Chapter Summary: Mixed Solved Problems on Complex Algebra

Transcript

0:00Introduction

Hello everyone! Let's start with one quick question. Mod z equals one, and we look at z over one minus z square. Is this number real, purely imaginary, or neither? Most students put z equals x plus iota y, and fill a full page. Toppers finish it in two lines. How? Today is our chapter summary: three tools that crack mixed problems fast, and two solved JEE problems. So, let's begin.

0:30Tool 1, the topper's trick: the conjugate test

So, tool number one, the topper's trick of this chapter. Quick reminder: if z is x plus iota y, then z bar, the little bar on top, is x minus iota y. The double arrow means implies. We just flip the sign of the imaginary part. Clear? Now think. Suppose a number E is real, so its imaginary part is zero. What happens when you take its conjugate? Exactly, nothing changes! E bar equals E itself. And it works both ways: this double-headed arrow means if and only if. So, to show a number is real, show that its conjugate is the same number. Alright, now the other case. If E is purely imaginary, it has no real part at all. What does the bar do to it? It flips the sign, correct! E bar equals minus E, okay? Zero also passes this test, so the line says: purely imaginary, or zero. Why is this test so fast? The bar goes inside products, quotients and powers. Here z one and z two are any two complex numbers; the small one and two are just labels. So, conjugate of z one z two is z one bar, z two bar; of the whole bracket, z one over z two, it is z one bar over z two bar; and of z to the n, it is z bar, to the n. So you never need to put z equals x plus iota y and expand. In JEE, this saves you real minutes. Got it?

2:04Tool 2: on the circle |z| = r, use z bar = r^2/z

Tool number two. Mod z, written with these two vertical bars, is the distance of z from the origin. And z times z bar equals mod z, square. This one is always true, okay? Now suppose mod z equals r, so z lies on a circle of radius r, centred at the origin. Can you write z bar using only z? Yes, exactly! z times z bar is r square, so divide by z. z bar equals r square over z. Now every bar can become plain z, and the algebra gets easy. And the most common case: mod z equals one. Then z bar is simply one over z. Remember this one! Alright, one more step. Mod z is one, so mod z square is one. So z square is also one, right? No, no, careful! This is the trap. In general, mod z square is not z square; the crossed equals sign means not equal. Look: mod iota, whole square, is one, but iota square is minus one. Only z times z bar equals mod z square. Clear?

3:14Tool 3: the triangle inequality

Tool number three: the triangle inequality. Read it with me. Mod of z one plus or minus z two is at most mod z one plus mod z two, and at least the mod of, mod z one minus mod z two. This sign with a line under it means less than or equal to. Look at the figure. Go along z one, then along z two, and the sum, z one plus z two, closes the triangle. So think: can one side of a triangle be longer than the other two sides together? No, never! One side is at most the sum of the other two. That is the right half of our line. And one side is at least the difference of the other two; this sign means greater than or equal to. That is the left half. That's it! In Example two, we use it to find the greatest and least mod z.

4:02Example 1: show z/(1-z^2) is purely imaginary if |z|=1

Now, on to Example one, and it is our question from the start. If mod z equals one, and z is not equal to plus or minus one, show that z over one minus z square is purely imaginary. That not equal condition just keeps the denominator away from zero. Pause the video, and try it with our tools. Done? Okay, let's check. Let's call the expression E, okay? Mod z is one, so by tool two, z bar is one over z. Now tool one: take the conjugate, and let the bar go inside. E bar equals z bar, over one minus z bar square. Put z bar equals one over z, and we get one over z, over one minus one over z square. Fractions inside a fraction, right? So multiply the top and the bottom by z square. What do you get? z over z square minus one, perfect! Now compare with E, which was z over one minus z square. Same thing, with the sign flipped. So E bar equals minus E, okay? So, by the conjugate test, E is purely imaginary. And it is never zero, because z itself is not zero. See, two lines, and no x plus iota y anywhere. That is how toppers do it!

5:25Example 2: greatest and least |z| if |z - 4/z| = 2

Now a harder one, JEE level. Mod of, z minus four over z, equals two. Find the greatest and the least values of mod z. Pause the video and try it. Hint: tool three. Done? Okay, let's check. Many students write: mod of z minus four over z is just mod z, minus four over mod z. Then they solve one equation and get one answer. Looks fine, right? No, no, careful! Mod does not split over a minus sign. Tool three gives only inequalities. Let r be mod z; then mod of four over z is four over r. So two is at least r minus four over r, and two is also at least four over r minus r. With me? Take the first one, and multiply by r, which is positive: r square minus two r minus four is at most zero. That happens only between the roots of this quadratic, and r is positive, so r is at most one plus root five. So the first inequality gave us the greatest value. Now where will the least value come from? Right, from the second inequality! Four over r minus r is at most two. Multiply by r: r square plus two r minus four is at least zero. So r is at least root five minus one. But wait. Inequalities only give bounds. Can mod z actually reach these values? Yes! Check on the real axis. Take z equals one plus root five. Then four over z is root five minus one, so z minus four over z is exactly two. And take z equals root five minus one. Now four over z is root five plus one, so z minus four over z is minus two, and its mod is two. Both bounds are reached. So the greatest mod z is root five plus one, and the least is root five minus one. The trap would have given you just one value, right? Mod does not simply split over a minus sign. Got it?

7:32Quick revision: conjugate test, z bar on a circle, bounds

So, quick revision. Real means E bar equals E, and purely imaginary means E bar equals minus E, okay? On the circle mod z equals r, z bar is r square over z. And mod z square is always z z bar, not z square. For the greatest or least mod z: triangle inequality, then check that the bounds are reached. Now go and try the practice problems on this page. All the best!