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Mathematics Session Complex Numbers Session Objectives Session Objective 1. Polar form of a complex number 2. Euler form of a complex number 3. Representation of z1+z2, z1-z2 4. Representation of z1.z2, z1/z2 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 i 2 2 x2 y2 x y 2 2 Y z(x,y) x2 y2 y O x X This you have learnt in the first session Representation of complex number in Polar or Trigonometric form x2 y2 r x x cos , r x2 y2 y x2 y2 y sin r z = r (cos + i sin ) where r | z | and arg z z(x,y) Y r Examples: 1 = cos0 + isin0 -1 = cos + i sin i = cos /2 + i sin /2 -i = cos (-/2) + i sin (-/2) O x = rcos y =rsin X Eulers form of a complex number z = x + iy z = r (cos + i sin ) z rei where ei cos i sin | ei | cos2 sin2 1 | z | | r || ei | r z r cos i sin rei Examples: i0 i i 2 1 e , 1 e ,i e , i e i 2 Express 1 – i in polar form, and then in euler form i 1 1 i 2 2 cos 4 isin 4 2 2 i 2 cos i sin 2e 4 4 4 Properties of eulers form ei ei cos i sin | ei | cos2 sin2 1 z | z |ei arg z remember this cos a i sin a cosb i sinb eiaeib i a b e cos a b i sin a b eia eib cos a i sin a cosb i sinb ab ab ab 2 cos cos i sin 2 2 2 a b ab i 2 2 cos e 2 ab i e e 2i sin e 2 ia ib a b 2 Illustrative Problem The value of ii is ____ a) 2 b) e-/2 c) d) 2 Solution: i = cos(/2) + i sin(/2) = ei/2 ii = (ei/2)i = e-/2 Now find i Ans : i i i Illustrative Problem Find the value of loge(-1). Solution: -1 = cos + i sin = ei loge(-1) = logeei = i General value: i(2n+1), nZ 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 z w is equal to a) i b) –i c) 1 d) –1 Solution By eulers form z | z | eiArg(z) iArg zw zw | zw | e | zw | | z || w | | z || w | | zw | 1(given) Arg zw Arg z Arg w Arg z Arg w zw e i 2 i 2 Representation of z1+z2 z1 = x1 + iy1, z2 = x2 + iy2 z = z1 + z2 = x1 + x2 + i(y1 + y2) z2(x2,y2) z(x1+x2,y1+y2) Note OAz2 z1Bz z1(x1,y1) O A B Oz1 + z1z Oz ie |z1| + |z2| |z1 + z2| Representation of z1-z2 z1 = x1 + iy1, z2 = x2 + iy2 z = z1 - z2 = x1 - x2 + i(y1 - y2) z2(x2,y2) z1(x1,y1) O -z2(-x2,-y2) Oz + z1z Oz1 ie |z1-z2| + |z2| |z1| also |z1-z2| + |z1| |z2| |z1-z2| ||z1| - |z2|| z(x1-x2,y1-y2) Representation of z1.z2 z1 = r1ei1, z2 = r2ei2 z = z1.z2 = r1r2ei(1+ 2) | z1z2 | r1r2 | z1 || z2 | Arg z1z2 1 2 Arg z1 Arg z2 Y r1r2ei(1+ 2) 1+ 2 r2ei2 O 2 1 r1ei1 x Representation of z1.ei and z1.e-i z1 = r1ei1 z = z1. ei = r1ei(1+ ) | z | r1 and Arg z 1 Y r1ei(1+ ) O 1 r1ei1 r1ei(1- ) What about z1e-i x Representation of z1/z2 z1 = r1ei1, z2 = r2ei2 z = z1/z2 = r1/r2ei(1- 2) z1 r1 z1 z z 2 r2 z 2 z1 Arg z Arg 1 2 Arg z1 Arg z 2 z2 2 r1ei1 1 r2ei2 r1/r2ei(1- 2) 1- 2 Illustrative Problem If z1 and z2 be two roots of the equation z2 + az + b = 0, z being complex. Further, assume that the origin, z1 and z2 form an equilateral triangle, then a) a2 = 3b b) a2 = 4b c) a2 = b d) a2 = 2b Solution z2 = z1ei /3 /3 z1 Sum z1 z2 z1 z1ei / 3 z1 ei0 ei / 3 i 6 2z1 cos e a 6 i Pr oduct z1z2 z12e 3 .......I b i Squaring I we get 3z12e 3 a2 Hence a2 = 3b De Moivre’s Theorem 1) n Z, cos i sin n cosn i sinn cos i sin cosn i sinn n cos a i sin a m cos a i sin a n cos b i sinb m cosb i sinb cos ma nb i sin ma nb n cos ma nb i sin ma nb De Moivre’s Theorem 2) n Q, cos n + i sin n is one of the values of (cos + i sin )n n p /q cos i sin p/q cosp i sinp 1/q cos p 2k i sin p 2k 1/ q 2k p 2k p cos i sin ,k 1,2,..., q 1 q q Particular case cos i sin 1/n cos 2k 2k i sin ,k 0,1,2,...,n 1 n n Illustrative Problem 7 2 2 cos5 i sin 5 cos i sin 7 7 Simplify 3 1 2 2 cos 4 i sin 4 4 cos i sin 3 3 2 5 Solution: cos i sin 2 5 5 cos i sin cos i sin cos3 isin3 4 7 2 7 cos i sin 1 2 3 4 3 cos i sin 2 2 1 2 cos i sin 3 Illustrative Problem Pr ove that : 1 i 1 i 2 n n n 1 2 n cos 4 Solution: 1 i 2 cos i sin 4 4 1 i 2 cos isin 4 4 1 i n 1 i n n n 2 cos i sin 4 4 n 2 1 i n 1 i n n n 2 cos i sin 4 4 n 1 n n 2 2 2cos 2 cos 4 4 n 2 n 2 Cube roots of unity 1 3 x 1 or x3 1 x3 1 0 or x 1 x2 x 1 0 1 i 3 1 i 3 , 2 2 2 or x 1, , or 1, 2 , x 1, 2 i 1 i 3 1 3 2 2 i cos i sin e3 2 2 2 3 3 2 i 1 i 3 1 3 2 2 i cos i sin e 3 2 2 2 3 3 2 Find using (cos0 + isin0)1/3 2 1 i 3 1 i 3 Note : 2 2 2 1 i 3 1 i 3 and 2 2 Properties of cube roots of unity 1 2 0 1..2 3 1 | | | 2 | 1 1 2 2 , , 2 4 2 3 3 1,, 2 are the vertices of equilateral triangle and lie on unit circle |z| = 1 Why so? O 2 1 Illustrative Problem If is a complex number such that 2++1 = 0, then 31 is a) 1 b) 0 c) 2 d) Solution 2 1 0 1 i 3 or 2 2 let , 31 31 or , 2 31 2 31 62 2 Nth roots of unity 1 n x 1 cos 0 i sin0 1 n 2k 2k cos isin ,k 0,1,...,n 1 n n k 0, x 1 ei0 2 i 2 2 k 1, x cos i sin e n n n 4 i 4 4 k 2, x cos i sin e n 2 n n . . k n 1, x cos 2 n 1 n i sin 2 n 1 n i e 2n1 n n1 Properties of Nth roots of unity a) 1 ..... 2 b) 1.. 2...... n1 n1 n 1 0 1 n1n 2 1 n1 1 n is odd 1 n is even c) Roots are in G.P d) Roots are the vertices of n sided regular polygon lying on unit circle |z| = 1 3 2 4/n 1 2/n n-1 Illustrative Problem Find fourth roots of unity. Solution: 1 4 x 1 cos 0 isin 0 1 4 2k 2k isin , k 0,1, 2,3 4 4 k 0, x cos 0 isin 0 1 cos k 1, x cos isin i 2 2 k 2, x cos isin 1 3 3 k 3, x cos i sin i 2 2 i -1 1 -i Illustrative Problem 1 Find 3 4 in eulers form. Solution: 1 3 4 cos i sin 3 4 cos3 isin 3 1 4 2k 3 2k 3 cos i sin , k 0,1, 2,3 4 4 e i 3 4 ,e i 5 4 ,e i 7 4 ,e i 9 4 e i 4 Class Exercise Class Exercise - 1 Express each of the following complex numbers in polar form and hence in eulers form. (a) 6 2 i (b) –3 i Solution 1. (a) r | 6 tan1 2i | 2 + 2 6 6 i 7 7 6 2i 2 2 cos isin 2 2 e 6 6 62 2 7 6 (–6, –2 ) 7 = 6 6 Solution Cont. b) –3i r 3i 0 9 3 3 0 2 3 i 3 3 –3i 3 cos isin 3e 2 2 2 tan 1 3 2 –3i Class Exercise - 2 If z1 and z2 are non-zero complex numbers such that |z1 + z2| = |z1| + |z2| then arg(z1) – arg (z2) is equal to (a) – (b) (c) 0 (d) 2 2 Solution | z z | | z | | z | (triangle inequality) 1 2 1 2 | z | | z | | z z | 1 2 1 2 z1 and z2 are in the same line z1 and z2 have same argument or their difference is multiple of 2 arg (z1) – arg (z2) = 0 or 2n in general Class Exercise - 3 Find the area of the triangle on the argand diagram formed by the complex numbers z, iz and z + iz. 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° 1 Area of PQR area of square OPQR 2 1 1 OP2 | z |2 2 2 Y Q (Z + iZ) R (iZ) P (Z) O X Class Exercise - 4 50 3 i 3 325 (x iy), where If 2 2 x and y are real, then the ordered pair (x, y) is given by ___. 3 1 a) , 2 2 1 3 c) , 2 2 1 3 b) , 2 2 3 1 d) , 2 2 Solution 50 50 50 3 3 i 3 3 i 50 325 i ( 3) 2 2 2 2 50 50 25 3 i 325 3 cos isin 2 2 6 6 25 25 25 25 3 cos 8 isin 8 3 cos isin 3 3 3 3 325 cos 8 isin 8 3 3 325 cos isin 3 3 = 325 (x + iy) x cos 1 3 2 y sin 3 3 2 Class Exercise - 5 If cos cos cos 0 sin sin sin then prove that cos3 cos3 cos3 3cos( ) sin3 sin3 sin3 3sin Solution Let x cos isin y cos isin z cos isin x + y + z = (cos cos cos ) i(sin sin sin ) = 0 + i0 = 0 x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx) = 0 x3 + y3 + z3 = 3xyz Solution Cont. x3 + y3 + z3 = 3xyz (cos isin )3 (cos isin )3 (cos isin )3 3 cos( ) isin( ) comparing real part cos3 cos3 cos3 3cos( ) comparing imaginary part sin3 sin3 sin3 3 sin Class Exercise - 6 1 2cos and x yn y If x 1 2cos , then y xm is equal to ___. yn xm a)2cos b)2cos(m n) c)2cos(m – n) d) 0 Solution x 1 2cos x x2 2cos x 1 0 2cos 4cos2 4 x 2 2cos 2isin 2 Take any one of the values say x cos isin cos isin Solution Cont. x cos isin Similarly y cos isin xm yn (cos isin )m (cos isin )n yn xm (cos isin )n (cos isin )m cos m – n isin m – n cos n m isin n m 2cos m n Class Exercise - 7 The value of the expression 1(2 )(2 2 ) ... (n 1)(n )(n 2 ), where is an imaginary cube root of unity is ___. Solution n 2 (r 1)(r )(r ) r2 n (r 1)(r 2 ( 2 )r 3 ) r2 n n 3 2 (r 1)(r r 1) (r 1) r2 r2 n 3 n n 3 n r 1 r – 1 r 1 r 1 r2 r2 2 n2 n 1 4 n Class Exercise - 8 If are the cube roots of p, p < 0, then for any x, y, z, x y z is equal to x y z (a) 1 (b) (c) 2 (d) None of these Solution 1 x p3 , p 0 1 1 1 p 3 , p 3 , p 3 2 1 say p 3 , , 2 x y z x y z2 x y z x y2 z 2 2 2 z x y x y2 z Class Exercise - 9 6 2k 2k The value of sin – icos 7 7 k 1 (a) –1 (b) 0 (c) i (d) –i is Solution: 6 2k 2k – icos sin 7 7 k 1 6 2k 2k –i cos isin 7 7 k 1 ...(i) roots of x7 – 1 = 0 are cos 2k 2k isin , k = 0, 1, …, 6 7 7 6 2k 2k isin 0 cos 7 7 k 0 Solution Cont. 6 2k 2k isin 0 cos 7 7 k 0 6 2k 2k 1 cos isin 0 7 7 k 1 6 2k 2k isin –1 ...(ii) cos 7 7 k 1 From (i) and (ii), we get 6 2k 2k – icos (–i)(–1) i sin 7 7 k 1 Class Exercise - 10 If 1, , 2,..., n 1 are the roots of the equation xn – 1 = 0, then the argument of 2 is 2 8 (a) (b) 4 (c) 6 (d) n n n n Solution: Y 2 2 n O 2 n X 1 As nth 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