now the equality cases, for the plus sign, that is, z one plus z two. Equality on the right only when z one and z two point the same way; the triangle goes flat. Equality on the left, when they point in opposite directions. Really simple, right?

4:02Topper's trick: nearest and farthest points

Now a topper's trick. Mod of z minus c is the distance of z from c. So mod of z minus c equals r means z stays at distance r from c: a circle, centre c, radius r, okay? Now fix any point, call it a, at distance d from the centre: d equals mod of a minus c. As z moves on the circle, where is it nearest to a, and where is it farthest? On the line through a and the centre c, exactly! So the maximum of mod z minus a is d plus r, and the minimum is mod of d minus r. No x plus iota y, no squaring. One line! Clear? For a whole disc, mod of z minus c less than or equal to r, the same idea works. Only, if a is inside, the minimum is zero, since z can sit right on a. So first check: inside or outside?

4:55Example 1: modulus of -3+4i, (2-i)^5 and quotients

Let's do Example one. Find the modulus of each. Part one: minus three plus four iota. Part two: one plus iota, over one minus iota. Part three: two minus iota, to the power five. Part four: three plus four iota, times one minus iota, over five minus twelve iota. Pause the video, and try. Done? Okay, let's check. Part one: root of nine plus sixteen, that is five. Part two: mod of a quotient is the quotient of the mods, root two over root two. Just one! No division needed. Part three: two minus iota, to the power five. So first we expand it by the binomial theorem, right? No, no, why work so hard? Mod of z to the power n is mod z to the power n. Mod of two minus iota is root five, and root five to the power five is twenty five root five, okay? Part four: product and quotient together. The mods of three plus four iota, one minus iota and five minus twelve iota are five, root two and thirteen; the dot means times. So, five root two by thirteen. Very quick.

6:11Example 2: least value of |2z-6+5i| on a disc

Now a harder one, a real JEE question. If mod of z minus three minus two iota is at most two, find the minimum of mod of two z minus six plus five iota. Pause and try it. Hint: use the topper's trick. Done? Okay, let's check. The condition: mod of z minus, three plus two iota, at most two. So z lies in a disc, with centre c equals three plus two iota, and radius two. Now take the two outside: two times mod of z minus, three minus five by two iota. Call this point a, okay? So we need the least distance of z from a, and then double it. Alright, it is a disc, and for a disc the minimum is zero, right? No, no, careful! Zero only if the point a is inside. So check. c minus a is nine by two iota, so d is nine by two, more than two. So the point a is outside the disc. Now the trick. Since a is outside, mod of d minus r is just d minus r. So the minimum of mod z minus a is nine by two minus two, that is five by two. Okay? But don't stop here! We took a two outside. So the answer is two times five by two: five. Forget that two, and you lose the question. Got it?

7:33Quick revision: modulus, z times z bar, nearest and farthest

So, quick revision. Mod z, root of x square plus y square, is the distance from the origin; mod of z one minus z two is a distance too. Mod z square is z z bar, not z square in general. Mod of a sum is at most the sum of the mods. On a circle, the farthest is d plus r, the nearest is mod of d minus r. Now try the practice problems on this page. All the best!