If z 1 = 1 z 1 ≠ - 1 and z 2 = z 1 - 1 z 1 + 1 , then show that the real part of z 2 is zero.

Let us assume z 1 = x + y i Applying modulus we obtain, z 1 = x 2 + y 2 = 1 [given that z 1 = 1 ] ⇒ x 2 + y 2 = 1 So z 2 = z 1 - 1 z 1 + 1 = x + y i - 1 x + y i + 1 Rationalizing the denominator we get, = ( x - 1 ) + y i ( x + 1 ) + y i × x + 1 - y i x + 1 - y i ⇒ ( x - 1 ) ( x + 1 ) - y ( x - 1 ) i + y ( x + 1 ) i - y 2 i 2 ( x + 1 ) 2 - y 2 i 2 = x 2 - 1 + y i ( x + 1 - x + 1 ) + y 2 x 2 + y 2 + 2 x + 1 From above as x 2 + y 2 = 1 real part will be zero = ( 1 - 1 ) x 2 + y 2 + 2 x + 1 + 2 i y x 2 + y 2 + 2 x + 1 = 0 + 2 i y x + 1 2 + y 2 Hence proved.

If z 1 = z 2 = … = z n = 1 , then show that z 1 + z 2 + z 3 + … + z n = 1 z 1 + 1 z 2 + 1 z 3 + … + 1 z n .

Given z 1 = z 2 = … = z n = 1 ⇒ z 1 2 = z 2 2 = … = z n 2 = 1 . . . . . ( i ) ⇒ z 1 z ¯ 1 = z 2 z ¯ 2 = … = z n z n ¯ = 1 ∵ z z ¯ = | z | 2 ⇒ z 1 = 1 z 1 , z 2 = 1 z 2 = … = z n = 1 z n L.H.S. = z 1 + z 2 + z 3 + … + z n = z 1 z 1 ¯ z 1 + z 2 z 2 ¯ z 2 + z 3 z 3 ¯ z 3 + … + z x z n ¯ z n z 1 2 z ¯ 1 + z 2 2 z ¯ 2 + z 3 2 z ¯ 3 + … + z n 2 z ¯ n ∵ z z ¯ = | z | 2 = 1 z - 1 + 1 z - 2 + 1 z - 3 + … + 1 z - n [using ( i ) ] = 1 z 1 + 1 z 2 + 1 z 3 + … + 1 z n [ ∵ z 1 ¯ + z 2 ¯ = z 1 + z 2 ¯ ] = 1 z 1 + 1 z 2 + 1 z 3 + … + 1 z n [ ∵ | z | = | z ¯ | ] ∴ L.H.S. = R.H.S.

Find the complex number satisfying the equation z + 2 | ( z + 1 ) | + i = 0 .

We have z + 2 | ( z + 1 ) | + i = 0 Putting z = x + i y we get, ⇒ ( x + y i ) + 2 | ( x + y i + 1 ) | + i = 0 ⇒ x + ( y + 1 ) i + 2 | ( x + 1 ) + y i ) | = 0 ⇒ x + ( y + 1 ) i + 2 ( x + 1 ) 2 + y 2 = 0 ⇒ x + ( y + 1 ) i + 2 x 2 + 2 x + 1 + y 2 = 0 + 0 i Separating real and imaginary parts we obtain, ⇒ x = - 2 x 2 + 2 x + 1 + y 2 and y = - 1 ⇒ x 2 = 2 x 2 + 2 x + 1 + y 2 ⇒ x 2 = 2 x 2 + 4 x + 2 + 2 y 2 ⇒ x 2 + 4 x + 2 + 2 y 2 = 0 ⇒ x 2 + 4 x + 2 + 2 ( - 1 ) 2 = 0 [ ∵ y = - 1 ] ⇒ x 2 + 4 x + 4 = 0 ⇒ ( x + 2 ) 2 = 0 ⇒ x + 2 = 0 ⇒ x = - 2 ∴ z = x + y i = - 2 - i .

If z and w are two complex numbers such that | z w | = 1 and arg ( z ) - arg ( w ) = π 2 , then show that z ¯ w = - i .

Let us take z = r 1 cos θ 1 + i sin θ 1 and w = r 2 cos θ 2 + i sin θ 2 z w = r 1 r 2 cos θ 1 + isin θ 1 cos θ 2 + isin θ 2 Given | z w | = r 1 r 2 = 1 Also a r g ( z ) - arg ( w ) = π 2 ⇒ θ 1 - θ 2 = π 2 ⇒ arg z w = π 2 Now z ¯ w = r 1 cos θ 1 - i sin θ 1 r 2 cos θ 2 + i sin θ 2 = 1 = r 1 r 2 cos θ 1 cos θ 2 + i cos θ 1 sin θ 2 - i sin θ 1 cos θ 2 - i 2 sin θ 1 sin θ 2 = r 1 r 2 cos θ 1 cos θ 2 + sin θ 1 sin θ 2 + i cos θ 1 sin θ 2 - sin θ 1 cos θ 2 = r 1 r 2 c o s θ 2 - θ 1 + isin θ 2 - θ 1 = r 1 r 2 cos - π 2 + i sin - π 2 = r 1 r 2 cos π 2 - i sin π 2 = 1 . [ 0 - i ] ∴ z ¯ w = - i Hence, proved.