Loci and Regions in the Argand Plane

Transcript

0:00Introduction

Hello everyone! Okay, one small challenge before we start. Mod of z minus iota equals mod of z minus one. What curve is this? Most students put z equals x plus iota y, square, expand, simplify, and two minutes are gone, right? Today you will see the answer in seconds, without any algebra. Our topic: loci and regions in the Argand plane. And look, JEE has asked these again and again. So, let's begin.

0:33Read the condition, see the curve

So, first, a locus is the set of all points z that satisfy a condition. Now, the secret of this topic. Take mod of z minus A, where A is a fixed point; the two vertical bars are the modulus. What does it mean in geometry? Yes, exactly! It is the distance from z to the point A, okay? That is half of today's lesson. Now the other half: arg of, z minus A over z minus B, okay? Arg is the argument, an angle: the angle that the segment A B subtends at z. So, read every condition as a distance or an angle, and the curve appears. Got it?

1:14Distance conditions: circles and lines

Alright, distance conditions first. Mod of z minus A equals r: every point at the same distance r from A, okay? Which curve is that? Yes, a circle! Centre A, radius r, with r positive. Less than r gives the inside, and greater than r the outside: regions, not curves. Clear? Now, mod of z minus A equals mod of z minus B: z is equally far from A and from B, okay? Where can such a point be? Exactly! On the perpendicular bisector of A B, the line through its midpoint at right angles to it. No squaring, no expanding, right? And if the distances are in a fixed ratio k? The big bars are the modulus, and the slashed equal sign says k is not equal to one. This is the Apollonius circle. Centre: A minus k square B, over one minus k square. Radius: k times mod of A minus B, over mod of one minus k square. At k equal to one, we are back to the bisector. With me?

2:22Sum and difference of distances: conics

Alright, now add the two distances. Mod of z minus A, plus mod of z minus B, equals two c. So the sum of the distances stays fixed. Which curve? Yes, an ellipse, with foci A and B, its two focus points. But only when two c is greater than mod of A minus B, the distance between them. Clear? Now, a JEE trap. What if two c is exactly equal to that distance? Still an ellipse, right? No, no, careful! In the triangle z A B, two sides together are never shorter than the third. So the sum equals that distance only when z sits on the segment itself. And if two c is smaller, no point works: the locus is empty. Now the difference: mod of, mod of z minus A, minus mod of z minus B, equals two c; see the outer bars. When two c is positive and less than mod of A minus B, it is a hyperbola, again with foci A and B, okay? Remove the outer bars, and only one branch remains.

3:30Angle conditions: rays, arcs and lines

Now, the angle conditions. Arg of z minus A equals theta; theta, this Greek letter, is a fixed angle. The direction from A to z is fixed, so z moves on a ray from A, with the point A itself left out; that is the not equal sign, okay? Next, arg of, z minus A over z minus B, equals theta, where mod theta lies strictly between zero and pi. In the left picture, the fixed points are called z one and z two. Their segment subtends the same angle theta at every z on the blue arc. So the locus is an arc through both points, without the points themselves. Now, the trap. This arc is part of a circle. So the locus is the whole circle, right? No, no! Only one arc. See the dashed arc below the points? It is not part of the locus. There the arg is theta minus pi, or theta plus pi; that is the minus plus sign, and pi is this Greek letter. So, test one convenient point, like the top of the circle, to choose the arc. Got it? Special case: this ratio is purely imaginary. Then its arg is plus or minus pi by two, a right angle, like the right picture. The angle in a semicircle is a right angle, isn't it? So the locus is the circle on A B as diameter, again without both points. And if the ratio is real, the locus is the line through A and B, with only B left out, where the denominator is zero. Clear?

5:04Topper's trick: before z=x+iy, distance or angle?

Now, the trick toppers use. Before you put z equals x plus iota y, stop and ask: distance or angle? See mod of z minus A? The arrow says: think distance, so a circle, a bisector, or a conic. See arg, or a ratio that is purely imaginary or real? Think angle: an arc or a line. This one habit saves you several minutes in JEE Main. Absolutely.

5:34Example 1: locus of |z - i| = |z - 1| (JEE Main 2019)

So, remember our challenge? This exact question came in JEE Main twenty nineteen. What does mod of z minus iota equals mod of z minus one represent? Pause the video and try it. Hint: distance or angle? Done? Okay, let's check. Distance, yes! Iota is the point zero comma one, and one is the point one comma zero. The two distances are equal, so z is equally far from both. Equidistant from two points means the perpendicular bisector. No algebra so far, right? And these two points are mirror images in the line y equals x. So the locus is the line y equals x, through O, the origin, with slope one. See? Challenge solved in seconds!

6:25Example 2: arc locus of arg((z - 1)/(z + 1)) = pi/4

Now a harder one. Find the locus of z, if arg of, z minus one over z plus one, equals pi by four. Pause and try. Hint: it is an angle condition. Done? Okay, let's check. Here A is one, and B is minus one. So the segment from minus one to one, of length two, subtends pi by four at z: an arc. Got it? Radius? For a chord of length two subtending pi by four, R is two, over two sin pi by four; the slash means divided by. That simplifies to root two, the square root of two. The centre is on the perpendicular bisector of the chord, the y axis. Its distance from the chord is root of, R square minus one, where one is half the chord. That is one. So the centre is zero comma one, or zero comma minus one. Now, remember the trap. Two possible circles, and each has two arcs. Only one arc is our locus. How will you choose it? Yes, test one point! The top of the circle centred at iota: z equals one plus root two, times iota. Put it in, and the arg comes out exactly pi by four. Tick, it works! So the locus is the major arc of x square plus, y minus one whole square, equals two, above the real axis. That is, x square plus y square minus two y minus one equals zero, with y greater than zero. One picture, one test point, done!

8:04Quick revision: distance gives circles, angle gives arcs

So, quick revision. Mod of z minus A is a distance: circle, bisector, Apollonius circle, ellipse, hyperbola. Arg is an angle: ray, arc, circle on a diameter, line. And for an arc locus, test one point to choose the correct arc. Now go and try the practice problems on this page. With this habit, you will find them easy. All the best!