Find the modulus, argument, and the principal argument of the complex numbers. (i) z = 1 + cos 18 π 25 + i sin 18 π 25 (ii) z = - 2 cos 30 ° + i sin 30 ° (iii) ( tan 1 - i ) 2 (iv) i - 1 i 1 - cos 2 π 5 + sin 2 π 5

i z = 1 + cos 18 π 25 + i sin 18 π 25 ⇒ z = 2 cos 2 9 π 25 + 2 isin 9 π 25 cos 9 π 25 = 2 cos 9 π 25 cos 9 π 25 + isin 9 π 25 So, the modulus of the function = 2 cos 9 π 25 The principal argument is 9 π 25 and the principal argument is 9 π 25 + 2 k π , k ∈ I ii z = - 2 cos 30 ° + i sin 30 ° ⇒ z = 2 - 3 2 - i 2 ⇒ z = 2 cos 5 π 6 + i sin 5 π 6 So, the modulus of the function = 2 The principal argument is 5 π 6 and the argument is 5 π 6 + 2 k π , k ∈ I iii tan 1 - i 2 = tan 2 1 - 1 - 2 i tan 1 So, the modulus of the function is tan 2 1 - 1 2 - 4 tan 2 1 = sec 2 1 The argument of the function is tan - 1 2 tan 1 1 - tan 2 1 = tan - 1 tan 2 = 2 - π and the argument is 2 k π + 2 - π , k ∈ I iv z = i − 1 i 1 − cos 2 π 5 + sin 2 π 5 = i − 1 i 2 sin 2 π 5 + 2 sin π 5 cos π 5 = i − 1 2 sin π 5 cos π 5 + isin π 5 | z | = | i − 1 | 2 s i n π 5 c o s π 5 + i s i n π 5 = 2 2 sin π 5 cos π 5 + isin π 5 = 1 2 cosec π 5 argz = arg i − 1 2 sin π 5 cos π 5 + isin π 5 = arg ( − 1 + i ) − arg 2 sin π 5 − arg cos π 5 + isin π 5 = 3 π 4 − 0 − π 5 = 11 π 20 So, the argument of function is 11 π 20 + 2 k π , k ∈ I Hence, arg z = 2 kπ + 11 π 20 , principal arg = 11 π 20

If ( 3 + i ) 100 = 2 99 ( a + i b ) , then find (i) a 2 + b 2 (ii) b

We know, the cube root of unity ω = - 1 + i 3 2 So, ( 3 + i ) = 2 ω i = - 2 ω i Thus, ( 3 + i ) 100 = 2 100 × ω 100 × i 100 = 2 100 × ω = 2 100 × - 1 + i 3 2 = 2 99 × - 1 + i 3 Comparing with a & b we get, a = - 1 & b = 3 So, a 2 + b 2 = 1 + 3 = 4

Show that e i 2 m θ i cotθ + 1 i cotθ - 1 m = 1 .

e 2 mi θ icot θ + 1 icot θ − 1 m = e 2 mi θ i ( cot θ − i i ( cot θ + i m = e 2 mi θ cot θ − i cot θ + i m = e 2 mi θ cos θ − isin θ cos θ + isin θ m = e 2 mi θ e − i θ e i θ m = e 2 mi θ ( e − 2 iθ ) m = e 2 mi θ e − 2 mi θ = e 0 = 1

If x r = cos π 3 r + i sin π 3 r , prove that x 1 x 2 x 3 … … . up to infinity = i .

x r = cos π 3 r + i sin π 3 r = e i π 3 r So, x 1 x 2 x 3 . . . . up to infinity = e i π 3 1 e i π 3 2 e i π 3 3 . . . . . . up to infinity = e i π 3 1 + 1 3 + . . . ∞ = e i π 3 1 1 - 1 3 = e i π 2 = i Hence, proved.

If z = x + i y is a complex number such that z = ( a + i b ) 2 then (i) find z ¯ (ii) show that x 2 + y 2 = a 2 + b 2 2

z = x + iy So, the complex conjugate of z is z ¯ = x - i y Given, z = a + i b 2 ⇒ x + i y = a 2 - b 2 + 2 i a b Equating the real and imaginary part, we get x = a 2 - b 2 & y = 2 a b So, z ¯ = x - i y = a 2 - b 2 - 2 i a b = a - i b 2 Again, x 2 + y 2 = a 2 - b 2 2 + 2 a b 2 = a 2 + b 2 2 Hence, proved.