Complex Numbers XII – STANDARD MATHEMATICS

If n is a positive integer, prove that

If a = cos 2  + isin 2 , b = cos2  + isin 2 , c = cos 2  + isin 2 . Prove that (i) a = cos 2  + isin 2  b = cos 2  + isin 2  c = cos 2  + isin 2  abc = (cos 2  + isin 2  )(cos 2  + isin 2  ) (cos 2  + isin 2  ) = cos (2  + 2  + 2  ) + isin (2  + 2  + 2  )

= cos (  +  +  ) + isin (  +  +  ) ……..(1) = cos (  +  +  ) – i sin (  +  +  ) …….(2) adding (1) and (2) we get

= (cos 2  + isin 2  )(cos 2  + isin 2  )(cos 2  + isin 2  ) – 1 = (cos 2  + isin 2  )(cos 2  + isin 2  )(cos 2  – isin 2  ) = (cos 2  + isin 2  )(cos 2  + isin 2  )(cos (–2  ) + isin (–2  )) = cos(2  + 2  – 2  ) + isin(2  + 2  – 2  ) = cos 2(  +  –  ) + isin 2(  +  –  )………….(1) If a = cos 2  + isin 2 , b = cos2  + isin 2 , c = cos 2  + isin 2 . Prove that (ii)

= [cos 2(  +  –  ) + i sin 2(  +  –  )] -1 = cos 2(  +  –  ) – i sin 2(  +  –  )………(2) Adding (1) and (2), we get,

Simplify: (cos  + isin  ) 0 = 1

Simplify:

Prove that:

If cos  + cos  + cos  = 0 = sin  + sin  + sin  : prove that (i) cos 3  + cos 3  + cos 3  = 3cos(  +  +  ) (ii) sin 3  + sin 3  + sin 3  = 3sin(  +  +  ) Given: cos  + cos  + cos  = 0 and sin  + sin  + sin  = 0  (cos  + cos  + cos  ) + i(sin  + sin  + sin  ) = 0 (cos  + isin  )(cos  + isin  )(cos  + isin  ) = 0 Let a = cos  + isin  b = cos  + isin  c = cos  + isin 

If a + b + c = 0, then a 3 + b 3 + c 3 = 3abc (cos  + isin  ) 3 + (cos  + isin  ) 3 + (cos  + isin  ) 3 = 3(cos  + isin  )(cos  + isin  )(cos  + isin  ) (cos 3  + isin 3  ) + (cos 3  + isin 3  ) + (cos 3  + isin 3  ) = 3[cos (  +  +  ) + isin (  +  +  )] (cos 3  + cos 3  + cos 3  ) + i(sin 3  + sin 3  + sin 3  ) = 3cos (  +  +  ) + 3isin (  +  +  ) Equating real and imaginary parts cos 3  + cos 3  + cos 3  = 3cos (  +  +  ) sin 3  + sin 3  + sin 3  = 3sin (  +  +  )

If n is a positive integer, prove that (i) (1 + i) n + (1 – i) n = Let 1 + i = r(cos  + isin  )

Substituting i = - i Adding (1) and (2), we get

If n is a positive integer, prove that (ii) (  3 + i) n + (  3 – i) n = Let  3 + i = r(cos  + isin  )

Substituting i = –i Adding (1) and (2), we get

If n is a positive integer, prove that (1 + cos  + isin  ) n + (1 + cos  - isin  ) n (1 + cos  + isin  ) n + (1 + cos  - isin  ) n

If  and  are the roots of x 2 – 2x + 4 = 0 prove that  n –  n = i2 n+1 sin (n  /3) and deduce  9 –  9 x 2 – 2x + 4 = 0 a = 1, b = –2, c = 4 The two roots are

Let 1 + i  3 = r(cos  + isin  )

Substituting i = –i we get Subtracting (2) from (1) we get

If n = 9  9 –  9

If x + 1/x = 2cos  prove that (i)x n + 1/x n = 2cos n  (ii) x n – 1/x n = 2isin n 

If x + 1/x = 2cos , y + 1/y = 2cos , prove that one of the values of

similarly

Adding (1) and (2)

Solve: x 8 + x 5 – x 3 – 1 = 0 x 8 + x 5 – x 3 – 1 = 0 x 5 (x 3 + 1) – (x 3 + 1) = 0 (x 3 + 1)(x 5 – 1) = 0 x = 0, x 5 – 1 = 0 Case (1) x = 0 x 3 = – 1 = cos  + isin  = cos(2k  +  ) + isin(2k  +  ) x = {cos(2k  +  ) + isin(2k  +  )} 1/3

The three values are Case (2) x 5 – 1 = 0 x 5 = 1 = cos 0 + isin 0 = cos(2k  ) + isin(2k  ) x = {cos(2k  ) + isin(2k  )} 1/5

The five values are

Solve: x 7 + x 4 + x = 0 x 7 + x 4 + x = 0 x 4 (x 3 + 1) + (x 3 + 1) = 0 (x 3 + 1)(x 4 + 1) = 0 x = 0, x = 0 Case (1) x = 0 x 3 = – 1 = cos  + isin  = cos(2k  +  ) + isin(2k  +  ) x = {cos(2k  +  ) + isin(2k  +  )} 1/3

The three values are Case (2) x = 0 x 4 = – 1 = cos  + isin  = cos(2k  +  ) + isin(2k  +  ) x = {cos(2k  +  ) + isin(2k  +  )} 1/4

The four values are

Find all the values of (1 + i) 1/4 Let 1 + i = r(cos  + isin  )

For k = 0,1,2,3,4 we get all the values

Find all the values of And hence prove that the product of the values is 1 Since sin  is – ve and cos  is +ve  lies in the fourth quadrant

For k = 0,1,2,3, we get all the values

Their product = = 1

x 2 – 2px + (p 2 + q 2 ) = 0 a = 1, b = –2p, c = p 2 + q 2 The two roots are If  and  are the roots of x 2 – 2px + (p 2 + q 2 ) = 0 and tan  = q/ y+p prove that

P represents the variable complex number z, find the locus of P if Let z = x + iy be the variable complex number

x(x + 1) + y(y + 1) = x 2 + (y + 1) 2 x 2 + x + y 2 + y = x 2 + y 2 + 2y + 1 x + y = 2y + 1 x – 2y + y – 1 = 0 x – y – 1 = 0 The locus of P is x – y – 1 = 0

Find the square root of – 8 – 6i. Let square root of – 8 – 6i be x + iy – 8 – 6i = (x + iy) 2 = x 2 + 2ixy + i 2 y 2 = x 2 + 2ixy – y 2 – 8 – 6i = x 2 – y 2 + 2ixy Equating the real and imaginary parts x 2 – y 2 = – 8 ……..(1) 2xy = –6 ……..(2) xy = –3 y = –3/x Sub y = –3/x in (1)

x 4 – 9 = – 8x x 4 + 8x 2 – 9 = 0 x 4 + 9x 2 – x 2 – 9 = 0 x 2 (x 2 + 9) – 1(x 2 + 9) = 0 (x 2 + 9)(x 2 – 1) = 0 x 2 = –9, 1 x = ±3i, ±1 Since is real x = ±1 If x = 1, then y = –3 If x = –1, then y = 3 The square root = 1 – 3i and –1 + 3i