Mathematics

Complex Numbers Session

Session Objectives

Session Objective 1.Polar form of a complex number 2.Euler form of a complex number 3.Representation of z 1 +z 2, z 1 -z 2 4.Representation of z 1.z 2, z 1 /z 2 5.De-Moivre theorem 6.Cube roots of unity with properties 7.Nth root of unity with properties

Representation of complex number in Polar or Trigonometric form z = x + iy z(x,y) X Y O x y  This you have learnt in the first session

Representation of complex number in Polar or Trigonometric form z = r (cos  + i sin ) Examples: 1 = cos0 + isin0 -1 = cos  + i sin  i = cos /2 + i sin /2 -i = cos (-/2) + i sin (-/2) z(x,y) X Y O x = rcos y =rsin  r

Eulers form of a complex number z = x + iy z = r (cos  + i sin ) Examples: Express 1 – i in polar form, and then in euler form

Properties of eulers form e i

Illustrative Problem Solution: i = cos(/2) + i sin(/2) = e i/2 i i = (e i/2 ) i = e -/2 The value of i i is ____ a)2b) e -/2 c)d)  2

Illustrative Problem Find the value of log e (-1). Solution: -1 = cos  + i sin  = e i log e (-1) = log e e i = i General value: i(2n+1), nZ As cos(2n+1) + isin(2n+1) = -1

Illustrative Problem If z and w are two non zero complex numbers such that |zw| = 1, and Arg(z) – Arg(w) = /2, then is equal to a) ib) –i c) 1d) –1

Solution

Representation of z 1 +z 2 z 1 = x 1 + iy 1, z 2 = x 2 + iy 2 z = z 1 + z 2 = x 1 + x 2 + i(y 1 + y 2 ) O A z 1 (x 1,y 1 ) z 2 (x 2,y 2 ) z(x 1 +x 2,y 1 +y 2 ) B Oz 1 + z 1 z  Oz ie |z 1 | + |z 2 |  |z 1 + z 2 |

Representation of z 1 -z 2 z 1 = x 1 + iy 1, z 2 = x 2 + iy 2 z = z 1 - z 2 = x 1 - x 2 + i(y 1 - y 2 ) z 2 (x 2,y 2 ) z 1 (x 1,y 1 ) -z 2 (-x 2,-y 2 ) z(x 1 -x 2,y 1 -y 2 ) O Oz + z 1 z  Oz 1 ie |z 1 -z 2 | + |z 2 |  |z 1 |  |z 1 -z 2 |  ||z 1 | - |z 2 || also |z 1 -z 2 | + |z 1 |  |z 2 |

Representation of z 1.z 2 z 1 = r 1 e i1, z 2 = r 2 e i2 z = z 1.z 2 = r 1 r 2 e i(1+ 2) r1ei1r1ei1 r2ei2r2ei2 r 1 r 2 e i(1+ 2) O x Y 11 22 1+ 21+ 2

 Representation of z 1.e i and z 1.e -i z 1 = r 1 e i1 z = z 1. e i = r 1 e i(1+ ) r1ei1r1ei1 r 1 e i(1+ ) O x Y 11  r 1 e i(1- ) What about z 1 e -i

Representation of z 1 /z 2 z 1 = r 1 e i1, z 2 = r 2 e i2 z = z 1 /z 2 = r 1 /r 2 e i(1- 2) r1ei1r1ei1 r2ei2r2ei2 r 1 /r 2 e i(1- 2) 11 22  1 -  2

Illustrative Problem If z 1 and z 2 be two roots of the equation z 2 + az + b = 0, z being complex. Further, assume that the origin, z 1 and z 2 form an equilateral triangle, then a)a 2 = 3bb) a 2 = 4b c) a 2 = bd) a 2 = 2b

Solution /3 z1z1 z 2 = z 1 e i /3 Hence a 2 = 3b

De Moivre’s Theorem 1) n  Z,

De Moivre’s Theorem 2) n  Q, cos n + i sin n is one of the values of (cos  + i sin ) n Particular case

Illustrative Problem Solution:

Illustrative Problem Solution:

Cube roots of unity Find using (cos0 + isin0) 1/3

Properties of cube roots of unity 1,,  2 are the vertices of equilateral triangle and lie on unit circle |z| = 1 Why so? 1  22 O

Illustrative Problem If  is a complex number such that  2 ++1 = 0, then  31 is a)1b) 0 c)  2 d) 

Solution

Nth roots of unity

Properties of Nth roots of unity c) Roots are in G.P d) Roots are the vertices of n sided regular polygon lying on unit circle |z| = 1 2/n 4/n  n-1 1  22 33

Illustrative Problem Solution: Find fourth roots of unity. 1 -i i

Illustrative Problem Solution:

Class Exercise

Class Exercise - 1 Express each of the following complex numbers in polar form and hence in eulers form. (a) (b) –3 i Solution 1.(a)

Solution Cont. b) –3i

Class Exercise - 2 Solution z 1 and z 2 are in the same line z 1 and z 2 have same argument or their difference is multiple of 2 arg (z 1 ) – arg (z 2 ) = 0 or 2n in general (triangle inequality) If z 1 and z 2 are non-zero complex numbers such that |z 1 + z 2 | = |z 1 | + |z 2 | then arg(z 1 ) – arg (z 2 ) is equal to (a) –  (b) (c) 0(d)

Class Exercise - 3 Solution We have to find the area of PQR. Note that OPQR is a square as OP = |z| = |iz| = OR and all angles are 90° Area of area of square OPQR Find the area of the triangle on the argand diagram formed by the complex numbers z, iz and z + iz.

Class Exercise - 4 If where x and y are real, then the ordered pair (x, y) is given by ___.

Solution = 3 25 (x + iy)

Class Exercise - 5 Solution x + y + z = If then prove that = 0 + i0 = 0 x 3 + y 3 + z 3 – 3xyz = (x + y + z) (x 2 + y 2 + z 2 – xy – yz – zx) = 0 x 3 + y 3 + z 3 = 3xyz

Solution Cont. x 3 + y 3 + z 3 = 3xyz

Class Exercise - 6 Solution If and then is equal to ___. d) 0 Take any one of the values say

Solution Cont. Similarly

Class Exercise - 7 Solution The value of the expression where  is an imaginary cube root of unity is ___.

Class Exercise - 8 Solution If are the cube roots of p, p < 0, then for any x, y, z, is equal to (a) 1(b)  (c)  2 (d) None of these

Class Exercise - 9 Solution: The value of is (a)–1(b) 0 (c) i(d) –i...(i) roots of x 7 – 1 = 0 are k = 0, 1, …, 6

Solution Cont....(ii) From (i) and (ii), we get

Class Exercise - 10 Solution: If 1, are the roots of the equation x n – 1 = 0, then the argument of is (a) (b) (c) (d) As n th root of unity are the vertices of n sided regular polygon with each side making an angle of 2  /n at the centre,  2 makes an angle of 4  /n with x axis and hence, arg(  2 ) = 4  /n

Thank you