COMPLEX NUMBERS

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Hector Booth
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1 COMPLEX NUMBERS 1. Any number of the form x+iy where x, y R and i -1 is called a Complex Number.. In the complex number x+iy, x is called the real part and y is called the imaginary part of the complex number. 3. A complex number is said to be purely imaginary if its real part is zero and is said to be purely real if its imaginary part is zero. 4. (a) Two complex numbers are said to be equal if their real parts are equal and their imaginary parts are equal. (b) In the set of complex numbers, there is no meaning to the phrases one complex is greater than or less than another i.e. If two complex numbers are not equal, we say they are unequal. (c) a+ ib > c + id is meaningful only when b 0, d Two complex numbers are conjugate if their sum and product are both real. They are of the form a+ib, a-ib. 6. cisθ 1 cisθ cis(θ 1 + θ ), cis θ cis θ a + ib 1 1 a + ib 1 + i 1 1 i i, i 1 + i i ( a a + b b ) + ia ( b a b ) a + b 1 1 cis(θ 1 θ ), cosθ isinθ. cosθ + isinθ 9. x + y is called the modulus of the complex number x + iy and is denoted by r or x + iy 10. Any value of θ obtained from the equations cosθ x r, sinθ y r is called an amplitude of the complex number. 11.i) The amplitude lying between -π and π is called the principal amplitude of the complex number. Rule for choosing the principal amplitude. ii) If θ is the principal amplitude, then -π < θ π π-θ θ -(π-θ) -θ

2 1. If α is the principle amplitude of a complex number, general amplitude nπ + α where n Z. 13.i) Amp (Z 1 Z ) Amp Z 1 + AmpZ ii) Amp Z Z 1 AmpZ AmpZ 1 iii) Amp z + Amp z π (when z is a negative real number) 0 (otherwise) 14. (a) r(cosθ + i sinθ) is the modulus amplitude form of x+iy. 15. If the amplitude of a complex number is π, its real part is zero. 16. If the amplitude of a complex number is π, its real part is equal to its imaginary part. 4

3 Very Short Answer Questions 1. If z (, 1) z ( 6, 3) then z1 z 1 Solution: - Given z1 z i z 6 + 3i ( ) z1 z 4 4i 4, 4. If z ( 3, 5) and z (, 6) find z, z 1 Solution: - Given z i z + 6i ( )( ) 0 z1 z 3 + 5i + 6i C + 30i i 1 ( ) z. z 4 + 8i 4, 8 3. Write the additive inverse of the following (i) Additive inverse of ( 3, 5) Additive inverse is ( 3, 5) (ii) Additive Inverse of ( 6, 5) + ( 10, 4) Given complex no is (4,1) Additive inverse is ( 4, 1)

4 (iii) Additive inverse of (,1) ( 4, 6) Let z1 + i z 4 + 6i ( ) ( ) z1z + l 4 + 6i 8 + 8i 6i + 8i Additive inverse is (-, -8) 4. If z ( 6, 3) and z (, 1) then find z1 / z 1 Solution: - Given z i z i ( 6 + 3i )( + i) z i 8 + 1i + 3i z z i 4 i i 1, If z cosθ + i sinθ then find 1 z z Solution: - z cosθ + i sinθ 1 1 cosθ i sinθ z cosθ + i sinθ cosθ i sinθ cosθ i sinθ 1 z i sinθ z

5 6. Write the multiplicative inverse of the following complex numbers (i) ( 3, 4 ) (ii) ( sin θ, cosθ ) (iii) ( 7, 4 ) (iv (,1) Solution: - (i) Let z 3 + 4i Multiplicative inverse of z is 1 3 4i 3 + 4i 3+ 4i 3 3i ( )( ) 3 4i 3 4, (ii) Let z sinθ + i cosθ Multiplicative inverse of 1 sinθ i cosθ z sinθ + 1cosθ sinθ cosθ ( i ) sinθ i cosθ sinθ i cosθ sin θ + cos θ (iii) Let z 7 + 4i Multiplicative inverse of 1 7 4i z 7+ 4i 7 + 4i 7 4i ( )( ) 7 4i 65

6 7. Write the following complex numbers in the form A + ib. i) ( 3i)(3 + 4i) ii) (1 + i) 3 iii) a ib a + ib iv) 4 + 3i ( + 3i)(4-3i) v) i ( 5i) 8 + 5i 5i vi) + 3 i 3+ i Sol: i)z ( 3i)(3 + 4i) z 6 + 8i 9i + 1 z 18 i 18 + ( 1)i ii) z (1 + i) (1) (i) + 3(1)(i) + (i) i + 6i + 8i 1+ 6i 6 8i 5 i iii) z a ib a + ib a ib a ib z a + ib a ib (a ib) a (ib) a + (ib) (a)(ib) a + b a b ab i a + b a + b

7 iv) z 4 + 3i ( + 3i)(4-3i) 4 + 3i 8 6i + 1i i i 4 + 3i 17 6i i 17 6i ( ) + i( ) (17) (6i) i i i v) z i ( 5i) 8 5 z i 8 5 z + 0i 8 vi) z + 5i 5i + 3 i 3+ i ( + 5i)(3 + i) ( 5i)(3 i) z + (3 i)(3 + i) (3 i)(3 i) 6 + 4i + 15i + 10i 6 4i 15i + 10i z i 4 19i z 13 8 z + 0i 13

8 8. Write the conjugate of the following complex numbers. i) (3 + 4i) ii) ( + 5i)( 4 + 6i) Sol: i)z 3 + 4i z 3 4i ii)z ( + 5i)( 4 + 6i) z 8 + 1i 0i + 30i z 38 8i z i 9. Simplify i) 4 6 i + i + i +...(n +1) Terms ii) i 18 3.i 7 + i ( 1+ i 4 ) ( i) 6 Sol: i) 4 6 i + i + i +...(n +1) terms..3.(n+1) i + i + i +...i ( 1) +...n terms + i ii) i 18 3.i 7 + i ( 1+ i 4 ) ( i) (i ) 3i i + i (1 + i )(i) 1+ 3i + ( 1)(1 + 1)( 1) 3i 1+ 3i

9 10. Find a square root for the following complex numbers. i) 7 + 4i ii) 47 + i8 3 Sol: i) z 7 + 4i Let square root of z be a + ib a + ib 7 + 4i (a + ib) 7 + 4i a b + abi 7 + 4i a b 8, ab 4...(1) a + ib 7 + 4i Squaring on both sides, a + ib 7 + 4i a + b a + b () Adding a b 7 a + b 5 a a 16 a ± 4 b 5 7 b 9 b ±3 a + ib ±(4 + 3i). ii) 47 + i8 3 Let the square root of z be a + ib, (a + ib) 47 + i8 3 a b 47, ab 8 3 a + ib 47 + i8 3

10 Squaring on both sides, a + b ( 47) + (8 3) a + b 49 a b 47 a a 1 a ± 1 b 96 b 48 b ± 4 3 a + ib ± ( i) 11. Find the multiplicative inverse of the following complex numbers. i) 5 + 3i ii) i iii) Sol: i) z 5 + 3i i 35 Let a + ib be multiplicative inverse then (a + ib)z 1 1 z a + ib or 1 a + ib z z a + ib (zz) z 5 3i 1 a + ib ( 5 3i) z ii) z i Let a + ib be multiplicative inverse then (a + ib)z 1 1 a + ib z

11 1 i i i i a + ib i iii) z i 35 Let a + ib be multiplicative inverse then (a + ib)z a + ib i 35 z i (a + ib) i (i ).i i 1. Express the following complex numbers in modules amplitude form. i) 1 i ii) 1+ i 3 iii) 3 + i iv) 1 i 3 Sol: i) 1-i Let 1-i r cos θ + ri sin θ r cos θ 1 and rsin θ -1 r 1+1 r 1 1 sin θ And r 1 1 cos θ r Therefore, π θ 3 π π 1-i cos + i sin 3 3

12 ii)1+ i 3 r cos θ + ri sin θ r cos θ 1 r sin θ 3 r (cos θ + sin θ ) 1+ 3 r 4 4 ± π tan θ 3 θ 3 π π cos + isin 3 3 iii) 3 + i r cos θ + ri sin θ r cos θ 3 r sin θ 1 r (cos θ + sin θ ) r ± tan θ 1 3 5π 5π cos + isin 6 6 5π 5π cos + isin 6 6 iv) 1 3i r cos θ + ri sin θ r cos θ 1 r sin θ 3 r (cos θ + sin θ ) 4 r ± tan θ 3 π θ 3

13 π π Hence cos + i sin Simplify i(3 + i)( + 4i)(1 + i) and obtain the modulus of that complex number. Sol: z i(3 + i)( + 4i)(1 + i) i( + 14i)(1 + i) i( + i + 14i 14) i( i) 4i + 3 8(4 + 3i) z 64 5 z i) If z 0 find Arg z + Arg z. ii) If z 1 1 and z i then find Arg(z 1 z ) z1 iii) If z 1 1 and z i then find Arg. z Sol: i) let z x + iy z x iy 1 y 1 y Arg z tan and Argz tan x x 1 y 1 y Arg z + Argz tan + tan x x 1 y 1 y tan tan x x 0

14 ii) z 1 1 and z i Arg(z z ) Arg z + Arg z Tan + Tan 1 0 π π π iii)z 1 1 and z i z 1 Arg Arg z1 Arg z z Tan 0 1 Tan π π π 15. i) If (cos α + isin α)(cos β + isin β ) cos θ + isin θ then find the value of θ. ii) If 3 + i r(cos θ + isin θ ) then find the value of θ in radian measure. iii) If x + iy cisα. cis β then find the value of x + y. z z iv) If ;z z is an imaginary number then find the value of z z z 1 1. v) If ( 3 + i) (a + ib) then show that a + b 4. Sol: i) (cos α + isin α)(cos β + isin β ) cos θ + i sin θ cos α cos β + isin α cos β + cos αisin β + i sin αsin β cos α cos β sin αsin β + i(sin α cos β + cos αsin β) cos(( α + β )) + isin(( α + β)) cos θ + isin θ θ ( α + β)

15 ii) let 3 + i r(cos θ + isin θ) then r cos θ 3 r sin θ 1 r (sin θ + cos θ ) 4 r ± 1 π tan θ θ 3 6 iii) x + iy (cos α + isin α)(cosβ + isin β ) (cos α cosβ sin αsin β ) + i(cos αsinβ sin αsin β) x + iy cos( α + β ) + isin( α + β) x cos( α + β) y sin( α + β) x + y 1 z iv) Given ;z 1 0 is an imaginary number ki say z z + z + z z z z 1 z z ki 4 + k 1 ki 4 + k v) ( 3 + i) (a + ib) i a + ib ( 4) a + b a + b 4 a + b

16 16. i) If z x + iy and z 1, then find the locus of z. Sol: i) z x + iy z x + y 1 x + y 1, given Locus is circle. ii) If the amplitude of (z 1) is π/ then find the locus of z. z 1 (x 1) + iy Given Tan π x 1 1 y x 1 0, y 0 also y > i) Find the equation of the perpendicular bisector of the line segment joining the points 7 + 7i, 7 7i. Sol: i)z i, z 7 7i Let A (7, 7),B(7, 7) be the points which represents above complex nos. A(7,7) M B(7, 7) Midpoint of AB is M, M(7, 0) Slope of AB Slope of the line perpendicular to AB is zero. Therefore equation of the line passing through (7,0) and having slope 0 is y 0.

17 ii) Find the equation of the straight line joining the points 9 + 6i, 11 4i in the Argand plane. Sol. Given complex nos. are A( 9, 6) B(11, 4) Slope of AB Equation of line AB: 1 y 6 (x + 9) y 1 x 9 x + y If z x + iy and if the point P in the Argand plane represents z, then describe geometrically the locus of z satisfying the equations i) z 3i 5 ii) z z 1 iii)img z 4 z 1 π iv) Arg z Sol: i) z 3i 5 x+iy 3i 5 (x ) + (y 3) 5 (x ) + (y 3) 5 x + y 4x 6y x + y 4x 6y 1 0

18 ii) z z 1 Squaring on both sides 4 z z 1 4(z )(z ) (z 1)(z 1) 4zz 8z 8z + 16 zz z z + 1 3zz 7z 7z (x + y ) 7(x) iii)img z 4 z x + iy z (x + iy) z x y + xyi Im(z ) xy xy 4 xy rectangular hyperbola. z 1 π iv) Arg z z 1 z 1 (x 1) + iy z + 1 (x + 1) + iy [(x 1) + iy][(x + 1) iy] [(x + 1) + iy][(x + 1) iy] (x 1)(x + 1) + y + iy(x + 1 x + 1) (x + 1) + y Argz y π x + y Tan y 1 x + y 1 y x + y 1 x + y y 1 0

19 19. Show that the points in the Argand diagram represented by the complex numbers + i, i, i are the vertices of an equilateral triangle. Sol: Let A(, ), B(, ), C( 3, 3 argand plane. ) be points represents given complex numbers in the AB ( + ) + ( + ) 4 BC ( + 3) + ( 3) BC AC ( + 3) + ( 3) 4 AB AC BC ABC is equilateral. 0. Find the eccentricity of the ellipse whose equation is 1 z 4 + z 10 5 Sol. Given equation is of the form SP+S 1 P a Where ( ) S 1 4,0 S ',0 and a 10 a5 5 SS ae X 5e 10e e Find the real and imaginary parts of the complex number a + ib. a ib Sol: a + ib (a + ib) (a + ib) a ib (a ib) (a + ib) (a) + (ib) + a(ib) (a) (ib) a b + iab a + b

20 a b ab + i a + b a + b Real part a a b + b ab Imaginary part a + b.. If 4x + i(3x y) 3 6i where x and y are real numbers, then find the values of x and y. Sol: We have 4x + i(3x y) 3 + i( 6). Equating the real and imaginary parts in the above equation, we get 4x 3, 3x y 6. Upon solving the simultaneous equations, we get x 3/4 and y 33/4. 3. If z 3i, then show that z 4z Sol: z 3i z 3i (z ) ( 3i) z + 4 4z 9 z 4z

21 Short Answer Questions 1. i) If (a + ib) x + iy, find x + y. Sol: i) (a + ib) x + iy a b + abi x + iy a b x ab y Now x (a b ) y 4a b x + y (a b ) + 4a b (a + b ) ii) If 3 x + iy + cosθ + isinθ then show that x + y 4x 3. 3 x + iy + cosθ + isinθ rationalizing the dr. 3( + cos θ i sin θ) ( + cos θ) i sin θ 3( + cosθ i sinθ) 4 + cos θ + 4 cosθ + sin θ cosθ 3 i sin θ cos θ cos θ 3 i sin θ cos θ cos θ 6 3 cos 3 sin x + θ, y θ cos θ cos θ L.H.S cos θ 3 sin θ x + y cos θ cos θ cos θ + 36 cos θ + 9sin θ (5 + 4cos θ)

22 cos θ (5 + 4cos θ) 9(5 + 4cos θ) (5 + 4cos θ) x + y cos θ R.H.S. 4 (6 + 3cos θ) 4x cos θ cos θ 15 1 cos θ 5 + 4cos θ cos θ x + y 4x 3. iii)if 1 x + iy then show that 4x cosθ + isinθ 1 x + iy 1+ cosθ + isinθ 1+ cos θ isin θ x + iy (1 + cos θ) i sin θ 1+ cos θ 1+ cos θ x + cos θ (1 + cos θ) 1 x x 1 4x 1 4x 1 0

23 iv) If + i u + iv and z x + iy find u, v. z i u + iv z i (x + 3) + iy ( + i)(x + 3 iy) (x + 3) + y (x + 3) + y i(x + 3 y) + (x + 3) + y (x + 3) + y (x + 3) + u y, v x + 3 y (x + 3) + y (x + 3) + y. i) If z 3 + 5i then show that z 3 10z + 58z Sol: i) z 3 + 5i (z 3) (5i) z 6z + 9 5i z 6z z 6z z 6z + 34z 0 3 (z 10z + 58z 136) + 4z 4z (z 10z + 58z 136) + 4(z 6z + 34) 0 3 z 10z + 58z ii) If z i 7 then show that 3z 3 4z + z z i 7

24 3 3 (z ) ( i 7) z 4z z 4z z 4z + 11z 0 3z 1z + 33z (3z 4z + z + 88) + ( 8z + 3z 88) 0 (3z 4z + z + 88) 8(z 4z + 11) 0 3z 4z + z i 11i iii) Show that and (1 i) 5 are conjugate to each other. i z (1 i) (if z a + ib, z a ib ) i (1 4) 4i i 3 4i ( i)( 3 + 4i) ( 3 4i)( 3 + 4i) 6 + 8i + 3i i i z [Conjugate to z is z ] 5 3. i) If 1/ 3 (x iy) a ib, then show that x a y b + 4(a b ). Sol: i) 1/ 3 (x iy) a ib x iy (a ib) x iy a + ib + 3a( ib)(a ib) 3 3 x a 3ab & iy ib 3a bi 3 3 x a 3ab & y b 3a b

25 3 y b 3a b x a y a 3b and b 3a b ( ) x y 4 a b a b ii) Write a + ib a ib a ib a + ib in the form x + iy. a + ib a ib a ib a + ib (a + ib)(a + ib) (a ib)(a ib) z (a ib)(a + ib) (a + ib)(a ib) a b + abi a b abi a + b a + b ( 8ab(a b )i) 1 (a + b ) iii)if x and y are real numbers such that (1 + i)x i ( 3i)y + + i i 3 + i 3 i values of x and y. then determine the (1 + i)x i ( 3i)y + i + i 3 + i 3 i [(1 + i)x i](3 i) [( 3i)y + i](3 + i) + i (3+ i)(3 i) (3 i)(3 + i) x(3 i + 3i + 1) + y(6 + i 9i + 3) 6i + 3i i i(x + 7y 3) 4x + 9y 3 + i x + 9y 3 0 x 7y 3 10 Solving we get, x 3, y 1.

26 4. i) Find the least positive integer n, satisfying n 1+ i 1. 1 i 1+ i Sol: i) 1 1 i n 4 n n (1 + i)(1 + i) (1 i)(1 + i) i 1 i 1 n 1 n 4 ( i i i i ) n 4 i 1 Least value of n i 1 i ii) If x + iy 1 i 1 + i find x and y i 1 i x + iy 1 i 1 + i 3 3 (1 + i)(1 + i) (1 i)(1 i) x + iy (1 i)(1 + i) (1 + i)(1 i) 3 3 i i i 1 x + iy x 0 y

27 iii)find the real values of θ in order that 3 + isin θ 1 i sin θ is a (a) Real Number (b) Purely Imaginary Number. z 3 + isin θ 1 i sin θ (3+ i sin θ )(1 + i sin θ) z (1 i sin θ )(1 + i sin θ) 3 4sin θ + 8i sin θ z 1+ 4sin θ Z is purely real imaginary part 0 8sin θ sin θ sin θ 0 θ n π, n I z is purely imaginary Real part zero 3 4sin θ sin θ sin 3 θ 4 3 sin θ ± π θ n π ±, n I 3 iv) Find the real value of x and y if x 1 y + 1 i. 3 + i 3 i ( )( ) ( )( ) ( )( ) ( )( ) x 1 y 1 x 1 3 i y 1 3+ i + i + i 3 + i 3 i 3 + i 3 i 3 i 3+ i

28 (x 1)(3 i) (y 1)(3 + i) + i (x 1) + 3(y 1) i(1 x + y 1) + i x + 3y 6 x + y + i x + 3y 6 0 y x 10 Solving we get, x 4, y Simplify the following complex numbers and find their modulus. ( + 4i)( 1+ i) ( 1 i)(3 i) Sol: z ( + 4i)( 1+ i) ( 1 i)(3 i) + 4i 4i + 8i i + i i + i 5( i) ( + i)( i) 5( i) i z i) If (1 i)( i)(3 i)...(1 ni) x iy then prove that Sol: i) (1 i)( i)(3 i)...(1 ni) x iy Taking modulus both sides. (1 1) ( i)... 1 ni x iy n x + y 5... (1 + n ) x + y (1 + n ) x + y.

29 ii) If the real part of z + 1 z + i is 1, then find the locus of z. z + 1 z + i k 1 + k i (x + 1) + iy k1 + ki (x) + (y + 1)i [(x + 1) + iy][x (y + 1)i] k1 + ki [x + (y + 1)i][x (y + 1)i] x(x + 1) + y(y + 1) + i(xy (x + 1)(y + 1)) Given real part 1 x + (y + 1) x + y + x + y x + (y + 1) x + y + x + y x + y + y + 1 x y 1 iii)if z 3 + i 4 determine the locus of z. z 3 + i 4 (x 3) + i(y + 1) 4 (x 3) + (y + 1) 16 x + y 6x + y x + y 6x + y 6 0 iv) If z + ai z ai then find the locus of z. z + ai z ai (x) + (y + a)i x + (y a)i x + (y + a) x + (y a) y 0

30 7. If z x + iy and if the point P in the Argand plane represents z, then describe geometrically the locus of P satisfying the equations i) z 3 7 Sol: i) z 3 7 (x) 3+ yi 7 (x 3) + 4y 7 4x 1x y 49 4x + 4y 1x 40 0 x + y 3x 10 0 Centre 3,0, radius 7 ii) z 4 Re (z + ) z 4 Re(z + ) x + y 4(x + ) x + y 4x 8 0 Circle centre (, 0). Radius 1 3. iii) z + i z i x + (y + 1) x (y 1) 4y y 1 y 1 0 Line parallel to x-axis. iv) z + 4i + z 4i 10 Sol: Given z + 4i + z 4i 10. (x + (y + 4)i) + (x + (y 4)i) 10

31 x + (y + 4) + x + (y 4) 10 ( ) x + (y + 4) 10 x (y 4) x + (y + 4) x + (y 4) 0 x + (y 4) Solving we get 5x + 9y 5 is ellipse. Centre (0, 0) Eccentricity e a b a e If z 1, z are two non-zero complex numbers satisfying i) z 1 + z z 1 + z then show that Arg z 1 Arg z 0. Sol: i) z 1 + z z 1 + z Squaring both sides ( ) z z z z (z + z )(z + z ) z + z + z z z z + z z + z z + z z z + z + z z z z + z z z z (x + iy )(x iy ) + (x + iy )(x iy ) x + y x + y Squaring on both sides we get

32 (x x y y ) (x y )(x y ) (x y y x ) 0 y x 1 1 y x 1 1 Arg z Arg z 0 1 ii) If z x + iy and the point P represents z in the Argand plane and Re(a) 0 then find the locus of P. Sol. z a 1 z + a z a z + a Squaring on both sides (x a) + y (x + a) + y 4xa 0 x 0 Parallel to y-axis. z a z + a 1 z3 z1 9. If is a real number, show that the points represented by the complex numbers z 1, z, z z 1 z 3 are collinear. Sol: z 1 θ z 3 z O z1 z3 Arg 0 z1 z θ 0 z1 z3 then z z 1 is real. z 1, z, z 3 are collinear.

33 10. Show that the four points in the Argand plane represented by the complex numbers + i, 4 + 3i, + 5i, 3i are the vertices of a square. Sol: Let the four points represented by the given complex numbers be A, B, C, D then A (, 1), B (4, 3), C (, 5), D (0, 3) AB BC AC BD (4 ) + (3 1) (4 ) + (3 5) ( ) + (5 1) (4 0) + (3 3) AB BC and AC BD Given Complex number are the vertices of a square. 11. Show that the points in the Argand diagram represented by the complex numbers + 7i, i, 4 3i, (1 + i) are the vertices of a rhombus. Sol: A(, 7), B,,C(4, 3),D, 3 1 AB BC CD

34 7 7 AD Slope of AC Slope of BD AC BD ABCD is rhombus. 1. Show that the points in the Argand diagram represented by the complex numbers z 1, z, z 3 are collinear if and only if there exists real numbers p, q, r not all zero, satisfying pz 1 + qz + rz 3 0 and p + q + r 0 Sol: pz 1 + qz + rz 3 0 pz 1 + qz rz 3 pz + qz p + q 1 Now p + q r (p + q) rz 3 pz + qz p + q 1 z 3 z 3 divides z 1 and z is q : p ratio. z 1, z, z 3 are collinear.

35 13. The points P, Q denote the complex numbers z 1, z in the Argand diagram. O is origin. If z1z + z1z 0 then show that POQ 90. Sol: z1z + z1z 0 z1z + z1z z1 z zz z z z 1 Real of 0 z z1 Or z is purely imaginary z1 Let Imaginary part of z is k. z 1 z ki. z1 π Arg. z 14. Determine the locus of z, z i, such that z 4 0. z i Sol: Let z x + iy, then z 4 x + iy 4 (x 4) + iy z i x + iy i x + i(y ) (x 4) + iy x i(y ) x + i(y ) x i(y ) (x 4x + y y) + i(x + 4y 8) x + (y ) Hence, real part of z 4 z i x + y 4x y x + (y ) The ratio on the RHS is zero. i.e. x 4x + y y 0 if and only if (x ) + (y 1) 5.

36 Therefore z i and z 4 Re 0 z i (x, y) (0, ) and (x ) + (y 1) 5. Hence the locus of the given point representing the complex number is the circle with (, 1) as centre and 5 units as radius except the point (0, ). 15. Write z 7 + i 1 in the polar form. Sol: If z 7 + i 1 x + iy Then x 7, y 1, y 1 tan θ 3 x 7 r x + y Since the given point lies in the second quadrant, we look for a solution of tan θ 3 which π lies in, π. We find that π θ is such a solution. 3 π π 7 + i 1 7 cos + i sin If the amplitude of z π, find its locus. z 6i Sol: Let z x + iy Then z x + iy z 6i x + i(y 6) [(x ) + iy][x (i(y 6)] [x + i(y 6)][x i(y 6)] x(x ) + y(y 6) xy (x ) (y 6) + i x + (y 6) x + (y 6) a + ib (say)

37 x(x ) + y(y 6) Then a x + (y 6) xy (x ) (y 6) b x + (y 6) π But by the hypothesis, amplitude of a + ib. Hence a 0 and b 0 x(x ) + y(y 6) 0 or x + y x 6y 0 and 3x +y 6 0 (1) () The points satisfying (1) and () constitute the arc of the circle x + y x 6y 0 intercepted by the diameter 3x + y 6 0 not containing the origin and excluding the points (0, 6) and (, 0). Hence this arc is the required locus. 17. Show that the equation of any circle in the complex plane is of the form zz + bz + bz + c 0,(b C;c R). Sol: Assume the general form of the equation of a circle in Cartesian coordinates as x + y + gx + fy + c 0,(g, f R)... (1) To write this equation in the complex variable form, Let (x, y) z. Then z + z z z i(z z) x, y i x + y z zz Substituting these results in equation (1), we obtain zz + g(z + z) + f (z z)( i) + c 0 i.e. zz + (g if )z + (g + if )z + c 0... () If g + if b, then equation () can be written as: zz + bz + bz + c 0..

38 18. Show that the complex numbers z satisfying z + z constitute a hyperbola. Sol. Substituting z x + iy in the given equation z + z, we obtain the Cartesian form of the given equation. (x + iy) + (x iy) i.e., x y + ixy + x y ixy Or x y i.e., x y 1. Since this equation denotes a hyperbola all the complex numbers satisfying z + z constitute the hyperbola x y Show that the points in the Argand diagram represented by the complex numbers 1+ 3i, 4 3i, 5 5i are collinear. Sol: Let P (1 + 3i) represented by (1, 3) Q 4 3i represented by (4, 3) R 5 5i represented by (5, 5) If slope of PQ slope QR then P, Q, R are collinear. Slope of PQ Slope of QR P, Q, R are collinear. 0. Find the equation of straight line joining the points represented by ( 4+3i), ( 3i) in the Argand plane. Sol: Take the given points as A 4 + 3i ( 4, 3) B 3i (, 3) Equation of the straight line AB is y 3 (x + 4) 4 i.e., x + y

39 1. The point P resent a complex number z in the Argand plane. If the amplitude of z is π/4, determine the locus of P. Sol: Let z x + iy By hypothesis, amplitude of z π. 4 Hence π z x z cos and 4 π z y z sin 4 Hence x 0, y 0 and x y. Clearly for any x [0, ), the point x + ix has amplitude π/4. The locus of P is the ray {( x, y ) R x 0, y 0, x y }. If the point P denotes the complex number z x + iy in the Argand plane and if z i z 1 purely imaginary number, find the locus of P. is a Sol: We note that z i z 1 is not defined if z 1. Since, z x + iy, ( ) z i x + iy i x + i y 1 z 1 x + iy 1 x 1+ iy z i z 1 i.e., ( ) ( ) ( ) ( ) x + i y 1 x 1 iy x 1 + iy x 1 iy x + y x y 1 x y + i ( ) ( ) x 1 + y x 1 + y Will be purely imaginary, if and z 1 and ( ) x y x y 0 + and ( ) ( ) The locus of P is the circle x, y 1,0. x + y x y 0. x 1 + y x + y x y 0 excluding the point (1, 0).

40 π 3. The complex no. z has argument θ, 0 < θ < and satisfy the equation z 3i 3. Then prove that Sol. Let z x+iy3 6 cot θ i. z z 3i 3 x + iy 3i 3 ( ) x i y 3 3 ( y ) x + x + y y + x + y y x y 6y + 6y 6y 1 1 x + y z 6y 6 1 zz z z y 6 x iy z y 6 x 6 i cot θ i z y z 6 cot θ i z

COMPLEX NUMBERS AND QUADRATIC EQUATIONS

Chapter 5 COMPLEX NUMBERS AND QUADRATIC EQUATIONS 5. Overview We know that the square of a real number is always non-negative e.g. (4) 6 and ( 4) 6. Therefore, square root of 6 is ± 4. What about the square