Chapter 1. Complex Numbers. Dr. Pulak Sahoo

Transcription

1 Chapter 1 Complex Numbers BY Dr. Pulak Sahoo Assistant Professor Department of Mathematics University Of Kalyani West Bengal, India sahoopulak1@gmail.com 1

2 Module-3: Straight Line and Circle in the Complex Plane 1 Equation of a Straight Line We now find out the equation of a straight line in the complex plane taking the real and imaginary axes as the axes of coordinates. The equation of any straight line in 2-dimensional cartesian coordinate system can be written as ax + by + c = 0, (1.1) where a, b and c are real numbers and a, b are not together zero. Let z = (x, y). Then z = (x, y). Therefore we can write z = x + iy and z = x iy. Putting x = z+z 2 and y = z z 2i From this we get in (1.1) we obtain z + z z z a + b + c = i (a ib)z + (a + ib)z + 2c = 0. Taking α = a + ib and 2c = c 1 we obtain from above that αz + αz + c 1 = 0, which is the equation of a straight line in the complex plane. Next we consider an equation of the form βz + βz + k = 0, (1.2) 2

3 where β is an arbitrary nonzero complex number and k is any real constant. Now if z = (x, y) and β = (a, b) then we obtain (a ib)(x + iy) + (a + ib)(x iy) + k = 0. From this we get 2ax + 2by + k = 0. That is ax + by + k = 0. Hence (1.2) represent a straight line in the complex plane. Equation of a straight line passing through two given points We consider the straight line passing through two points A(z 1 ) and B(z 2 ). Let P (z) be any other point on the line (see Fig. 1.1). Then Fig. 1.1: Therefore z z 1 z 1 z 2 is purely real and hence arg z z 1 z 1 z 2 = 0 or π. i.e. z z1 z 1 z 2 (z z 1 ) (z 1 z 2 ) = z z 1 z 1 z 2 = z z 1 z 1 z 2 i.e. (z z 1 )(z 1 z 2 ) = (z 1 z 2 )(z z 1 ) i.e. (z z 1 )(z 1 z 2 ) = (z 1 z 2 )(z z 1 ) i.e. z(z 1 z 2 ) z(z 1 z 2 ) + (z 1 z 2 z 2 z 1 ) = 0, which is the desired equation of the straight line. 3

4 2 Equation of a Circle We consider a circle in the complex plane having center at C(a) and radius r. If P (z) is any point on its circumference, then CP = z a. Therefore, CP = z a = r and hence z a 2 = r 2 i.e. (z a)(z a) = r 2 i.e. (z a)(z a) = r 2 i.e. z z a z a z + a a = r 2 i.e. z z a z a z + ( a 2 r 2 ) = 0, which is the equation of a circle in complex plane. Equation of a Circle through three given points We consider the circle passing through three points P (z 1 ), Q(z 2 ) and R(z 3 ). Let S(z) be any other point on its circumference. Then P SQ = P RQ (see Fig. 1.2). If S is in Fig. 1.2: between P and Q, then P SQ P RQ = π. Thus in any case, That is P SQ P RQ = 0 or π. Arg z z 1 z z 2 Arg z 3 z 1 z 3 z 2 = 0 or π i.e. Arg (z z 1)(z 3 z 2 ) (z z 2 )(z 3 z 1 ) 4 = 0 or π.

5 Therefore (z z 1)(z 3 z 2 ) (z z 2 )(z 3 z 1 ) is real and hence i.e. (z z1 )(z 3 z 2 ) (z z 2 )(z 3 z 1 ) which is the equation of the circle. (z z 1 ) (z 3 z 2 ) (z z 2 ) (z 3 z 1 ) = (z z 1)(z 3 z 2 ) (z z 2 )(z 3 z 1 ) = (z z 1)(z 3 z 2 ) (z z 2 )(z 3 z 1 ), Example 1.1. Find the equation of the circle joining the points A(z 1 ) and B(z 2 ) as diameter of the circle. Solution. Let P (z) be any point on the circumference of the circle (see Fig. 1.3). Then the angle between the lines AP and BP is π 2. Therefore arg z z 1 z z 2 = π 2, and hence z z 1 z z 2 purely imaginary. So, which is the required equation. z z 1 z z 2 + z z1 z z 2 = 0 z z 1 i.e. + z z 1 = 0 z z 2 z z 2 i.e. (z z 1 )(z z 2 ) + (z z 2 )(z z 1 ) = 0, is Fig. 1.3: Example 1.2. Show that two lines with end points z 1, z 2 and z 3, z 4 are perpendicular if and only if z1 z 2 Arg = ± π z 3 z

6 Solution. Let AB and CD be the two lines and the numbers z 1, z 2 correspond to the end points of AB and the numbers z 3, z 4 correspond to the end points of CD (see Fig. 1.4). The angle made by the line AB with the positive real axis is α = arg(z 1 z 2 ), and the line CD makes an angle β = arg(z 4 z 3 ) with the positive real axis. Hence, the two lines AB and CD are perpendicular if and only if arg(z 1 z 2 ) arg(z 4 z 3 ) = ± π 2, and therefore we must have Arg z1 z 2 = ± π z 3 z 4 2 as the necessary and sufficient condition for the two lines being perpendicular. Fig. 1.4: Example 1.3. Verify that the sum of all the three interior angles of a triangle is equal to π. Solution. Labeling the three vertices of the triangle as P,Q, and R in anticlockwise order, (see Fig. 1.5) the three interior angles of the triangle PQR are given by ( ( ( z α = Arg 3 z 1 z z 2 z 1 ), β = Arg 1 z 2 z z 3 z 2 ), γ = Arg 2 z 3 z 1 z 3 ), where z 1, z 2, and z 3 are the complex numbers corresponding to the three points P,Q,R respectively. Therefore the sum of the three angles of the triangle is 6

7 Fig. 1.5: z3 z 1 z1 z 2 α + β + γ = Arg + Arg + Arg z 2 z 1 z 3 z 2 z3 z 1 z 1 z 2 z 2 z 3 = Arg z 2 z 1 z 3 z 2 z 1 z 3 = Arg ( 1) = π. z2 z 3 z 1 z 3 Example 1.4. Show that for any complex number z, z 1 z+1 = λ and arg ( z 1 z+1) = µ, represents orthogonal circles where λ and µ are constants. Solution. We have z 1 z + 1 = λ (x 1) + iy i.e. (x + 1) + iy = λ where g = λ2 +1. Clearly (1.3) represent a circle. λ 2 1 Again arg ( z 1 z+1) = µ implies that i.e. (x 1)2 + y 2 (x + 1) 2 + y 2 = λ 2 i.e. x 2 + y 2 + 2gx + 1 = 0, (1.3) arg (z 1) arg (z + 1) = µ i.e. arg (x 1 + iy) arg (x iy) = µ y y i.e. tan 1 tan 1 = µ x 1 x + 1 i.e. tan 1 2y = µ x 2 + y 2 1 2y i.e. x 2 + y 2 1 = tan µ = ν, say, i.e. x 2 + y 2 + 2fy 1 = 0, (1.4) 7

8 where f that is = 2. Clearly (1.4) represent a circle. Since the condition for orthogonality, ν 2g 1 g 2 + 2f 1 f 2 = c 1 + c 2 is satisfied by the circles (1.3) and (1.4), they are orthogonal. 8