Chapter Summary: Mixed Solved Problems on Complex Geometry

Transcript

0:00Introduction

Hello everyone! Alright, one puzzle before we start. Draw two discs in the Argand plane. The first has radius one, the second has radius two, and they sit some distance apart. Now pick one point from each disc. How close can these two points ever get? Most students put z equals x plus iota y, and fill a full page with algebra. Toppers just draw two circles. Today is our chapter summary on complex geometry: three tools that crack mixed problems, and two solved JEE style problems. So, let's begin.

0:38Tool 1: the modulus is a distance

So, tool number one, the most useful idea of this chapter. Mod z, written with these two vertical bars, is the distance of the point z from the origin, the point O on screen. Okay? Now take two points, z one and z two; the small one and two written below are just labels. So tell me: what does mod of z one minus z two measure? The distance between the two points, yes! Subtract them, take the mod, and you have the distance. As simple as that. Now fix one point a, and a positive number r. Collect all the points z whose distance from a is exactly r. What shape do they make? Circle, exactly! Centre a, and radius r. So mod of z minus a equals r is a circle, and we did not need x and y at all. And this sign, less than or equal to, means the distance is at most r. So we get the circle together with all the points inside it: a disc, like a coin lying flat. Our puzzle is made of two such discs, right?

1:47Tool 2, the topper's trick: distance or angle?

Tool number two, the topper's trick for mixed problems. Every condition here is about a distance, or about an angle. Mod of z minus a is a distance, and the arrow means, leads to: a circle, a perpendicular bisector, or a conic. And what if you see arg, the argument of a complex number, or a ratio that must be real or purely imaginary? What is that about? An angle, yes! And angle conditions lead to an arc, a circle, or a line. Careful: if z minus a, over z minus b, is purely imaginary, you get both arcs: the whole circle with the segment from a to b as diameter, leaving out only a and b themselves. You have seen all of these loci in this chapter, okay? So ask this question first, and draw the picture. Expand with z equals x plus iota y only if you really need to. Look, in JEE Main this saves you several minutes. Got it?

2:48Tool 3: the triangle inequality

Tool number three: the triangle inequality, from chapter one. Read it with me. Mod of z one plus or minus z two is at most mod z one plus mod z two. And it is at least the mod of, mod z one minus mod z two. Same sign as before: less than or equal to. Look at the figure. Go along z one, then along z two. The green arrow, z one plus z two, closes the triangle; the dashed lines complete the parallelogram. Now think: can one side of a triangle beat the other two put together? Never, of course! 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, opening the other way, means greater than or equal to. That is the left half, okay? For mixed problems, keep this form ready: mod of z one minus z two is at least mod z one, minus mod z two. It comes from the left half, because a real number is never bigger than its own mod. Example one needs exactly this line.

3:58Example 1: greatest |z| when |z - 3/z| = 2

Now, Example one. Mod of, z minus three over z, equals two. Find the maximum value of mod z. Pause the video, and try it with our tools. Done? Okay, let's check. Let r be mod z. The mod of a quotient is the mod of the top, over the mod of the bottom. So mod of three over z is three over r. Clear? Now, we want a maximum, so we need an upper limit on r. Which half of the triangle inequality gives it? The left half, the line we kept ready! Mod of z minus three over z is at least mod z, minus mod of three over z, that is, r minus three over r. And that mod is given to be two. So two is at least r minus three over r. Multiply by r; r is greater than zero, so the sign does not flip. r square minus two r minus three is at most zero. The double arrow means, implies. Factorise: r minus three, times r plus one, is at most zero. And r plus one is positive, so r is at most three. So the maximum is three, and we are finished, right? Not so fast! An inequality only gives a limit. We must check that three is really reached. Take z equals three. Then three over z is one, and z minus three over z is three minus one, that is, two. The tick says it works. So, the maximum of mod z is three. Remember, for the greatest or least mod z: triangle inequality first, then check that the value is reached. Got it?

5:38Example 2: least distance between points of two discs

Now a harder one, JEE level, and it is our puzzle from the start. Mod of z one minus eight minus two iota is at most one. Mod of z two minus two minus six iota is at most two. Find the minimum of mod of z one minus z two. Pause the video and try it. Hint: tool one. Done? Okay, let's check. First disc. Many students see minus eight and minus two iota, and say the centre is minus eight minus two iota. Looks fine, right? No, no, careful! Take the minus sign outside a bracket: mod of z one, minus the bracket, eight plus two iota. So it is a disc with centre eight plus two iota, and radius one. With me? Same way, the second disc: centre two plus six iota, and radius two. Now, how far apart are the two centres? Use tool one. Subtract, and take the mod, exactly! Eight plus two iota, minus the bracket, two plus six iota, gives six minus four iota. Its mod is the square root of thirty six plus sixteen, that is, root fifty two, which is two root thirteen. Two root thirteen is about seven point two; the wavy equals sign means, approximately. But the two radii add up to only one plus two, three. Seven point two is greater than three, so the discs do not meet. There is a gap between them. So, where are the two closest points, one on each disc? On the line joining the two centres, absolutely! Just like the nearest point of a circle, from chapter one. The segment between the centres is two root thirteen long. The first disc covers one unit of it, and the second covers two. So the minimum is two root thirteen minus one minus two, that is, two root thirteen minus three. Our puzzle is solved, with no x and no y, only a picture!

7:43Quick revision: triangle inequality, gap between two discs

So, quick revision. Ask first: distance or angle? Mod of z minus a equal to r is a circle, and at most r is a disc. For the greatest or least mod z: triangle inequality, then check that the value is reached. And when two discs do not meet, the minimum distance is the distance between the centres, minus the sum of the radii. That is our chapter summary on complex geometry! Now go and try the practice problems on this page. All the best!